JEE Wave Optics Questions

For any plane wave front, the light travels in the direction perpendicular (normal) to that plane.
A general plane can be written as $$ax + by + cz = \text{constant}$$, whose normal vector is $$\mathbf{n} = a\,\hat i + b\,\hat j + c\,\hat k$$.

The given wave fronts are $$x + y + z = \text{constant}$$.
Comparing with $$ax + by + cz = \text{constant}$$, we get the normal (and hence the direction of wave propagation)

$$\mathbf{n} = 1\,\hat i + 1\,\hat j + 1\,\hat k = \langle 1,\,1,\,1\rangle$$.

Let $$\theta$$ be the angle between the direction of propagation and the x-axis.
The x-axis is represented by the unit vector $$\hat i = \langle 1,\,0,\,0\rangle$$.

First, write the unit vector along the direction of propagation:

$$\hat u = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{\langle 1,1,1\rangle}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\langle 1,1,1\rangle}{\sqrt{3}} = \left\langle \frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right\rangle$$.

Using the dot-product formula $$\hat u \cdot \hat i = |\hat u|\,|\hat i| \cos\theta$$.
Since $$|\hat u| = |\hat i| = 1$$, we get

$$\cos\theta = \hat u \cdot \hat i = \left\langle \frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right\rangle \cdot \langle 1,0,0\rangle = \frac{1}{\sqrt{3}}$$.

Therefore, $$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right).$$

The correct choice is Option A.

In Young's double slit experiment, the intensity from each slit is proportional to the slit width. If one slit has width $$w$$ and the other has width $$w/2$$, then $$I_1 = I$$ and $$I_2 = I/2$$ (say).

The maximum intensity is $$I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{I} + \sqrt{I/2}\right)^2 = I\left(1 + \dfrac{1}{\sqrt{2}}\right)^2$$

The minimum intensity is $$I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{I} - \sqrt{I/2}\right)^2 = I\left(1 - \dfrac{1}{\sqrt{2}}\right)^2$$

The ratio is:

$$\dfrac{I_{max}}{I_{min}} = \dfrac{\left(1 + \dfrac{1}{\sqrt{2}}\right)^2}{\left(1 - \dfrac{1}{\sqrt{2}}\right)^2} = \dfrac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1\right)^2}$$

$$= \dfrac{2 + 2\sqrt{2} + 1}{2 - 2\sqrt{2} + 1} = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$$

Hence, the correct answer is Option B.

Young's double slit experiment in a liquid of refractive index $$\mu = 1.44$$.

Slit separation $$d = 1.5$$ mm, wavelength in air $$\lambda_0 = 690$$ nm, screen distance $$D = 0.72$$ m.

Wavelength in liquid: $$\lambda = \frac{\lambda_0}{\mu} = \frac{690}{1.44} = 479.17$$ nm.

Fringe width: $$\beta = \frac{\lambda D}{d} = \frac{479.17 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3}}$$

$$= \frac{345 \times 10^{-9}}{1.5 \times 10^{-3}} = 230 \times 10^{-6} = 0.23 \times 10^{-3}$$ m $$= 0.23$$ mm.

The correct answer is Option A: 0.23 mm.

In Young’s double-slit experiment, the fringe width $$\beta$$ is given by the standard formula

$$\beta = \frac{\lambda D}{d}$$

where
$$\lambda$$ = wavelength of light,
$$D$$ = distance between the slits and the screen,
$$d$$ = separation between the two slits.

Case 1: Initial slit separation $$d_1 = 0.2\ \text{mm}$$.
Fringe width $$\beta_1 = \frac{\lambda D}{d_1}$$ $$-(1)$$

Case 2: New slit separation $$d_2 = 0.4\ \text{mm} = 2d_1$$.
Fringe width $$\beta_2 = \frac{\lambda D}{d_2} = \frac{\lambda D}{2d_1} = \frac{1}{2}\left(\frac{\lambda D}{d_1}\right)$$

Using $$-(1)$$, $$\beta_2 = \frac{1}{2}\beta_1$$ $$-(2)$$

The percentage change in fringe width is

$$\frac{\beta_2 - \beta_1}{\beta_1}\times 100\% = \frac{\frac{1}{2}\beta_1 - \beta_1}{\beta_1}\times 100\%$$

$$= \left(-\frac{1}{2}\right)\times 100\% = -50\%$$

The negative sign shows a decrease. Hence, the fringe width decreases by $$50\%$$.

Therefore, the percentage change (magnitude) is $$50\%$$  ⇒  Option C.

Two light beams have intensities in the ratio $$I_1 : I_2 = 1 : 9$$. Let $$I_1 = I$$ and $$I_2 = 9I$$.

The maximum intensity is $$I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{I} + 3\sqrt{I}\right)^2 = 16I$$

The minimum intensity is $$I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{I} - 3\sqrt{I}\right)^2 = 4I$$

The ratio is $$\dfrac{I_{max}}{I_{min}} = \dfrac{16I}{4I} = \dfrac{4}{1}$$

Hence, the correct answer is Option D.

Assertion (A): In Young's double slit experiment, fringes produced by red light are closer compared to those produced by blue light.

Reason (R): The fringe width is directly proportional to the wavelength of light.

In Young's double slit experiment, the fringe width is given by $$\beta = \frac{\lambda D}{d}$$ where $$\lambda$$ is the wavelength, $$D$$ is the screen distance, and $$d$$ is the slit separation. From the formula, $$\beta \propto \lambda$$, showing that the fringe width is indeed directly proportional to the wavelength of light, so the Reason (R) is true.

Red light has a longer wavelength ($$\approx 620-750$$ nm) than blue light ($$\approx 450-490$$ nm). Since $$\beta \propto \lambda$$, red light produces wider fringes, not closer fringes. Therefore the Assertion (A) is false.

Hence, (A) is false but (R) is true, which corresponds to Option 4.

Let the electric field (amplitude) produced by a slit be directly proportional to its width.

Width of slit 1 = $$d$$     → amplitude $$E_1 \propto d$$

Width of slit 2 = $$x\,d$$  → amplitude $$E_2 \propto x\,d$$

Choose the proportionality constant equal for both slits and write

$$E_1 = E_0 d, \qquad E_2 = E_0 x d$$

Hence the ratio of the two amplitudes is

$$\frac{E_2}{E_1}=x.$$

For any point on the screen the resultant intensity is

$$I = \bigl(E_1 + E_2\bigr)^2 = E_1^2 + E_2^2 + 2E_1E_2\cos\phi,$$

where $$\phi$$ is the phase difference between the two waves at that point.

At an interference maximum, $$\cos\phi = +1$$, so

$$I_{\max} = (E_1 + E_2)^2.$$

At an interference minimum, $$\cos\phi = -1$$, so

$$I_{\min} = (E_1 - E_2)^2.$$

Given that

$$\frac{I_{\max}}{I_{\min}} = \frac{9}{4},$$

write this ratio in terms of the amplitudes:

$$\frac{(E_1 + E_2)^2}{(E_1 - E_2)^2} = \frac{9}{4}.$$

Substitute $$E_2 = xE_1$$:

$$\frac{(E_1 + xE_1)^2}{(E_1 - xE_1)^2} = \frac{(1 + x)^2}{(1 - x)^2} = \frac{9}{4}.$$

Take square roots on both sides (keeping the positive value for the ratio of two positive quantities):

$$\frac{1 + x}{|\,1 - x\,|} = \frac{3}{2}.$$

Now two cases arise.

Case 1: $$x \lt 1$$  ⇒ $$|1 - x| = 1 - x$$

$$\frac{1 + x}{1 - x} = \frac{3}{2} \;\; \Longrightarrow \;\; 2 + 2x = 3 - 3x \;\; \Longrightarrow \;\; 5x = 1 \;\; \Longrightarrow \;\; x = 0.2.$$

This contradicts the statement that the second slit is wider than the first (because $$x d$$ would then be narrower). Therefore discard this case.

Case 2: $$x \gt 1$$  ⇒ $$|1 - x| = x - 1$$

$$\frac{1 + x}{x - 1} = \frac{3}{2} \;\; \Longrightarrow \;\; 2 + 2x = 3x - 3.$$

Rearrange: $$2 + 2x - 3x = -3 \;\; \Longrightarrow \;\; 2 - x = -3 \;\; \Longrightarrow \;\; x = 5.$$

The physically acceptable value is therefore

$$x = 5.$$

Hence the width of the second slit must be five times the width of the first slit.

Option B is correct.

Let the incident beam on $$P_1$$ be un-polarised with intensity $$I_0$$.

Step 1 (Polariser $$P_1$$)
An ideal polariser transmits one-half of the incident intensity. Hence, after $$P_1$$ the light is plane-polarised with

$$I_1 = \frac{I_0}{2}$$ $$-(1)$$

Step 2 (Geometry of the three axes)
$$P_1$$ and $$P_2$$ are crossed, so the angle between their transmission axes is $$90^\circ$$ (i.e. $$\frac{\pi}{2}$$).
Insert $$P_3$$ between them such that the angle between $$P_2$$ and $$P_3$$ is $$\theta$$. Therefore, the angle between $$P_1$$ and $$P_3$$ is $$\frac{\pi}{2} - \theta$$.

Step 3 (Intensity after $$P_3$$)
Using Malus’ law (transmitted intensity $$I = I_{\text{in}}\cos^2\alpha$$, where $$\alpha$$ is the angle between the light’s polarisation and the polariser’s axis), we have

$$I_2 = I_1 \cos^2\!\left(\frac{\pi}{2}-\theta\right) = \frac{I_0}{2}\,\sin^2\theta$$ $$-(2)$$

Step 4 (Intensity after $$P_2$$)
Now the light emerging from $$P_3$$ makes an angle $$\theta$$ with the axis of $$P_2$$, so again applying Malus’ law,

$$I_3 = I_2 \cos^2\theta = \frac{I_0}{2}\,\sin^2\theta\,\cos^2\theta$$

Combine the trigonometric factors:

$$\sin^2\theta\,\cos^2\theta = \left(\sin\theta\cos\theta\right)^2 = \left(\tfrac{1}{2}\sin2\theta\right)^2 = \frac{1}{4}\sin^2 2\theta$$

Hence

$$I_3 = \frac{I_0}{2}\times\frac{1}{4}\sin^2 2\theta = \frac{I_0}{8}\,\sin^2 2\theta$$ $$-(3)$$

Step 5 (Maximising the transmitted intensity)
The factor $$\sin^2 2\theta$$ attains its maximum value of $$1$$ when

$$2\theta = \frac{\pi}{2},\,\frac{3\pi}{2},\ldots$$ Taking the first positive solution inside $$0 \lt \theta \lt \frac{\pi}{2}$$ gives

$$\theta = \frac{\pi}{4}$$.

Conclusion
For maximum transmitted intensity through the three polarisers, the angle between $$P_2$$ and $$P_3$$ must be $$\frac{\pi}{4}$$.

Therefore, Option A is correct.

In Young's double slit experiment, the position of the $$n^{th}$$ bright fringe is given by $$y_n = \frac{n \lambda D}{d}$$, where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation. To determine the least number of bright fringes of 480 nm light that will coincide with those of 600 nm light, one sets the positions of the $$n_1^{th}$$ bright fringe of $$\lambda_1 = 480$$ nm and the $$n_2^{th}$$ bright fringe of $$\lambda_2 = 600$$ nm equal. Equating their expressions, $$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$$, simplifies to $$n_1 \lambda_1 = n_2 \lambda_2$$, or $$n_1 \times 480 = n_2 \times 600$$.

This equation can be written as $$\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}$$, meaning the smallest integer values satisfying the ratio are $$n_1 = 5$$ and $$n_2 = 4$$. Verification shows $$5 \times 480 = 2400$$ nm and $$4 \times 600 = 2400$$ nm, confirming that the 5th bright fringe of 480 nm light coincides with the 4th bright fringe of 600 nm light.

Therefore, the least number of bright fringes of 480 nm light required for the first coincidence is 5.

The correct answer is Option 1: 5.

We need to evaluate the Assertion and Reason about Young's double slit experiment in a denser medium.

Assertion (A): If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer.

Reason (R): The speed of light reduces in an optically denser medium than air while its frequency does not change.

The fringe width in Young's double slit experiment is given by:

$$\beta = \frac{\lambda D}{d}$$

where $$\lambda$$ is the wavelength of light, D is the distance to screen, and d is the slit separation.

In a denser medium, the wavelength decreases: $$\lambda_{medium} = \frac{\lambda_{air}}{n}$$, where n is the refractive index (n > 1).

Since $$\lambda$$ decreases, the fringe width $$\beta$$ decreases, meaning fringes come closer. So Assertion (A) is TRUE.

In a denser medium, the speed of light is $$v = c/n$$, which is less than c. The frequency remains unchanged ($$f = f_0$$). This is correct because the relationship $$v = f\lambda$$ means that when v decreases and f stays the same, $$\lambda$$ must decrease. So Reason (R) is TRUE.

The reason that fringes come closer is that $$\lambda$$ decreases in the denser medium. The wavelength decreases precisely because the speed decreases while frequency stays constant (as stated in R). Therefore, R correctly explains A.

The correct answer is Option 2: Both (A) and (R) are true and (R) is the correct explanation of (A).

For sustained interference in Young’s double-slit experiment, the two light waves reaching a point on the screen must satisfy two basic conditions:
  • They must have the same (or almost the same) frequency / wavelength.
  • They must maintain a constant phase difference.

If the source is white light and both slits are uncovered, every wavelength from the same source reaches both slits, so corresponding components of each wavelength are coherent and an interference pattern is obtained (central white fringe, coloured side fringes).

In the present situation one slit is covered with a red filter while the other is covered with a green filter. Hence:

  • The light emerging from slit $$S_1$$ is almost monochromatic red with wavelength say $$\lambda_r$$.
  • The light emerging from slit $$S_2$$ is almost monochromatic green with wavelength say $$\lambda_g$$, where $$\lambda_g \neq \lambda_r$$.

Therefore the two beams that superpose on the screen have different frequencies $$\nu_r$$ and $$\nu_g$$. The phase difference between them changes continuously with time because

$$\text{Phase difference} = 2\pi\bigl(\nu_r - \nu_g\bigr)t + \text{(constant path term)}$$

Since $$\nu_r \neq \nu_g$$, the term $$2\pi\bigl(\nu_r - \nu_g\bigr)t$$ varies rapidly; the phase relation is not stable. Time-averaged (detector) intensity at any point becomes simply the sum of individual intensities:

$$I = I_r + I_g$$

No position on the screen receives a consistently larger or smaller resultant amplitude, so bright and dark bands cannot form. Each slit only produces its own single-slit diffraction envelope; the two envelopes just overlap without producing any alternating maxima and minima.

Hence no interference fringes are observed.

Correct option: Option B

In Young's double-slit interference experiment, the position of the nth bright fringe from the central maximum is given by the formula:

$$ x = \frac{n \lambda D}{d} $$

where:
- $$n$$ is the fringe order,
- $$\lambda$$ is the wavelength of light,
- $$D$$ is the distance from the slits to the screen,
- $$d$$ is the separation between the slits.

For the first light source:
Wavelength, $$\lambda_1 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$$,
Position of the 10th bright fringe, $$x_1 = 10 \text{ mm} = 0.01 \text{ m}$$,
Fringe order, $$n = 10$$.

Substituting into the formula:

$$ x_1 = \frac{10 \lambda_1 D}{d} = 0.01 \quad ...(1) $$

For the second light source:
Wavelength, $$\lambda_2 = 660 \text{ nm} = 660 \times 10^{-9} \text{ m}$$,
Fringe order, $$n = 10$$ (same as before),
We need to find the new position $$x_2$$.

The formula becomes:

$$ x_2 = \frac{10 \lambda_2 D}{d} \quad ...(2) $$

Since the experimental setup ($$D$$ and $$d$$) remains unchanged, we can take the ratio of equations (2) and (1):

$$ \frac{x_2}{x_1} = \frac{\frac{10 \lambda_2 D}{d}}{\frac{10 \lambda_1 D}{d}} = \frac{\lambda_2}{\lambda_1} $$

Therefore:

$$ x_2 = x_1 \cdot \frac{\lambda_2}{\lambda_1} $$

Substituting the given values (keeping units consistent in mm and nm for simplicity):
$$x_1 = 10 \text{ mm}$$,
$$\lambda_1 = 600 \text{ nm}$$,
$$\lambda_2 = 660 \text{ nm}$$.

So:

$$ x_2 = 10 \times \frac{660}{600} $$

Simplifying the fraction:

$$ \frac{660}{600} = \frac{66}{60} = \frac{11}{10} = 1.1 $$

Thus:

$$ x_2 = 10 \times 1.1 = 11 \text{ mm} $$

The distance of the 10th bright fringe from the central maximum for the second wavelength is 11 mm.

For a thin film of refractive index $$n$$ on both sides by air (refractive index 1), the net phase difference between the two reflected rays is given by the formula:

$$ \Delta \phi = \frac{4\pi n t}{\lambda} + \pi \quad\text{-(1)} $$

The extra $$\pi$$ arises from the phase inversion at the top surface.

Transmission minima correspond to reflection maxima, which require constructive interference of the reflected waves, so the condition is

$$ \Delta \phi = 2\,m\,\pi,\quad m=1,2,3,\dots \quad\text{-(2)} $$

Equating (1) and (2) gives

$$ \frac{4\pi n t}{\lambda} + \pi = 2\,m\,\pi. $$

Simplifying yields the film thickness at the $$m^{\text{th}}$$ minimum:

$$ t_m = \frac{(2m - 1)\,\lambda}{4\,n} \quad\text{-(3)} $$

The change in thickness between successive minima ($$m$$ to $$m+1$$) is

$$ \Delta t = t_{m+1} - t_m = \frac{\lambda}{2\,n} \quad\text{-(4)} $$

Substituting $$\lambda = 560 \times 10^{-9}\,\text{m}$$ and $$n = 1.4$$ gives

$$ \Delta t = \frac{560\times10^{-9}}{2\times1.4} = 2.0\times10^{-7}\,\text{m}. $$

The film is supported on a circular ring of radius $$r = 1.8\,\text{cm} = 0.018\,\text{m}$$, so its area is

$$ A = \pi\,r^2 = \pi\,(0.018)^2 = 3.24\times10^{-4}\,\pi~\text{m}^2. $$

When the film thickness decreases by $$\Delta t$$, the volume of liquid evaporated is

$$ \Delta V = A\,\Delta t = (3.24\times10^{-4}\,\pi)\,(2.0\times10^{-7}) = 6.48\times10^{-11}\,\pi~\text{m}^3. $$

This occurs every $$\Delta t_{\text{time}} = 12\,\text{s}$$.

The evaporation rate is

$$ \frac{\Delta V}{\Delta t_{\text{time}}} = \frac{6.48\times10^{-11}\,\pi}{12} = 5.4\times10^{-12}\,\pi~\text{m}^3/\text{s}. $$

Expressing this in the form $$\pi\times10^{-13}$$ gives

$$ 5.4\times10^{-12} = 54\times10^{-13}. $$

Hence the evaporation rate is

$$ 54\,\pi\times10^{-13}~\text{m}^3/\text{s}. $$

$$54\,\pi\times10^{-13}\,\text{m}^3/\text{s}$$

For a single-slit Fraunhofer diffraction pattern the angular position of the $$m^{\text{th}}$$ minimum is given by the condition

$$a \sin\theta_m = m\lambda \qquad (m = \pm 1, \pm 2, \dots) \; -(1)$$

Here $$a$$ is the slit width and $$\lambda$$ the wavelength of light.

The second minimum on the left of the central maximum corresponds to $$m = -2$$, while the third minimum on the right corresponds to $$m = +3$$. Their angular positions (taking magnitudes) are therefore

$$\theta_2 = \arcsin\!\left( \frac{2\lambda}{a} \right), \quad \theta_3 = \arcsin\!\left( \frac{3\lambda}{a} \right) \; -(2)$$

The measured angular separation between these two minima is given to be $$30^\circ$$:

$$\theta_3 + \theta_2 = 30^\circ = \frac{\pi}{6}\ \text{rad} \; -(3)$$

Both angles are well below $$20^\circ$$, so the small-angle approximation $$\sin\theta \approx \theta$$ (in radians) is justified. Using this in (2):

$$\theta_3 \approx \frac{3\lambda}{a}, \qquad \theta_2 \approx \frac{2\lambda}{a} \; -(4)$$

Add the two expressions from (4) and set them equal to (3):

$$\frac{3\lambda}{a} + \frac{2\lambda}{a} = \frac{\pi}{6}$$

$$\Rightarrow \frac{5\lambda}{a} = \frac{\pi}{6}$$

$$\Rightarrow a = \frac{5\lambda \times 6}{\pi} \; -(5)$$

Given $$\lambda = 628\ \text{nm} = 0.628\ \mu\text{m}$$, substitute in (5):

$$a = \frac{30 \times 0.628}{\pi}\ \mu\text{m} \approx \frac{18.84}{3.142}\ \mu\text{m} \approx 6.0\ \mu\text{m}$$

Hence, the width of the slit is $$\mathbf{6\ \mu\text{m}}$$.

Let the intensities of the two coherent beams be $$I_1 = 4I$$ and $$I_2 = 9I$$.

For two coherent sources, the resultant intensity at any point is given by the interference formula
$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\,\cos\phi$$
where $$\phi$$ is the phase difference between the beams.

Maximum intensity ($$\phi = 0$$, constructive interference):
$$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$

Minimum intensity ($$\phi = \pi$$, destructive interference):
$$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$$

The required quantity is the difference between these two:
$$I_{\max} - I_{\min} = \left(I_1 + I_2 + 2\sqrt{I_1 I_2}\right) \;-\; \left(I_1 + I_2 - 2\sqrt{I_1 I_2}\right)$$

Simplifying gives
$$I_{\max} - I_{\min} = 4\sqrt{I_1 I_2}$$

Substitute $$I_1 = 4I$$ and $$I_2 = 9I$$:
$$\sqrt{I_1 I_2} = \sqrt{(4I)(9I)} = \sqrt{36I^2} = 6I$$

Hence
$$I_{\max} - I_{\min} = 4 \times 6I = 24I$$

The problem states that this difference equals $$xI$$, so
$$x = 24$$

Therefore, the value of $$x$$ is $$24$$.

For a single slit of width $$a$$, the first diffraction minima occur at angles $$\theta$$ that satisfy

$$a \sin \theta = \pm \lambda \qquad -(1)$$

Hence the central diffraction maximum extends from $$-\theta_1$$ to $$+\theta_1$$ where

$$\sin \theta_1 = \frac{\lambda}{a} \qquad -(2)$$

For the Young’s double-slit interference, bright fringes (maxima) are obtained at

$$d \sin \theta = m \lambda ,\; m = 0, \pm 1, \pm 2, \ldots \qquad -(3)$$

The extreme interference maximum that can still lie inside the central diffraction maximum is obtained by equating the conditions $$(2)$$ and $$(3)$$:

$$d \sin \theta_1 = m_{\text{max}}\lambda$$
$$\Rightarrow d \left(\frac{\lambda}{a}\right) = m_{\text{max}}\lambda$$
$$\Rightarrow m_{\text{max}} = \frac{d}{a} \qquad -(4)$$

The problem states that a total of 20 interference maxima are contained inside the central diffraction maximum. These 20 maxima are the bright fringes on both sides of the central bright fringe but excluding the central one itself. Therefore

Number of maxima on one side $$= \frac{20}{2}=10$$
$$\Rightarrow m_{\text{max}} = 10 \qquad -(5)$$

Substituting $$(5)$$ into $$(4)$$ gives the required slit width:

$$a = \frac{d}{m_{\text{max}}} = \frac{1.5 \text{ mm}}{10} = 0.15 \text{ mm}$$

Converting to centimetres,

$$0.15 \text{ mm} = 0.015 \text{ cm} = 15 \times 10^{-3} \text{ cm}$$

Thus, the width of each slit is $$15 \times 10^{-3}$$ cm, so the required value of $$x$$ is $$15$$.

Unpolarised light of intensity $$I_0$$ passes through polaroid A, then polaroid B at $$45°$$.

After polaroid A (first polarizer), intensity becomes $$I_1 = \frac{I_0}{2}$$ (Malus's law for unpolarised light through a polarizer).

After polaroid B at angle $$45°$$ to A. By Malus's law: $$I_2 = I_1\cos^2 45° = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$.

The correct answer is Option 1: $$\frac{I_0}{4}$$.

Find the linear width of the central maximum in single-slit diffraction.

In single-slit diffraction the first minima occur at angles where $$a\sin\theta = \pm\lambda$$. For small angles ($$\sin\theta \approx \theta$$) the distance from the center to the first minimum on the screen is given by $$y = \frac{\lambda f}{a}$$, where $$f$$ is the focal length of the lens, so the total width of the central maximum is $$2y = \frac{2\lambda f}{a}$$.

The wavelength is $$\lambda = 6000\text{ A} = 6000 \times 10^{-10}\text{ m} = 6 \times 10^{-7}\text{ m}$$, the slit width is $$a = 0.01\text{ mm} = 0.01 \times 10^{-3}\text{ m} = 1 \times 10^{-5}\text{ m}$$, and the focal length is $$f = 20\text{ cm} = 0.2\text{ m}$$.

Substituting these values into the expression for the width gives $$2y = \frac{2 \times 6 \times 10^{-7} \times 0.2}{1 \times 10^{-5}} = \frac{2.4 \times 10^{-7}}{10^{-5}} = 2.4 \times 10^{-2}\text{ m} = 24\text{ mm}$$.

The correct answer is Option B: 24 mm.

In Young's double slit experiment, when one slit has width 4 times that of the other, the amplitudes are related to the slit widths.

The intensity is proportional to the slit width (since wider slits allow more light through). So if one slit has width $$w$$ and the other has width $$4w$$:

$$I_1 : I_2 = 1 : 4$$

Since amplitude is proportional to the square root of intensity:

$$a_1 : a_2 = 1 : 2$$

The maximum intensity occurs when the waves interfere constructively:

$$I_{max} = (a_1 + a_2)^2 = (1 + 2)^2 = 9$$

The minimum intensity occurs when the waves interfere destructively:

$$I_{min} = (a_1 - a_2)^2 = (1 - 2)^2 = 1$$

Therefore, the ratio of maximum to minimum intensity is:

$$\frac{I_{max}}{I_{min}} = \frac{9}{1} = 9 : 1$$

The answer is Option C: $$9 : 1$$.

When a polaroid is placed between two crossed polaroids (at 90° to each other), and the middle polaroid makes angle $$\theta$$ with the first:

Intensity after first polaroid: $$I_0/2$$

After middle polaroid: $$\frac{I_0}{2}\cos^2\theta$$

After last polaroid (at $$90° - \theta$$ to middle): $$\frac{I_0}{2}\cos^2\theta \cos^2(90°-\theta) = \frac{I_0}{2}\cos^2\theta\sin^2\theta = \frac{I_0}{8}\sin^2 2\theta$$

Maximum when $$\sin^2 2\theta = 1$$, i.e., $$2\theta = 90°$$, i.e., $$\theta = 45°$$.

The answer is $$45°$$, which corresponds to Option (4).

We need to find the angle of refraction when unpolarised light is reflected with complete polarisation.

Key Concept: Brewster's Law

When unpolarised light strikes a surface at the Brewster angle ($$i_B$$), the reflected light is completely polarised. At this angle:

$$\tan i_B = \mu$$

Also, at Brewster's angle, the reflected and refracted rays are perpendicular:

$$i_B + r = 90°$$

Angle of incidence = 60° (this is the Brewster angle).

Finding the angle of refraction:

$$r = 90° - i_B = 90° - 60° = 30°$$

Verification: $$\tan 60° = \sqrt{3} = \mu$$. By Snell's law: $$\sin 60° = \sqrt{3} \sin r$$, so $$\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$$, giving $$r = 30°$$. ✓

The correct answer is Option 1: 30°.

A microwave of wavelength $$\lambda = 2.0$$ cm falls normally on a slit of width $$a = 4.0$$ cm, and we need the angular spread of the central maximum.

In single‐slit diffraction the first minimum occurs where $$a\sin\theta = \lambda$$, so that $$\sin\theta = \frac{\lambda}{a} = \frac{2.0}{4.0} = 0.5$$ and hence $$\theta = 30°$$.

The central maximum extends from $$-\theta$$ to $$+\theta$$ on either side, giving a total angular spread of $$2\theta = 2 \times 30° = 60°$$.

The correct answer is Option C) 60 degrees.

Path diff=7λ/4. Phase diff=2π×7/4=7π/2. cos(7π/2)=cos(3π+π/2)=0.

I=I_max×cos²(δ/2)=I_max×cos²(7π/4)=I_max×(1/√2)²=I_max/2. Ratio=1/2. Option (1).

In single slit diffraction, the positions of the minima are given by:

$$a \sin\theta = n\lambda$$

where $$a$$ is the slit width, $$\lambda$$ is the wavelength, and $$n$$ is the order of the minimum.

For small angles, $$\sin\theta \approx \tan\theta = \frac{y}{f}$$, where $$y$$ is the position on the focal plane and $$f$$ is the focal length of the lens.

The position of the $$n$$-th minimum is:

$$y_n = \frac{n\lambda f}{a}$$

The width of the 1st secondary maximum is the distance between the 1st and 2nd minima:

$$\Delta y = y_2 - y_1 = \frac{2\lambda f}{a} - \frac{\lambda f}{a} = \frac{\lambda f}{a}$$

Substituting the given values: $$\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$$, $$f = 100 \text{ cm} = 1 \text{ m}$$, $$a = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$$:

$$\Delta y = \frac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = \frac{400 \times 10^{-9}}{2 \times 10^{-4}} = 2 \times 10^{-3} \text{ m} = 2 \text{ mm}$$

The correct answer is $$2 \text{ mm}$$.

Let $$I_A$$ and $$I_B$$ be the intensities at points $$A$$ and $$B$$ before inserting the polaroids.
Given $$\dfrac{I_A}{I_B}=2 \; \Rightarrow \; I_A = 2\,I_B$$ $$-(1)$$

Now place two polaroids in front of point $$B$$. Call the initial unpolarized intensity reaching the first polaroid $$I_B$$.

Step 1: First polaroid
Transmission of unpolarized light through a single polaroid is $$\dfrac{1}{2}$$ of the incident intensity.
Therefore the intensity emerging from the first polaroid is
$$I_1 = \dfrac{I_B}{2}$$ $$-(2)$$
The light is now plane-polarized along the pass-axis of the first polaroid.

Step 2: Second polaroid
The pass-axis of the second polaroid is at $$45^{\circ}$$ to that of the first. For plane-polarized light, Malus’ law states
$$I = I_{\text{in}} \cos^{2}\theta$$
where $$\theta$$ is the angle between the light’s polarization direction and the pass-axis of the analyzer (second polaroid). Here $$\theta = 45^{\circ}$$, so $$\cos^{2}45^{\circ} = \left(\dfrac{1}{\sqrt 2}\right)^2 = \dfrac{1}{2}$$.

Hence the intensity after the second polaroid is
$$I_B' = I_1 \cos^{2}45^{\circ} = \dfrac{I_B}{2} \times \dfrac{1}{2} = \dfrac{I_B}{4}$$ $$-(3)$$

Step 3: New intensity ratio
Point $$A$$ is unaffected, so its intensity remains $$I_A$$. The new ratio of intensities is
$$r_{\text{new}} = \dfrac{I_A}{I_B'} = \dfrac{I_A}{\dfrac{I_B}{4}} = 4 \dfrac{I_A}{I_B}$$

Using $$(1)$$, $$\dfrac{I_A}{I_B}=2$$, therefore
$$r_{\text{new}} = 4 \times 2 = 8$$

Hence the new value of $$r$$ is $$8$$.

For single slit diffraction, the first minimum occurs at:

$$a\sin\theta = \lambda$$

where $$a = 0.001$$ mm $$= 10^{-6}$$ m and $$\lambda = 5000 \; \mathring{A} = 5 \times 10^{-7}$$ m.

$$\sin\theta = \frac{\lambda}{a} = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5$$

$$\theta = 30°$$

The answer is $$\boxed{30}$$ degrees.

Find the angular divergence for the second-order minima in single-slit diffraction.

$$a\sin\theta = n\lambda$$

where $$a$$ is the slit width, $$n$$ is the order, and $$\lambda$$ is the wavelength.

$$\sin\theta = \frac{2\lambda}{a} = \frac{2 \times 600 \times 10^{-9}}{0.4 \times 10^{-3}} = \frac{1200 \times 10^{-9}}{4 \times 10^{-4}} = 3 \times 10^{-3}$$

Since $$\sin\theta$$ is very small, $$\theta \approx \sin\theta = 3 \times 10^{-3}$$ rad.

The angular divergence is the total angle between the 2nd order minima on both sides of the central maximum:

$$\text{Angular divergence} = 2\theta = 2 \times 3 \times 10^{-3} = 6 \times 10^{-3} \text{ rad}$$

The correct answer is 6.

In a single slit diffraction pattern, the position of the $$n$$th minimum is given by $$y_n = \frac{n\lambda D}{a}$$, where the wavelength $$\lambda = 6000$$ Angstrom $$= 6000 \times 10^{-10}$$ m $$= 6 \times 10^{-7}$$ m, the distance from slit to screen $$D = 50$$ cm $$= 0.50$$ m, and $$a$$ is the slit width.

The distance between the first and third minima is $$y_3 - y_1 = 3$$ mm $$= 3 \times 10^{-3}$$ m; substituting into the expression gives $$y_3 - y_1 = \frac{3\lambda D}{a} - \frac{1 \cdot \lambda D}{a} = \frac{2\lambda D}{a}.$$

Rearranging yields $$a = \frac{2\lambda D}{y_3 - y_1},$$ and thus $$a = \frac{2 \times 6 \times 10^{-7} \times 0.50}{3 \times 10^{-3}} = \frac{6 \times 10^{-7}}{3 \times 10^{-3}} = 2 \times 10^{-4} \text{ m}.$$

Expressed as $$\_\_\_ \times 10^{-4}$$ m, the numerical result is 2.

We need to find the position of the first order minima in a single slit diffraction experiment.

Recall the condition for minima in single slit diffraction

For a single slit of width $$d$$, the condition for the $$n$$-th order minima is:

$$d \sin\theta = n\lambda$$

where $$\lambda$$ is the wavelength and $$n = 1, 2, 3, \ldots$$

Use the small angle approximation

For small angles (which is valid when the slit width is much larger than the wavelength), $$\sin\theta \approx \tan\theta = \frac{y}{D}$$, where $$y$$ is the distance of the minima from the central maximum and $$D$$ is the distance to the screen.

Substituting into the minima condition:

$$d \cdot \frac{y}{D} = n\lambda$$

$$y = \frac{n\lambda D}{d}$$

Substitute the given values for the first order minimum ($$n = 1$$)

Given: $$\lambda = 550$$ nm $$= 550 \times 10^{-9}$$ m, $$D = 100$$ cm $$= 1$$ m, $$d = 0.20$$ mm $$= 0.20 \times 10^{-3}$$ m $$= 2 \times 10^{-4}$$ m.

$$y = \frac{1 \times 550 \times 10^{-9} \times 1}{2 \times 10^{-4}}$$

$$y = \frac{550 \times 10^{-9}}{2 \times 10^{-4}} = 275 \times 10^{-5} \text{ m}$$

Therefore, $$x = 275$$.

The answer is 275.

Find the distance from center where intensity becomes half of maximum for the first time in Young's double slit experiment.

In Young's experiment, the intensity at a point on the screen is:

$$ I = I_0 \cos^2\left(\frac{\phi}{2}\right) $$

where $$\phi$$ is the phase difference.

$$ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right) $$

$$ \cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2} $$

$$ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} $$

$$ \frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2} $$

$$ \phi = \frac{2\pi}{\lambda} \times \Delta x $$

where $$\Delta x$$ is the path difference. So:

$$ \frac{\pi}{2} = \frac{2\pi}{\lambda} \times \Delta x \implies \Delta x = \frac{\lambda}{4} $$

Path difference $$\Delta x = \frac{yd}{D}$$ where $$y$$ is distance from center, $$d$$ is slit separation, $$D$$ is screen distance.

$$ \frac{\lambda}{4} = \frac{yd}{D} \implies y = \frac{\lambda D}{4d} $$

$$\lambda = 5000$$ A $$= 5 \times 10^{-7}$$ m, $$d = 1.0$$ mm $$= 10^{-3}$$ m, $$D = 1.0$$ m.

$$ y = \frac{5 \times 10^{-7} \times 1.0}{4 \times 10^{-3}} = \frac{5 \times 10^{-7}}{4 \times 10^{-3}} = 1.25 \times 10^{-4} \text{ m} = 125 \times 10^{-6} \text{ m} $$

The answer is 125 $$\times 10^{-6}$$ m.

We need to find the difference between maximum and minimum intensities when two coherent beams of intensities $$I$$ and $$4I$$ are superimposed.

For two coherent light beams with intensities $$I_1$$ and $$I_2$$, the maximum intensity for constructive interference is given by $$I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2$$ and the minimum intensity for destructive interference is given by $$I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2$$.

Substituting $$I_1 = I$$ and $$I_2 = 4I$$ into the expression for the maximum intensity gives $$I_{\max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$$, while substituting into the minimum intensity yields $$I_{\min} = (\sqrt{4I} - \sqrt{I})^2 = (2\sqrt{I} - \sqrt{I})^2 = (\sqrt{I})^2 = I$$.

The difference between these extremes is $$I_{\max} - I_{\min} = 9I - I = 8I$$, which shows that $$x = 8$$.

The correct answer is 8.

Given: slit separation $$d = 1 \text{ mm} = 10^{-3} \text{ m}$$, screen distance $$D = 1 \text{ m}$$, wavelength $$\lambda = 500 \text{ nm} = 5 \times 10^{-7} \text{ m}$$.

Width of central maximum of single-slit diffraction pattern:

The angular half-width of the central maximum is $$\theta_0 = \frac{\lambda}{w}$$. The linear half-width on the screen is $$y_0 = \frac{D\lambda}{w}$$. The full width of the central maximum is $$2y_0 = \frac{2D\lambda}{w}$$.

Fringe width of double-slit interference pattern:

The fringe spacing (distance between consecutive maxima) is $$\beta = \frac{D\lambda}{d}$$.

Number of interference maxima within the central diffraction maximum:

The number of double-slit fringes that fit within the central diffraction maximum is:

$$N = \frac{\text{Width of central maximum}}{\text{Fringe width}} = \frac{2D\lambda / w}{D\lambda / d} = \frac{2d}{w}$$

We are told $$N = 10$$:

$$\frac{2d}{w} = 10$$

$$w = \frac{2d}{10} = \frac{d}{5}$$

$$w = \frac{10^{-3}}{5} = 2 \times 10^{-4} \text{ m}$$

The answer is $$2$$.

Two wavelengths $$\lambda_1 = 450$$ nm and $$\lambda_2 = 650$$ nm are used in Young's double slit experiment. We need to find the minimum order of fringe produced by $$\lambda_2$$ that overlaps with a fringe of $$\lambda_1$$.

Recall the condition for bright fringes.

The position of the $$n$$-th bright fringe is $$y_n = \frac{n\lambda D}{d}$$, where $$D$$ is the screen distance and $$d$$ is the slit separation.

Set up the overlap condition.

For fringes to overlap: $$n_1\lambda_1 = n_2\lambda_2$$

$$ n_1 \times 450 = n_2 \times 650 $$

$$ \frac{n_1}{n_2} = \frac{650}{450} = \frac{13}{9} $$

Find the minimum integer solution.

Since 13 and 9 are coprime (no common factors), the minimum values are $$n_1 = 13$$ and $$n_2 = 9$$.

The minimum order of fringe produced by $$\lambda_2$$ that overlaps is $$n = \boxed{9}$$.

We need to find the thickness of a glass plate that shifts the central maximum to the position of the 4th bright fringe in Young's double slit experiment.

Wavelength: $$\lambda = 500 \text{ nm}$$

Refractive index of glass plate: $$\mu = 1.5$$

Central maximum shifts to the position of the 4th bright fringe.

When a glass plate of thickness $$t$$ and refractive index $$\mu$$ is placed in front of one slit, it introduces an additional optical path. Inside the glass, the wavelength becomes $$\lambda/\mu$$, so the optical path through the glass is $$\mu t$$ instead of $$t$$. The extra path difference introduced is:

$$\Delta = \mu t - t = (\mu - 1)t$$

The central maximum originally occurs where the path difference from both slits is zero. With the glass plate, the central maximum shifts to the point where the extra path difference due to the glass exactly compensates. For the central maximum to shift to the position of the $$n^{th}$$ bright fringe, the extra path difference must equal $$n\lambda$$:

$$(\mu - 1)t = n\lambda$$

Here, $$n = 4$$ (shift to 4th bright fringe position):

$$(1.5 - 1) \times t = 4 \times 500 \text{ nm}$$

$$0.5 \times t = 2000 \text{ nm}$$

$$t = \frac{2000}{0.5} = 4000 \text{ nm} = 4 \text{ } \mu m$$

The answer is 4 $$\mu$$m.

The position of the $$m^{\text{th}}$$ bright fringe on the screen (measured from the central point $$O$$) in a Young’s double-slit experiment is given by the condition for constructive interference: $$y_m = \dfrac{m \, \lambda \, D}{d}$$, where

$$m$$ = fringe order, here $$m = 8$$,
$$\lambda$$ = wavelength of light, $$6000 \text{ Å} = 6000 \times 10^{-10}\,\text{m} = 6 \times 10^{-7}\,\text{m}$$,
$$D$$ = slit-screen distance, $$1\,\text{m}$$,
$$d$$ = instantaneous slit separation.

The slit separation oscillates as $$d = \bigl(0.8 + 0.04 \sin \omega t\bigr)\,\text{mm}$$ with $$\omega = 0.08\,\text{rad s}^{-1}$$. Hence

Maximum separation (when $$\sin \omega t = +1$$): $$d_{\max} = (0.8 + 0.04)\,\text{mm} = 0.84\,\text{mm} = 0.84 \times 10^{-3}\,\text{m}$$.
Minimum separation (when $$\sin \omega t = -1$$): $$d_{\min} = (0.8 - 0.04)\,\text{mm} = 0.76\,\text{mm} = 0.76 \times 10^{-3}\,\text{m}$$.

Because $$y_m$$ is inversely proportional to $$d$$, the fringe lies farthest from $$O$$ when $$d$$ is minimum and closest when $$d$$ is maximum.

Extreme positions of the 8th bright fringe:

$$y_{\max} = \dfrac{8 \times 6 \times 10^{-7} \times 1}{0.76 \times 10^{-3}} = \dfrac{4.8 \times 10^{-6}}{0.76 \times 10^{-3}} = \left( \dfrac{4.8}{0.76} \right) \times 10^{-3}\,\text{m} \approx 6.3158 \times 10^{-3}\,\text{m} = 6.3158\,\text{mm}$$.

$$y_{\min} = \dfrac{8 \times 6 \times 10^{-7} \times 1}{0.84 \times 10^{-3}} = \dfrac{4.8 \times 10^{-6}}{0.84 \times 10^{-3}} = \left( \dfrac{4.8}{0.84} \right) \times 10^{-3}\,\text{m} \approx 5.7143 \times 10^{-3}\,\text{m} = 5.7143\,\text{mm}$$.

Separation between the two extreme positions:

$$\Delta y = y_{\max} - y_{\min} = (6.3158 - 5.7143)\,\text{mm} = 0.6015\,\text{mm}$$.

Converting to micrometres: $$0.6015\,\text{mm} = 0.6015 \times 10^{3}\,\mu\text{m} = 601.5\,\mu\text{m}$$.

Hence, the required separation is 601.50 µm.

For the 8th bright fringe in Young’s double-slit experiment, the instantaneous position on the screen is

$$y_m(t)=\frac{m\lambda D}{d(t)}$$

with $$m=8,\;D=1\text{ m},\;\lambda =6000\text{ \AA}=6\times10^{-7}\text{ m}$$ and

$$d(t)=\bigl(0.8+0.04\sin\omega t\bigr)\text{ mm} =(8\times10^{-4}+4\times10^{-5}\sin\omega t)\text{ m},\qquad \omega =0.08\text{ rad s}^{-1}$$

The speed of this fringe is obtained by differentiating:

$$v(t)=\frac{dy_m}{dt} =-\frac{m\lambda D}{\bigl[d(t)\bigr]^2}\,\frac{dd}{dt}$$

First derivative of the slit separation:

$$\frac{dd}{dt}=4\times10^{-5}\;\omega\cos\omega t =4\times10^{-5}\times0.08\cos\omega t =3.2\times10^{-6}\cos\omega t\;(\text{m s}^{-1})$$

Hence

$$v(t)=\frac{8\,(6\times10^{-7})(3.2\times10^{-6})\, |\cos\omega t|}{\bigl(8\times10^{-4}+4\times10^{-5}\sin\omega t\bigr)^2}$$

To find the maximum value, put $$\theta=\omega t$$ and maximise

$$f(\theta)=\frac{|\cos\theta|} {\bigl(a+b\sin\theta\bigr)^2},\qquad a=8\times10^{-4},\;b=4\times10^{-5}$$

Within $$0\le\theta\lt2\pi$$, choose the region $$\cos\theta\gt0$$ (same magnitude repeats elsewhere).
Set $$\displaystyle\frac{d}{d\theta}\ln f(\theta)=0$$:

$$-\tan\theta-\frac{2b\cos\theta}{a+b\sin\theta}=0$$

After simplification, using $$x=\sin\theta$$:

$$b x^{2}-a x-2b=0$$

Substituting $$a=8\times10^{-4},\;b=4\times10^{-5}$$ gives

$$x^{2}-20x-2=0\; \Longrightarrow\; x=\sin\theta\approx-0.0995$$

With this $$\cos\theta=\sqrt{1-x^{2}}\approx0.995$$ and

$$d_{\text{max}}=a+b\sin\theta =8\times10^{-4}+4\times10^{-5}(-0.0995) \approx7.960\times10^{-4}\text{ m}$$

Finally, the maximum fringe speed is

$$v_{\max} =\frac{8(6\times10^{-7})(3.2\times10^{-6})(0.995)} {(7.960\times10^{-4})^{2}} \approx2.4\times10^{-5}\text{ m s}^{-1}$$

Expressing in micrometres per second:

$$v_{\max}\approx2.4\times10^{-5}\text{ m s}^{-1} =24\text{ }\mu\text{m s}^{-1}$$

Therefore, the maximum speed of the 8th bright fringe is 24 μm/s.

We need to find the ratio of intensities at points P and Q on the screen in Young’s double slit experiment, where the phase differences are $$\frac{\pi}{3}$$ and $$\frac{\pi}{2}$$ respectively.

In Young’s double slit experiment, when two coherent waves of equal amplitude $$a$$ interfere with a phase difference $$\phi$$, the resultant intensity at that point is given by $$I = 4I_0\cos^2\left(\frac{\phi}{2}\right)$$, where $$I_0$$ is the intensity due to a single slit (i.e., $$I_0 \propto a^2$$). This formula follows from the superposition principle: if the two waves are $$y_1 = a\sin(\omega t)$$ and $$y_2 = a\sin(\omega t + \phi)$$, then the resultant amplitude is $$A = 2a\cos(\phi/2)$$, and since intensity is proportional to the square of amplitude, we obtain $$I = 4I_0\cos^2(\phi/2)$$.

We begin by calculating the intensity at point P where the phase difference is $$\phi_P = \frac{\pi}{3}$$. Substituting into the formula gives $$I_P = 4I_0\cos^2\left(\frac{\pi/3}{2}\right) = 4I_0\cos^2\left(\frac{\pi}{6}\right)$$. We know that $$\cos\left(\frac{\pi}{6}\right) = \cos 30° = \frac{\sqrt{3}}{2}$$, and therefore $$I_P = 4I_0 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \times \frac{3}{4} = 3I_0$$.

Next, we calculate the intensity at point Q where the phase difference is $$\phi_Q = \frac{\pi}{2}$$. Substituting into the same formula yields $$I_Q = 4I_0\cos^2\left(\frac{\pi/2}{2}\right) = 4I_0\cos^2\left(\frac{\pi}{4}\right)$$. Since $$\cos\left(\frac{\pi}{4}\right) = \cos 45° = \frac{1}{\sqrt{2}}$$, it follows that $$I_Q = 4I_0 \times \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \times \frac{1}{2} = 2I_0$$.

Finally, the ratio of the intensities is $$\frac{I_P}{I_Q} = \frac{3I_0}{2I_0} = \frac{3}{2}$$, which shows that $$I_P : I_Q = 3 : 2$$. The correct answer is Option 4: 3 : 2.

Three polaroids A, B, and C are arranged such that:

- Pass axes of A and B are perpendicular to each other

- C is placed between A and B at 45° to both

The unpolarized light of intensity $$I_0$$ passes through in sequence:

To begin, after Polaroid A,

$$ I_1 = \frac{I_0}{2} $$

(Unpolarized light through a polaroid gives half the intensity)

Next, after Polaroid C (at 45° to A),

By Malus's law:

$$ I_2 = I_1 \cos^2 45° = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4} $$

From this, after Polaroid B (at 45° to C, or 90° to A),

$$ I_3 = I_2 \cos^2 45° = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8} $$

We are given two statements about Brewster's angle and need to verify their truth.

First, recall the formula for Brewster's angle. When light travels from a medium with refractive index $$n_1$$ to a medium with refractive index $$n_2$$, Brewster's angle $$\theta_B$$ is given by:

$$\tan \theta_B = \frac{n_2}{n_1}$$

Now, analyze Statement I: If Brewster's angle for light propagating from air to glass is $$\theta_B$$, then Brewster's angle for light propagating from glass to air is $$\frac{\pi}{2} - \theta_B$$.

For light propagating from air to glass:

Refractive index of air, $$n_1 \approx 1$$, and refractive index of glass, $$n_2 = \mu_g$$.

So, Brewster's angle $$\theta_B$$ satisfies:

$$\tan \theta_B = \frac{\mu_g}{1} = \mu_g$$

Thus, $$\theta_B = \tan^{-1}(\mu_g)$$.

For light propagating from glass to air:

Refractive index of glass, $$n_1 = \mu_g$$, and refractive index of air, $$n_2 \approx 1$$.

Let Brewster's angle be $$\theta_B'$$. Then:

$$\tan \theta_B' = \frac{n_2}{n_1} = \frac{1}{\mu_g}$$

Thus, $$\theta_B' = \tan^{-1}\left(\frac{1}{\mu_g}\right)$$.

Now, consider $$\frac{\pi}{2} - \theta_B$$:

$$\tan\left(\frac{\pi}{2} - \theta_B\right) = \cot \theta_B = \frac{1}{\tan \theta_B} = \frac{1}{\mu_g}$$

Therefore, $$\frac{\pi}{2} - \theta_B = \tan^{-1}\left(\frac{1}{\mu_g}\right)$$.

This matches $$\theta_B'$$, so Statement I is true.

Now, Statement II: The Brewster's angle for light propagating from glass to air is $$\tan^{-1}(\mu_g)$$.

From above, Brewster's angle for glass to air is $$\theta_B' = \tan^{-1}\left(\frac{1}{\mu_g}\right)$$, not $$\tan^{-1}(\mu_g)$$.

Note that $$\tan^{-1}(\mu_g)$$ is $$\theta_B$$, the angle for air to glass, not glass to air.

Thus, Statement II is false.

Therefore, Statement I is true, but Statement II is false.

The correct option is B.

In Young's double slit experiment the position of the 5th bright fringe from the central maximum is $$y_5 = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$$, the distance between the slits and the screen is $$D = 1 \text{ m}$$, and the wavelength is $$\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$$.

The position of the $$n$$th bright fringe is given by the relation $$y_n = \frac{n \lambda D}{d}$$, where $$d$$ is the separation between the slits.

Solving this relation for $$d$$ gives $$d = \frac{n \lambda D}{y_n} = \frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}}$$. Further simplification leads to $$d = \frac{3000 \times 10^{-9}}{5 \times 10^{-2}} = \frac{3 \times 10^{-6}}{5 \times 10^{-2}} = 6 \times 10^{-5} \text{ m}$$ and finally $$d = 60 \times 10^{-6} \text{ m} = 60 \; \mu\text{m}$$.

The correct answer is Option A: 60 $$\mu$$m.

We have $$n$$ polarizing sheets arranged such that each makes an angle of $$45°$$ with the preceding sheet. Unpolarized light of intensity $$I$$ is incident, and the output intensity is $$\frac{I}{64}$$.

When unpolarized light passes through the first polarizer, the intensity becomes:

$$I_1 = \frac{I}{2}$$

Now, by Malus's law, when polarized light passes through a polarizer at angle $$\theta$$, the transmitted intensity is $$I\cos^2\theta$$. Each subsequent sheet is at $$45°$$ to the previous one, so after each sheet the intensity is multiplied by $$\cos^2 45° = \frac{1}{2}$$.

After $$n$$ sheets, the output intensity is:

$$I_{out} = \frac{I}{2} \times \left(\cos^2 45°\right)^{n-1} = \frac{I}{2} \times \left(\frac{1}{2}\right)^{n-1} = \frac{I}{2^n}$$

Setting this equal to $$\frac{I}{64}$$:

$$\frac{I}{2^n} = \frac{I}{64}$$

$$2^n = 64 = 2^6$$

$$n = 6$$

So, the answer is $$6$$.

In a double slit experiment, the fringe width is given by:

$$\beta = \frac{\lambda D}{d}$$

where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation. Since $$D$$ and $$d$$ remain constant, we have $$\beta \propto \lambda$$.

Now:

$$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$$

$$\beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} = 2 \times \frac{600}{400} = 2 \times 1.5 = 3 \text{ mm}$$

Hence, the correct answer is Option 4.

For single slit diffraction, the condition for first minimum is:

$$a\sin\theta = \lambda$$

Given: $$\lambda = 600$$ nm, $$\theta = 30°$$

$$a = \frac{\lambda}{\sin\theta} = \frac{600 \text{ nm}}{\sin 30°} = \frac{600}{0.5} = 1200 \text{ nm} = 1.2 \ \mu\text{m}$$

This matches option 1: 1.2 μm.

Assertion A: The phase difference of two light waves changes if they travel through different media having the same thickness but different indices of refraction.

This is correct. The optical path length is $$\mu t$$ where $$\mu$$ is the refractive index and $$t$$ is the thickness. Different $$\mu$$ values lead to different optical paths and hence a phase difference.

Reason R: The wavelengths of waves are different in different media.

This is correct. In a medium with refractive index $$\mu$$, the wavelength becomes $$\lambda/\mu$$.

R correctly explains A: since the wavelength changes in different media, the same physical thickness corresponds to different numbers of wavelengths, causing a phase difference between the two waves.

The correct answer is Both A and R are correct and R is the correct explanation of A.

In Young’s double slit experiment, transparent slabs of thickness $$t = 0.1$$ mm with refractive indices $$\mu_1 = 1.51$$ and $$\mu_2 = 1.55$$ are placed in front of slits $$S_1$$ and $$S_2$$ respectively. The path difference introduced by the slab at $$S_1$$ is $$(\mu_1 - 1)t$$, and that introduced by the slab at $$S_2$$ is $$(\mu_2 - 1)t$$. Consequently, the net extra path difference becomes $$\Delta = (\mu_2 - \mu_1)t = (1.55 - 1.51) \times 0.1 \times 10^{-3} = 0.04 \times 10^{-4} = 4 \times 10^{-6}$$ m.

Dividing this by the wavelength gives the fringe shift as $$\frac{\Delta}{\lambda} = \frac{4 \times 10^{-6}}{4000 \times 10^{-10}} = \frac{4 \times 10^{-6}}{4 \times 10^{-7}} = 10$$, so the central bright fringe shifts by $$\boxed{10}$$ fringes.

We need to find the value of $$\theta$$ such that the angle of incidence for total polarization (Brewster's angle) equals $$\tan^{-1}\left(1 + \dfrac{10}{\theta}\right)^{1/2}$$.

At Brewster's angle, the reflected and refracted rays are perpendicular to each other, giving the condition $$\tan i_B = \dfrac{n_2}{n_1}$$.

For dielectric media with $$\mu_r = 1$$, the refractive index is $$n = \sqrt{\varepsilon_r}$$ (since $$n = \sqrt{\mu_r \varepsilon_r}$$).

Thus, $$n_1 = \sqrt{2.8}$$ and $$n_2 = \sqrt{6.8}$$.

It follows that $$\tan i_B = \dfrac{\sqrt{6.8}}{\sqrt{2.8}} = \sqrt{\dfrac{6.8}{2.8}} = \sqrt{\dfrac{68}{28}} = \sqrt{\dfrac{17}{7}}$$.

The given expression for the Brewster angle is $$i_B = \tan^{-1}\left(1 + \dfrac{10}{\theta}\right)^{1/2}$$.

Therefore, $$\tan i_B = \sqrt{1 + \dfrac{10}{\theta}}$$.

Squaring both sides yields $$\dfrac{17}{7} = 1 + \dfrac{10}{\theta}$$.

From this, $$\dfrac{10}{\theta} = \dfrac{17}{7} - 1 = \dfrac{10}{7}$$ and hence $$\theta = 7$$.

The value of $$\theta$$ is $$\boxed{7}$$.

Given: Unpolarised light intensity $$I_0 = 32$$ W/m², final intensity $$I = 3$$ W/m², three polaroids with last perpendicular to first.

After the first polaroid: $$I_1 = \frac{I_0}{2} = 16$$ W/m²

Let the angle between first and second polaroid be $$\theta$$.

After second polaroid: $$I_2 = I_1 \cos^2\theta = 16\cos^2\theta$$

The angle between second and third polaroid is $$(90° - \theta)$$ (since third is perpendicular to first).

After third polaroid: $$I_3 = I_2\cos^2(90° - \theta) = 16\cos^2\theta \cdot \sin^2\theta$$

$$16\cos^2\theta\sin^2\theta = 3$$

$$4\sin^2 2\theta = 3$$ (using $$\sin 2\theta = 2\sin\theta\cos\theta$$)

$$\sin^2 2\theta = \frac{3}{4}$$

$$\sin 2\theta = \frac{\sqrt{3}}{2}$$

$$2\theta = 60°$$

$$\theta = 30°$$

The angle is 30°.

In Young's double slit experiment with wavelengths $$\lambda_1 = 7000 \, \mathring{A}$$ and $$\lambda_2 = 5500 \, \mathring{A}$$, slit separation $$d = 2.5$$ mm, and screen distance $$D = 150$$ cm, find the least distance from the central fringe where bright fringes coincide.

Find fringe widths.

$$\beta_1 = \frac{\lambda_1 D}{d} = \frac{7000 \times 10^{-10} \times 1.5}{2.5 \times 10^{-3}} = 4.2 \times 10^{-4} \text{ m}$$

$$\beta_2 = \frac{\lambda_2 D}{d} = \frac{5500 \times 10^{-10} \times 1.5}{2.5 \times 10^{-3}} = 3.3 \times 10^{-4} \text{ m}$$

Condition for coincidence.

Bright fringes coincide when $$n_1 \beta_1 = n_2 \beta_2$$:

$$n_1 \times 4.2 \times 10^{-4} = n_2 \times 3.3 \times 10^{-4}$$

$$\frac{n_1}{n_2} = \frac{3.3}{4.2} = \frac{33}{42} = \frac{11}{14}$$

Find the least distance.

The smallest integers satisfying this ratio are $$n_1 = 11$$ and $$n_2 = 14$$.

$$y = n_1 \beta_1 = 11 \times 4.2 \times 10^{-4} = 46.2 \times 10^{-4} = 4.62 \times 10^{-3} \text{ m}$$

$$y = 462 \times 10^{-5} \text{ m}$$

Therefore $$n = 462$$.

The correct answer is 462.

Two light beams with intensities in the ratio $$I_1 : I_2 = 9 : 4$$. The resultant intensity in interference is given by $$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$, where $$\phi$$ is the phase difference.

Maximum intensity occurs when $$\cos\phi = 1$$: $$I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2$$.

Minimum intensity occurs when $$\cos\phi = -1$$: $$I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2$$.

Since $$I_1 : I_2 = 9 : 4$$, we have $$\sqrt{I_1} : \sqrt{I_2} = 3 : 2$$, and hence $$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \frac{(3 + 2)^2}{(3 - 2)^2} = \frac{25}{1}$$. The ratio of intensity of maxima to minima is $$25 : 1$$. The correct answer is Option D.

We need to find the resolving power of a telescope with aperture $$D = 24.4$$ cm and wavelength $$\lambda = 2440$$ Å. Converting units gives $$D = 24.4$$ cm $$= 0.244$$ m and $$\lambda = 2440$$ Å $$= 2440 \times 10^{-10}$$ m $$= 2.44 \times 10^{-7}$$ m.

The angular limit of resolution (Rayleigh criterion) is $$\theta = \frac{1.22\lambda}{D}$$, and the resolving power is $$\frac{1}{\theta} = \frac{D}{1.22\lambda}$$. Substituting into this expression gives

$$\frac{0.244}{1.22 \times 2.44 \times 10^{-7}}$$

$$=\frac{0.244}{2.9768 \times 10^{-7}}$$

$$=\frac{2.44 \times 10^{-1}}{2.9768 \times 10^{-7}}$$

$$=8.196 \times 10^5$$

$$\approx 8.2 \times 10^5$$

The correct answer is Option C.

Given the intensity ratio of two coherent sources is $$4:1$$.

Let $$I_1 = 4I$$ and $$I_2 = I$$.

The amplitudes are proportional to the square root of intensities:

$$A_1 = 2\sqrt{I}, \quad A_2 = \sqrt{I}$$

Maximum intensity (constructive interference):

$$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (2\sqrt{I} + \sqrt{I})^2 = 9I$$

Minimum intensity (destructive interference):

$$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (2\sqrt{I} - \sqrt{I})^2 = I$$

Now computing the required ratio:

$$\frac{I_{max} + I_{min}}{I_{max} - I_{min}} = \frac{9I + I}{9I - I} = \frac{10I}{8I} = \frac{10}{8} = \frac{5}{4}$$

Comparing with $$\frac{5}{x}$$, we get $$x = 4$$.

Hence, the correct answer is Option B.

The problem states that light whose electric field vectors are completely removed by a good polaroid is incident on a prism at Brewster's angle.

A polaroid removes one component of the electric field. If the electric field vectors that are removed correspond to the component that lies in the plane of incidence, then the transmitted light is polarized perpendicular to the plane of incidence (s-polarized).

However, re-reading the problem: "electric field vectors are completely removed" means the polaroid transmits light polarized in the plane of incidence (p-polarized light), since the perpendicular component has been removed.

At Brewster's angle, p-polarized light (electric field in the plane of incidence) has zero reflection. All of the p-polarized light is transmitted (refracted) into the prism.

Since the incident light is purely p-polarized (the other component was removed by the polaroid), there will be no reflection at Brewster's angle, and total transmission occurs.

Hence, the correct answer is Option D.

A microscope is moved from air (refractive index 1) to oil (refractive index 2). We need to find the change in resolving power.

The resolving power of a microscope is:

$$RP = \frac{2n \sin\theta}{1.22\lambda}$$

where $$n$$ is the refractive index of the medium, $$\theta$$ is the half-angle of the cone of light, and $$\lambda$$ is the wavelength of light in vacuum (or air).

In air: $$RP_{\text{air}} = \frac{2 \times 1 \times \sin\theta}{1.22\lambda}$$

In oil: $$RP_{\text{oil}} = \frac{2 \times 2 \times \sin\theta}{1.22\lambda}$$

$$\frac{RP_{\text{oil}}}{RP_{\text{air}}} = \frac{2 \times 2}{2 \times 1} = 2$$

The resolving power in oil is twice that in air.

The correct answer is Option B: Resolving power will be twice in the oil than it was in the air.

In Young's double slit experiment, a glass plate of refractive index $$\mu = 1.5$$ and thickness $$x\lambda$$ is introduced in one of the beams. The intensity at the original central maximum position remains unchanged.

Calculate the extra path difference introduced by the glass plate.

$$\Delta = (\mu - 1)t = (1.5 - 1)(x\lambda) = 0.5 x\lambda$$

Apply the condition for unchanged intensity.

For the intensity at the central maximum to remain unchanged, the phase difference introduced must be an integer multiple of $$2\pi$$. This means the extra path difference must be an integer multiple of $$\lambda$$:

$$\Delta = n\lambda \quad (n = 1, 2, 3, ...)$$

$$0.5 x\lambda = n\lambda$$

$$x = 2n$$

Find the minimum value of $$x$$.

For $$n = 1$$: $$x = 2$$

The correct answer is Option B.

In Young's double slit experiment, the fringe width in air is $$\beta = 12 \text{ mm}$$. The entire arrangement is placed in water of refractive index $$\mu = \dfrac{4}{3}$$. We need to find the new fringe width.

The fringe width in Young's double slit experiment is given by $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation.

Since the setup is placed in a medium of refractive index $$\mu$$, the wavelength changes to $$\lambda' = \frac{\lambda}{\mu}$$ while the distances $$D$$ and $$d$$ remain unchanged.

Next, the new fringe width is $$\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$$. Substituting gives $$\beta' = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9 \text{ mm}$$.

The correct answer is Option B: $$9$$ mm.

We are given two beams with intensities $$I$$ and $$9I$$, and we need to find the difference in their resultant intensities at points $$P$$ and $$Q$$, where the phase difference is $$\frac{\pi}{2}$$ and $$\pi$$ respectively.

Recall that when two waves of intensities $$I_1$$ and $$I_2$$ interfere with a phase difference $$\delta$$, the resultant intensity is given by:
$$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta$$

At point $$P$$, since $$\delta = \frac{\pi}{2}$$, one finds
$$I_P = I + 9I + 2\sqrt{I \times 9I}\cos\frac{\pi}{2}$$
Because $$\cos\frac{\pi}{2}=0$$, it follows that
$$I_P = 10I + 0 = 10I$$

Similarly, at point $$Q$$ where $$\delta = \pi$$, we have
$$I_Q = I + 9I + 2\sqrt{I \times 9I}\cos\pi$$
Using $$\cos\pi = -1$$ and $$\sqrt{9I^2} = 3I$$ yields
$$I_Q = 10I + 2(3I)(-1) = 10I - 6I = 4I$$

Therefore, the desired difference is
$$I_P - I_Q = 10I - 4I = 6I$$

The correct answer is Option B: $$6I$$.

Two coherent sources have an intensity ratio of $$1:4$$, and we are required to determine $$\frac{\alpha}{\beta}$$ given that $$\frac{I_{max}+I_{min}}{I_{max}-I_{min}} = \frac{2\alpha+1}{\beta+3}$$.

Let us denote the intensities by $$I_1 = I$$ and $$I_2 = 4I$$; since amplitude is proportional to the square root of intensity, we set $$a_1 = a$$ and $$a_2 = 2a$$.

For constructive interference, the amplitudes add, giving

$$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I} + 2\sqrt{I})^2 = 9I\,. $$

Conversely, destructive interference yields

$$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{I} - 2\sqrt{I})^2 = I\,. $$

Substituting these expressions into the given ratio gives

$$\frac{I_{max}+I_{min}}{I_{max}-I_{min}} = \frac{9I + I}{9I - I} = \frac{10I}{8I} = \frac{5}{4}\,. $$

Therefore,

$$\frac{2\alpha+1}{\beta+3} = \frac{5}{4}\,. $$

By equating numerators and denominators separately, we obtain

$$2\alpha + 1 = 5 \implies \alpha = 2$$

and

$$\beta + 3 = 4 \implies \beta = 1\,. $$

Hence, the required ratio is

$$\frac{\alpha}{\beta} = \frac{2}{1} = 2\,. $$

Answer: Option B: 2

We have an unpolarised light beam of intensity $$2I_0$$ passing through polaroid P. When unpolarised light passes through a polaroid, its intensity is halved. So the intensity after P is $$\frac{2I_0}{2} = I_0$$.

Now this polarised light passes through polaroid Q, whose passing axis makes an angle of 30° with that of P. By Malus's law, the transmitted intensity is: $$I = I_0 \cos^2 30° = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4} = \frac{3I_0}{4}$$

Hence, the correct answer is Option 3.

We have two light beams of intensities $$4I$$ and $$9I$$ interfering on a screen. The resultant intensity when two coherent sources of intensities $$I_1$$ and $$I_2$$ interfere with a phase difference $$\phi$$ is given by $$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$.

At point A, the phase difference is zero ($$\phi = 0$$), so $$\cos 0 = 1$$, and the resultant intensity is $$I_A = 4I + 9I + 2\sqrt{4I \cdot 9I} = 13I + 2\sqrt{36I^2} = 13I + 12I = 25I$$.

At point B, the phase difference is $$\pi$$, so $$\cos\pi = -1$$, and the resultant intensity is $$I_B = 4I + 9I - 2\sqrt{4I \cdot 9I} = 13I - 12I = I$$.

The difference in resultant intensities is $$I_A - I_B = 25I - I = 24I$$.

Hence, the correct answer is 24.

We need to find the wavelength of light in a double slit experiment from the change in fringe width when the screen is moved.

Recall the fringe width formula $$\beta = \frac{\lambda D}{d}$$ where $$\lambda$$ is wavelength, D is the distance from slits to screen, and d is the slit separation.

When the screen is moved towards the slits by $$\Delta D = 5 \times 10^{-2}$$ m, the fringe width decreases by $$\Delta\beta = 3 \times 10^{-3}$$ cm $$= 3 \times 10^{-5}$$ m. $$\Delta\beta = \frac{\lambda \cdot \Delta D}{d}$$

d = 1 mm = 10^{-3} m. $$\lambda = \frac{\Delta\beta \times d}{\Delta D} = \frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}}$$

$$= \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 6 \times 10^{-7} \text{ m}$$

$$= 600 \text{ nm}$$

The answer is 600 nm.

The angular width of the fringe in air is $$\theta = 0.35°$$, the distance to the screen is $$D = 2$$ m, the wavelength is $$\lambda = 450$$ nm, and the refractive index of the medium is $$\mu = \frac{7}{5}$$.

In Young's double slit experiment, the angular width of a fringe is given by $$\theta = \frac{\lambda}{d}$$ where $$d$$ is the slit separation. When the system is immersed in a medium of refractive index $$\mu$$, the effective wavelength becomes $$\lambda' = \frac{\lambda}{\mu}$$, so the new angular width is $$\theta' = \frac{\lambda'}{d} = \frac{\lambda}{\mu d} = \frac{\theta}{\mu}.$$

Substituting the values, $$\theta' = \frac{0.35°}{\frac{7}{5}} = 0.35° \times \frac{5}{7} = 0.25° = \frac{1}{4}°.$$

Since the angular width is $$\frac{1}{\alpha}$$ degrees, we have $$\frac{1}{\alpha} = \frac{1}{4} \quad\Longrightarrow\quad \alpha = 4.$$

Hence, the value of $$\alpha$$ is 4.

In Young's double slit experiment, the slit separation is 0.6 mm, the screen distance is 80 cm, and the first dark fringe appears directly opposite one of the slits.

The central bright fringe is at the midpoint of the two slits. If the first dark fringe is directly opposite one slit, its distance from the central maximum is:

$$y = \frac{d}{2} = \frac{0.6}{2} = 0.3 \text{ mm}$$

The position of the first dark fringe (m = 0) in YDSE is:

$$y = \frac{(2m+1)\lambda D}{2d} = \frac{\lambda D}{2d}$$

Equating these expressions yields

$$\frac{d}{2} = \frac{\lambda D}{2d}$$

This simplifies to

$$d^2 = \lambda D$$

Therefore, the wavelength is given by

$$\lambda = \frac{d^2}{D} = \frac{(0.6 \times 10^{-3})^2}{0.8}$$

Evaluating this gives

$$\lambda = \frac{0.36 \times 10^{-6}}{0.8} = 0.45 \times 10^{-6} \text{ m} = 450 \text{ nm}$$

The wavelength of light is 450 nm.

We use sodium light of wavelengths $$\lambda_1 = 650$$ nm and $$\lambda_2 = 655$$ nm with a single slit of aperture $$a = 0.5$$ mm $$= 0.5 \times 10^{-3}$$ m. The screen is at distance $$D = 2.0$$ m.

Concept: In single-slit diffraction, the first secondary maximum occurs approximately midway between the first and second minima. The minima are located at $$a \sin\theta = m\lambda$$. The first secondary maximum is approximately at $$a \sin\theta = \dfrac{3\lambda}{2}$$.

The position of the first secondary maximum on the screen (using small angle approximation $$\sin\theta \approx \tan\theta = \dfrac{y}{D}$$) is given by:

$$y = \dfrac{3\lambda D}{2a}$$

For $$\lambda_1 = 650$$ nm, we get

$$y_1 = \dfrac{3 \times 650 \times 10^{-9} \times 2.0}{2 \times 0.5 \times 10^{-3}} = \dfrac{3900 \times 10^{-9}}{10^{-3}} = 3900 \times 10^{-6} \text{ m}$$

For $$\lambda_2 = 655$$ nm, we get

$$y_2 = \dfrac{3 \times 655 \times 10^{-9} \times 2.0}{2 \times 0.5 \times 10^{-3}} = \dfrac{3930 \times 10^{-9}}{10^{-3}} = 3930 \times 10^{-6} \text{ m}$$

The separation between the positions of the first maxima is

$$\Delta y = y_2 - y_1 = (3930 - 3900) \times 10^{-6} = 30 \times 10^{-6} \text{ m} = 3 \times 10^{-5} \text{ m}$$

The answer is 3 $$\times 10^{-5}$$ m.

In Young's double slit experiment, the fringe width (separation between consecutive bright fringes) is given by:

where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation.

Given data:

For the first light: $$\lambda_1 = 560$$ nm, $$\beta_1 = 7.2$$ mm

For the second light: $$\beta_2 = 8.1$$ mm, $$\lambda_2 = ?$$

Since the experimental setup (D and d) remains the same:

Solving for $$\lambda_2$$:

Therefore, the wavelength of the second light is 630 nm.

Two beams of intensities $$I$$ and $$4I$$ interfere. We need to find the difference in resultant intensities at points $$A$$ (phase difference $$\pi/2$$) and $$B$$ (phase difference $$\pi/3$$).

The resultant intensity when two beams of intensities $$I_1$$ and $$I_2$$ interfere with phase difference $$\phi$$ is given by:

$$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$

Here $$I_1 = I$$ and $$I_2 = 4I$$, so:

$$I_R = I + 4I + 2\sqrt{I \times 4I}\cos\phi = 5I + 4I\cos\phi$$

At point A (phase difference $$\pi/2$$), the intensity is:

$$I_A = 5I + 4I\cos\frac{\pi}{2} = 5I + 4I(0) = 5I$$

At point B (phase difference $$\pi/3$$), the intensity is:

$$I_B = 5I + 4I\cos\frac{\pi}{3} = 5I + 4I\left(\frac{1}{2}\right) = 5I + 2I = 7I$$

The difference between these intensities is:

$$I_B - I_A = 7I - 5I = 2I$$

Since the difference is $$xI$$, we get $$x = 2$$.

In single-slit (or pinhole) diffraction, the angular width of the central maximum is inversely proportional to the diameter of the aperture. Specifically, for a circular pinhole of diameter $$d$$, the angular radius of the first dark ring (Airy disk) is given by $$\sin\theta \approx 1.22\frac{\lambda}{d}$$.

When the diameter of the pinhole is slightly increased, the ratio $$\frac{\lambda}{d}$$ decreases, which means the angular spread of the diffraction pattern decreases. Therefore, the size of the diffraction pattern (the central bright spot and the surrounding rings) decreases.

At the same time, increasing the diameter of the pinhole allows more light to pass through it. The amount of light collected by the pinhole is proportional to its area, which is proportional to $$d^2$$. Therefore, the intensity of the diffraction pattern increases when the pinhole diameter is increased.

Hence, when the pinhole diameter is slightly increased, the size of the diffraction pattern decreases but the intensity increases.

In Young’s double-slit experiment the fringe width is governed by the well-known relation $$\beta=\dfrac{\lambda D}{d}$$ where $$\lambda$$ is the wavelength of the light, $$D$$ is the distance from the slits to the screen and $$d$$ is the separation between the two slits.

In this problem the slit separation is not constant; it varies with time according to $$d(t)=d_0+a_0\sin\omega t,$$ where the symbols $$d_0,\;a_0,\;\omega$$ are fixed constants and the condition $$a_0<d_0$$ is implied so that the distance between the slits never becomes negative.

Because $$\beta=\dfrac{\lambda D}{d},$$ the fringe width will attain its maximum value when $$d(t)$$ is minimum and will attain its minimum value when $$d(t)$$ is maximum.

The minimum value of $$d(t)$$ occurs when $$\sin\omega t=-1$$, giving

$$d_{\text{min}}=d_0-a_0.$$

Correspondingly, the maximum value of $$d(t)$$ occurs when $$\sin\omega t=+1$$, giving

$$d_{\text{max}}=d_0+a_0.$$

Now the largest possible fringe width (call it $$\beta_{\text{large}}$$) is obtained by substituting $$d_{\text{min}}$$ into the formula for $$\beta$$:

$$\beta_{\text{large}}=\dfrac{\lambda D}{d_{\text{min}}}=\dfrac{\lambda D}{d_0-a_0}.$$

Similarly, the smallest possible fringe width (call it $$\beta_{\text{small}}$$) arises when $$d=d_{\text{max}}$$:

$$\beta_{\text{small}}=\dfrac{\lambda D}{d_{\text{max}}}=\dfrac{\lambda D}{d_0+a_0}.$$

The problem asks for the difference between these two extreme values:

$$\Delta\beta=\beta_{\text{large}}-\beta_{\text{small}}=\dfrac{\lambda D}{d_0-a_0}-\dfrac{\lambda D}{d_0+a_0}.$$

We combine the two fractions by taking a common denominator:

$$\Delta\beta=\lambda D\left[\dfrac{1}{d_0-a_0}-\dfrac{1}{d_0+a_0}\right]$$

$$=\lambda D\left[\dfrac{d_0+a_0-(d_0-a_0)}{(d_0-a_0)(d_0+a_0)}\right]$$

$$=\lambda D\left[\dfrac{d_0+a_0-d_0+a_0}{d_0^2-a_0^2}\right]$$

$$=\lambda D\left[\dfrac{2a_0}{d_0^2-a_0^2}\right].$$

Simplifying, we obtain

$$\boxed{\Delta\beta=\dfrac{2\lambda D a_0}{\,d_0^2-a_0^2\,}}.$$

This expression matches Option B in the list provided.

Hence, the correct answer is Option 2.

In Young's double slit experiment, the fringe width is given by $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength of light, $$D$$ is the distance from the slits to the screen, and $$d$$ is the separation between the slits.

When the source is changed from red to violet light, the wavelength decreases since violet light has a shorter wavelength (approximately 400 nm) compared to red light (approximately 700 nm).

Since the fringe width $$\beta$$ is directly proportional to $$\lambda$$, a decrease in wavelength causes a decrease in fringe width. This means the consecutive fringe lines come closer together on the screen.

The central bright fringe remains bright regardless of wavelength change, the brightness of fringes depends on intensity not wavelength, and the intensity of minima remains zero for perfectly coherent sources.

The correct answer is that consecutive fringe lines will come closer.

Let the intensities of the two coherent light sources be $$I_1 = 2x$$ and $$I_2 = 1$$ (ratio $$2x : 1$$).

For an interference pattern, the maximum and minimum intensities are:

$$I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{2x} + 1\right)^2$$

$$I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{2x} - 1\right)^2$$

Computing $$I_{max} - I_{min}$$:

$$I_{max} - I_{min} = \left(\sqrt{2x} + 1\right)^2 - \left(\sqrt{2x} - 1\right)^2 = 4\sqrt{2x}$$

Computing $$I_{max} + I_{min}$$:

$$I_{max} + I_{min} = \left(\sqrt{2x} + 1\right)^2 + \left(\sqrt{2x} - 1\right)^2 = 2(2x + 1)$$

Therefore the required ratio is:

$$\frac{I_{max} - I_{min}}{I_{max} + I_{min}} = \frac{4\sqrt{2x}}{2(2x + 1)} = \frac{2\sqrt{2x}}{2x + 1}$$

The correct answer is Option (3): $$\frac{2\sqrt{2x}}{2x + 1}$$.

We are told that one slit has width three times the other. Since amplitude is proportional to slit width, if the amplitude from the narrower slit is $$A$$, then the amplitude from the wider slit is $$3A$$.

In a double-slit experiment, the maximum intensity occurs when the two waves interfere constructively, and the minimum intensity occurs when they interfere destructively.

The maximum intensity is $$I_{\max} = (A_1 + A_2)^2 = (A + 3A)^2 = (4A)^2 = 16A^2$$.

The minimum intensity is $$I_{\min} = (A_2 - A_1)^2 = (3A - A)^2 = (2A)^2 = 4A^2$$.

The ratio of maximum to minimum intensity is $$\frac{I_{\max}}{I_{\min}} = \frac{16A^2}{4A^2} = \frac{4}{1}$$.

So the ratio is $$4 : 1$$.

Hence, the correct answer is Option D.

For two coherent light waves, the superposition principle gives the resultant intensity as

$$I \;=\; I_1\;+\;I_2\;+\;2\sqrt{I_1I_2}\cos\phi,$$

where $$I_1$$ and $$I_2$$ are the individual intensities and $$\phi$$ is the phase difference between the waves.

We are told that both sources have equal intensity $$I_1 = I_2 = I_0$$. Substituting these values, we obtain

$$I \;=\; I_0 \;+\; I_0 \;+\; 2\sqrt{I_0\cdot I_0}\cos\phi.$$

Since $$\sqrt{I_0\cdot I_0} = I_0,$$ this simplifies to

$$I \;=\; 2I_0 \;+\; 2I_0\cos\phi.$$

Now, the problem statement mentions that at a minimum the intensity is zero. A minimum occurs when the phase difference is an odd multiple of $$\pi$$, giving $$\cos\phi = -1$$. Substituting $$\cos\phi = -1$$ verifies the condition:

$$I_{\text{min}} = 2I_0 + 2I_0(-1) = 2I_0 - 2I_0 = 0,$$

which is consistent with the information provided.

A maximum occurs when the phase difference is an even multiple of $$\pi$$, so $$\cos\phi = 1$$. Substituting $$\cos\phi = 1$$ into the expression for $$I$$ gives

$$I_{\text{max}} = 2I_0 + 2I_0(1) = 2I_0 + 2I_0 = 4I_0.$$

Therefore, the intensity of light at the maxima is $$4I_0$$.

Hence, the correct answer is Option C.

We begin with the well-known formula for the Young’s double slit experiment that gives the linear separation between two consecutive bright (or dark) fringes on the screen, usually called the fringe width. The formula is stated as

$$\beta \;=\;\dfrac{\lambda\,D}{d}$$

where

$$\lambda$$ is the wavelength of the monochromatic light used,

$$D$$ is the distance between the double-slit plane and the screen, and

$$d$$ is the separation between the two slits.

In the present question, the experimental arrangement is not altered; that means $$D$$ and $$d$$ stay exactly the same when we switch the colour of light. Therefore, the only quantity in the expression for $$\beta$$ that is going to change is the wavelength $$\lambda$$.

Now, orange light has a wavelength toward the longer-wavelength part of the visible spectrum, typically about $$\lambda_\text{orange}\approx 600\text{ nm}$$, while blue light lies toward the shorter-wavelength side, roughly $$\lambda_\text{blue}\approx 450\text{ nm}$$. Hence we have a clear inequality

$$\lambda_\text{blue}\;<\;\lambda_\text{orange}$$

Substituting each wavelength separately into the fringe-width formula gives

$$\beta_\text{orange}\;=\;\dfrac{\lambda_\text{orange}\,D}{d}$$

and

$$\beta_\text{blue}\;=\;\dfrac{\lambda_\text{blue}\,D}{d}.$$

Because the constants $$D$$ and $$d$$ are identical in both cases, we can compare the two fringe widths simply by comparing the two numerators, i.e. the wavelengths. Since $$\lambda_\text{blue}$$ is smaller than $$\lambda_\text{orange}$$, we get

$$\beta_\text{blue}\;<\;\beta_\text{orange}.$$

Thus, when the source of light is changed from orange (longer wavelength) to blue (shorter wavelength), the value of $$\beta$$ decreases. A smaller $$\beta$$ means that the linear distance between two successive fringes on the screen shrinks. In plain words, the fringes crowd closer together.

No other qualitative change among the listed options—such as the central bright fringe turning dark or an increase in the intensity of minima—follows directly from the basic theory when only the wavelength is altered. The uniquely correct consequence is the decrease in fringe spacing.

Hence, the correct answer is Option B.

We have two slits whose widths are unequal. Let the smaller slit have width $$a$$. The problem states that the other slit is three times wider, so its width is $$3a$$.

The amplitude of the light emerging from a slit is directly proportional to its width. Hence, if the smaller slit contributes an amplitude $$A$$, then the wider slit contributes an amplitude $$3A$$, because $$3a$$ is three times $$a$$.

Intensity and amplitude are related by the formula

$$I \propto (\text{amplitude})^{2}.$$

Therefore, the individual intensities from the two slits are

$$I_{1}=A^{2}, \qquad I_{2}=(3A)^{2}=9A^{2}.$$

When the two waves interfere, the resultant intensity depends on the vector sum of the amplitudes. For constructive interference (bright fringes), the amplitudes add; for destructive interference (dark fringes), they subtract.

Maximum intensity

The formula for the maximum resultant amplitude is the algebraic sum of the individual amplitudes:

$$A_{\max}=A+3A=4A.$$

So the maximum intensity is

$$I_{\max}=A_{\max}^{2}=(4A)^{2}=16A^{2}.$$

Minimum intensity

The formula for the minimum resultant amplitude is the absolute value of the difference of the individual amplitudes:

$$A_{\min}=|A-3A|=|-2A|=2A.$$

Thus the minimum intensity is

$$I_{\min}=A_{\min}^{2}=(2A)^{2}=4A^{2}.$$

Ratio of minimum to maximum intensity

$$\frac{I_{\min}}{I_{\max}}=\frac{4A^{2}}{16A^{2}}=\frac{4}{16}=\frac{1}{4}.$$

This ratio is given in the statement as $$x:4$$, so

$$x=1.$$

So, the answer is $$1$$.

The fringe width in Young's double slit experiment is given by $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength of light, $$D$$ is the distance from the slits to the screen, and $$d$$ is the separation between the slits.

We are given $$\beta = 6$$ mm $$= 6 \times 10^{-3}$$ m, $$d = 1$$ mm $$= 1 \times 10^{-3}$$ m, and $$D = 10$$ m. Rearranging for $$\lambda$$, we get $$\lambda = \frac{\beta \cdot d}{D} = \frac{6 \times 10^{-3} \times 1 \times 10^{-3}}{10} = \frac{6 \times 10^{-6}}{10} = 6 \times 10^{-7}$$ m.

Converting to nanometres, $$\lambda = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}$$. Therefore, $$x = 600$$.

We have an unpolarized light beam passing through a polarizer and an analyzer. The intensity emerging from the analyzer is initially measured as 100 Lumens.

When unpolarized light of intensity $$I_0$$ passes through a polarizer, the transmitted intensity becomes $$\frac{I_0}{2}$$. When this polarized light then passes through the analyzer at an angle $$\theta$$ to the polarizer, by Malus's law, the transmitted intensity is $$I = \frac{I_0}{2}\cos^2\theta$$.

Initially, the polarizer and analyzer are aligned ($$\theta = 0°$$), so $$100 = \frac{I_0}{2}\cos^2 0° = \frac{I_0}{2}$$.

Now the analyzer is rotated by 30°. The new intensity is $$I' = \frac{I_0}{2}\cos^2 30°$$.

We know $$\cos 30° = \frac{\sqrt{3}}{2}$$, so $$\cos^2 30° = \frac{3}{4}$$.

$$I' = \frac{I_0}{2} \times \frac{3}{4} = 100 \times \frac{3}{4} = 75$$ Lumens.

So, the answer is $$75$$.

We have a Young’s double-slit experiment in which the distance between the slits is given as $$d = 0.3\ \text{mm}$$. First we convert this separation into metres so that every quantity is in SI units: $$0.3\ \text{mm} = 0.3 \times 10^{-3}\ \text{m} = 3 \times 10^{-4}\ \text{m}$$.

The screen is placed at a distance $$D = 1.5\ \text{m}$$ from the slits.

The statement says that the distance between the fourth bright fringes on the two opposite sides of the central bright fringe is $$2.4\ \text{cm}$$. Converting into metres gives $$\Delta y = 2.4\ \text{cm} = 2.4 \times 10^{-2}\ \text{m}$$.

In Young’s experiment, the position of the $$m^{\text{th}}$$ bright fringe (measured from the central maximum) is described by the formula

$$y_m = \frac{m \lambda D}{d},$$

where $$\lambda$$ is the wavelength of the light. Thus, the fourth bright fringe on the right of the centre is at

$$y_{+4} = \frac{4 \lambda D}{d},$$

and the fourth bright fringe on the left of the centre is at

$$y_{-4} = -\,\frac{4 \lambda D}{d}.$$

The separation between these two fringes is therefore

$$\Delta y = y_{+4} - y_{-4} = \frac{4 \lambda D}{d} - \Bigl(-\frac{4 \lambda D}{d}\Bigr) = \frac{8 \lambda D}{d}.$$

We substitute the known values now. Rearranging the above relation gives

$$\lambda = \frac{\Delta y\, d}{8D}.$$

Putting in $$\Delta y = 2.4 \times 10^{-2}\ \text{m},\quad d = 3 \times 10^{-4}\ \text{m},\quad D = 1.5\ \text{m},$$ we obtain

$$\lambda = \frac{(2.4 \times 10^{-2})(3 \times 10^{-4})}{8 \times 1.5}.$$

First, multiply the numerators:

$$(2.4 \times 3) \times 10^{-2-4} = 7.2 \times 10^{-6}\ \text{m}.$$

Next, multiply the denominators:

$$8 \times 1.5 = 12.$$

Hence,

$$\lambda = \frac{7.2 \times 10^{-6}}{12} = 0.6 \times 10^{-6}\ \text{m} = 6 \times 10^{-7}\ \text{m}.$$

We now need the frequency $$\nu$$, and we use the fundamental relation between speed of light, wavelength, and frequency:

$$c = \lambda \nu.$$

So,

$$\nu = \frac{c}{\lambda}.$$

Taking $$c = 3 \times 10^{8}\ \text{m s}^{-1}$$ and $$\lambda = 6 \times 10^{-7}\ \text{m},$$ we get

$$\nu = \frac{3 \times 10^{8}}{6 \times 10^{-7}} = \frac{3}{6} \times 10^{8+7}\ \text{Hz} = 0.5 \times 10^{15}\ \text{Hz} = 5 \times 10^{14}\ \text{Hz}.$$

This matches the form $$x \times 10^{14}\ \text{Hz}$$ with $$x = 5$$.

So, the answer is $$5$$.

Let the common thickness of the two columns be $$t$$.

For a light beam, the number of complete waves contained in a distance is obtained by dividing the distance by the wavelength in that medium. Hence, in a column of thickness $$t$$:

$$\text{Number of waves in vacuum} = N_{\text{vac}} = \dfrac{t}{\lambda_0}$$

where $$\lambda_0$$ is the wavelength of yellow light in vacuum. We are given

$$\lambda_0 = 6000\;\text{\AA} = 6000 \times 10^{-10}\;\text{m} = 6.0 \times 10^{-7}\;\text{m}.$$

Inside air the speed (and hence the wavelength) changes according to the refractive index. First we recall the basic relationship

$$n = \dfrac{\text{speed in vacuum}}{\text{speed in medium}} = \dfrac{\lambda_0}{\lambda_{\text{air}}}.$$

Re-arranging for the wavelength in air, we have

$$\lambda_{\text{air}} = \dfrac{\lambda_0}{n}.$$

The refractive index of air is supplied as $$n = 1.0003$$, so the wavelength in air becomes

$$\lambda_{\text{air}} = \dfrac{6.0 \times 10^{-7}\;\text{m}}{1.0003}.$$

Next we write the number of waves in the air column:

$$\text{Number of waves in air} = N_{\text{air}} = \dfrac{t}{\lambda_{\text{air}}} = \dfrac{t}{\lambda_0/n} = \dfrac{n\,t}{\lambda_0}.$$

The statement in the question says that the difference between the two wave counts equals one:

$$|N_{\text{air}} - N_{\text{vac}}| = 1.$$

Because air’s wavelength is slightly shorter, $$N_{\text{air}} > N_{\text{vac}}$$. Therefore we set

$$N_{\text{air}} - N_{\text{vac}} = 1.$$

Substituting the expressions we have derived:

$$\dfrac{n\,t}{\lambda_0} - \dfrac{t}{\lambda_0} = 1.$$

Factorising $$t/\lambda_0$$ gives

$$\dfrac{t}{\lambda_0}\,(n - 1) = 1.$$

Now we solve for the thickness $$t$$:

$$t = \dfrac{\lambda_0}{\,n - 1\,}.$$

Putting the numerical values,

$$t = \dfrac{6.0 \times 10^{-7}\;\text{m}}{1.0003 - 1} = \dfrac{6.0 \times 10^{-7}\;\text{m}}{0.0003}.$$

The division gives

$$t = 2.0 \times 10^{-3}\;\text{m}.$$

Converting metres to millimetres (since $$1\;\text{mm} = 10^{-3}\;\text{m}$$) we have

$$t = 2.0\;\text{mm}.$$

So, the answer is $$2\;\text{mm}$$.

We begin with an un-polarised beam of light that produces an intensity $$I$$ on the screen when nothing obstructs it. Un-polarised light contains vibrations in all possible planes perpendicular to the direction of propagation.

When such a beam passes through a single ideal Polaroid, only the vibrations lying in the transmission axis of the Polaroid are allowed to pass. As a result, exactly one-half of the incident intensity survives. Stating this result,

$$I_{\text{after first Polaroid}} \;=\;\dfrac{I}{2}.$$

We now place a second Polaroid $$P_2$$ after the first one $$P_1$$. Let the angle between their transmission axes be $$\theta$$. For a beam already plane-polarised by the first Polaroid, the intensity transmitted by the second Polaroid is governed by Malus’ law.

Malus’ Law (to be stated explicitly): If a plane-polarised light of intensity $$I_0$$ is incident on a Polaroid and the angle between the light’s plane of polarisation and the Polaroid’s transmission axis is $$\theta$$, the emerging intensity $$I$$ is

$$I \;=\; I_0 \cos^2\theta.$$

Here the incident intensity on $$P_2$$ is $$I_0 = \dfrac{I}{2}$$ and the emergent intensity is given to be $$\dfrac{I}{2}$$. Substituting these values in Malus’ law, we have

$$\dfrac{I}{2} \;=\; \left(\dfrac{I}{2}\right)\cos^2\theta.$$

Dividing both sides by $$\dfrac{I}{2}$$ gives

$$\cos^2\theta \;=\; 1.$$

So $$\cos\theta = \pm 1$$ and therefore $$\theta = 0^\circ \ (\text{or }180^\circ).$$ Physically this means that the two Polaroids were initially parallel (their axes coincided), which is the simplest arrangement that keeps the intensity unchanged at $$\dfrac{I}{2}$$.

Next, we rotate only the second Polaroid $$P_2$$ through an additional angle $$\phi$$ while keeping $$P_1$$ fixed. The new angle between their axes now is $$\phi$$. After this rotation the first Polaroid still transmits $$\dfrac{I}{2}$$, and the second Polaroid again follows Malus’ law. Therefore the new intensity on the screen becomes

$$I_{\text{new}} \;=\;\left(\dfrac{I}{2}\right)\cos^2\phi.$$

According to the problem this new intensity must equal $$\dfrac{3I}{8}$$. Equating the two expressions,

$$\left(\dfrac{I}{2}\right)\cos^2\phi \;=\; \dfrac{3I}{8}.$$

We now cancel the common factor $$I$$ from both sides:

$$\dfrac{1}{2}\cos^2\phi \;=\; \dfrac{3}{8}.$$

Multiplying every term by 8 to clear the denominators:

$$4\cos^2\phi \;=\; 3.$$

Dividing by 4,

$$\cos^2\phi \;=\; \dfrac{3}{4}.$$

Taking the positive square root (since we are looking for a small physical rotation, $$0^\circ \le \phi \le 90^\circ$$),

$$\cos\phi \;=\; \dfrac{\sqrt{3}}{2}.$$

The angle whose cosine equals $$\dfrac{\sqrt{3}}{2}$$ is

$$\phi \;=\; 30^\circ.$$

Hence, the correct answer is Option 30°.

We know that in Young’s double-slit experiment the position of the $$m^{\text{th}}$$ bright fringe from the central white fringe is given by the formula

$$y_m=\frac{m\lambda D}{d},$$

where

$$$y_m=\text{distance of the }m^{\text{th}}\text{ bright fringe from the centre},$$$

$$\lambda=\text{wavelength of light},$$

$$$D=\text{distance between the slits and the screen},$$$

$$$d=\text{separation between the two slits}.$$$

Here the first bright fringe for any colour corresponds to $$m=1$$, so for violet light we have

$$y_v=\frac{\lambda_v D}{d}$$

and for red light

$$y_r=\frac{\lambda_r D}{d}.$$

We are supplied with the following numerical data:

$$$D=1.5\ \text{m},\qquad d=0.3\ \text{mm}=0.3\times10^{-3}\ \text{m}=3.0\times10^{-4}\ \text{m},$$$

$$$y_v=2.0\ \text{mm}=2.0\times10^{-3}\ \text{m},\qquad y_r=3.5\ \text{mm}=3.5\times10^{-3}\ \text{m}.$$$

First we find the wavelengths individually. For violet light, substituting the known values gives

$$$\lambda_v=\frac{y_v d}{D}=\frac{2.0\times10^{-3}\,\text{m}\,\times\,3.0\times10^{-4}\,\text{m}}{1.5\ \text{m}}.$$$

Multiplying the numerators,

$$2.0\times10^{-3}\times3.0\times10^{-4}=6.0\times10^{-7},$$

and then dividing by $$1.5$$,

$$$\lambda_v=\frac{6.0\times10^{-7}}{1.5}=4.0\times10^{-7}\ \text{m}=400\ \text{nm}.$$$

In an exactly similar manner for red light we have

$$$\lambda_r=\frac{y_r d}{D}=\frac{3.5\times10^{-3}\,\text{m}\,\times\,3.0\times10^{-4}\,\text{m}}{1.5\ \text{m}}.$$$

Again multiplying the numerators,

$$$3.5\times10^{-3}\times3.0\times10^{-4}=10.5\times10^{-7}=1.05\times10^{-6},$$$

and dividing by $$1.5$$,

$$$\lambda_r=\frac{1.05\times10^{-6}}{1.5}=7.0\times10^{-7}\ \text{m}=700\ \text{nm}.$$$

We are asked for the difference in wavelengths, so we subtract:

$$\Delta\lambda=\lambda_r-\lambda_v.$$

Putting in the two values just obtained,

$$\Delta\lambda=700\ \text{nm}-400\ \text{nm}=300\ \text{nm}.$$

So, the answer is $$300\ \text{nm}.$$

We have a Young’s double-slit arrangement. For any two consecutive bright (or dark) fringes, the angular separation is given by the standard result of interference:

$$\Delta\theta \;=\;\frac{\lambda}{d}$$

where $$\lambda$$ is the wavelength of light and $$d$$ is the centre-to-centre separation of the two slits.

The wavelength given is 500 nm. First we convert it completely into SI units:

$$500\ \text{nm}=500\times10^{-9}\ \text{m}=5\times10^{-7}\ \text{m}$$

The slit separation is 0.05 mm, which we again express in metres:

$$0.05\ \text{mm}=0.05\times10^{-3}\ \text{m}=5\times10^{-5}\ \text{m}$$

Now we substitute these numerical values into the formula for $$\Delta\theta$$:

$$\Delta\theta \;=\;\frac{\lambda}{d}\;=\;\frac{5\times10^{-7}\ \text{m}}{5\times10^{-5}\ \text{m}}$$

The powers of ten and the numerical factors simplify very neatly:

$$\Delta\theta \;=\;\frac{5}{5}\times10^{(-7)-(-5)}\;=\;1\times10^{-2}\ \text{radian}$$

That is

$$\Delta\theta =0.01\ \text{radian}$$

To make comparison with the options easier, we convert this radian value to degrees, using the relation $$1\ \text{radian}=57.3^\circ$$ (approximately):

$$\Delta\theta =0.01\times57.3^\circ =0.573^\circ$$

On rounding to two significant figures, this becomes $$0.57^\circ$$, which matches the option provided.

Hence, the correct answer is Option B.

We are given that the incident beam is already plane-polarized with a uniform intensity $$I_0 = 3.3\;{\rm W\,m^{-2}}$$ and that it falls normally on a polarizer whose clear cross-sectional area is $$A = 3 \times 10^{-4}\;{\rm m^{2}}.$$

First we find the incident power. Power is related to intensity by the elementary relation $$P = I \,A.$$ Substituting the given numbers we have

$$P_0 = I_0\,A = \bigl(3.3\;{\rm W\,m^{-2}}\bigr)\,\bigl(3 \times 10^{-4}\;{\rm m^{2}}\bigr) = 9.9 \times 10^{-4}\;{\rm W}.$$

Now the transmission through a polarizer obeys Malus’ Law, which states

$$I = I_0 \cos^{2}\theta,$$

where $$\theta$$ is the angle between the incident light’s polarization direction and the transmission axis of the polarizer. Because the polarizer is rotating steadily, this angle changes uniformly from $$0$$ to $$2\pi$$ during one full revolution. Hence the transmitted intensity keeps oscillating, and we want the average value over an entire turn.

The time (or angular) average of $$\cos^{2}\theta$$ over one full cycle is

$$\langle\cos^{2}\theta\rangle = \frac{1}{2}.$$ Therefore the average transmitted intensity is one half of the incident intensity:

$$\langle I\rangle = \frac{I_0}{2} = \frac{3.3}{2}\;{\rm W\,m^{-2}} = 1.65\;{\rm W\,m^{-2}}.$$

Correspondingly, the average transmitted power becomes

$$\langle P\rangle \;=\; \langle I\rangle\,A = \bigl(1.65\;{\rm W\,m^{-2}}\bigr)\,\bigl(3 \times 10^{-4}\;{\rm m^{2}}\bigr) = 4.95 \times 10^{-4}\;{\rm W}.$$

Next we must find how long one revolution of the polarizer takes. The angular speed is given as $$\omega = 31.4\;{\rm rad\,s^{-1}}.$$ One complete turn corresponds to an angle of $$2\pi\;{\rm rad}.$$ Using the basic relation $$T = \dfrac{2\pi}{\omega}$$ we have

$$T = \frac{2\pi}{31.4} = \frac{6.28}{31.4} = 0.20\;{\rm s}.$$

Finally, energy transmitted in one revolution equals the average power multiplied by the time for one revolution:

$$E = \langle P\rangle\,T = \bigl(4.95 \times 10^{-4}\;{\rm W}\bigr)\,\bigl(0.20\;{\rm s}\bigr) = 9.9 \times 10^{-5}\;{\rm J}.$$ Numerically $$9.9 \times 10^{-5}\;{\rm J} \approx 1.0 \times 10^{-4}\;{\rm J}.$$

Hence, the correct answer is Option B.

Let the intensity of the unpolarised light incident on the polariser be $$I_0.$$

Because the statement tells us that the polariser-analyser system “does not absorb any light”, we treat the polariser as changing only the state of polarisation, not the intensity. Hence, after the light passes through the polariser its intensity is still $$I_0,$$ but it is now plane-polarised along the transmission axis of the polariser.

If the analyser’s transmission axis makes an angle $$\theta$$ with the polariser’s axis, Malus’ law gives the transmitted intensity

$$I \;=\; I_0 \,\cos^2\theta.$$

According to the question this intensity is 36 % of the original, so

$$I = 0.36\,I_0.$$

Substituting this value into Malus’ law, we have

$$0.36\,I_0 \;=\; I_0 \,\cos^2\theta.$$

Dividing both sides by $$I_0$$ eliminates the common factor:

$$\cos^2\theta = 0.36.$$

Taking the positive square root (the angle is between 0° and 90°),

$$\cos\theta = \sqrt{0.36} = 0.6 = \dfrac{3}{5}.$$

Hence

$$\theta = \cos^{-1}\!\left(\dfrac{3}{5}\right) = 53^\circ.$$

To make the transmitted intensity fall to zero, the analyser’s axis must become perpendicular to the polariser’s axis, i.e. the angle between them must be $$90^\circ.$$ At present the angle is $$\theta = 53^\circ.$$ Therefore the analyser must be rotated through an additional angle

$$\phi = 90^\circ - \theta = 90^\circ - 53^\circ = 37^\circ.$$

The numerical value is confirmed by the given relation $$\sin^{-1}\!\left(\dfrac{3}{5}\right) = 37^\circ.$$ So the analyser needs to be rotated through $$37^\circ$$ to extinguish the light completely.

Hence, the correct answer is Option B.

In a Young’s double slit experiment the distance between successive bright (or dark) fringes on the screen is called the fringe width. It is given by the well-known formula

$$\beta=\frac{\lambda D}{d},$$

where $$\lambda$$ is the wavelength of light, $$D$$ is the distance of the screen from the slits, and $$d$$ is the separation between the two slits.

Let the fixed length of the screen segment in which we are counting fringes be $$L$$. The number of fringes that fit into this segment is simply the total length divided by the width of one fringe:

$$n=\frac{L}{\beta}.$$

Substituting the expression for $$\beta$$, we get

$$n=\frac{L}{\dfrac{\lambda D}{d}} =\frac{L d}{\lambda D}.$$

Notice that $$L$$, $$d$$ and $$D$$ remain unchanged when we replace one light source by another. Hence the product $$\dfrac{L d}{D}$$ is a constant for the given apparatus. We can therefore write

$$n\propto\frac{1}{\lambda}.$$

This proportionality means that the number of fringes is inversely proportional to the wavelength.

Now, with wavelength $$\lambda_1=700\ \text{nm}$$, the observed number of fringes is

$$n_1=16.$$

If the wavelength is changed to $$\lambda_2=400\ \text{nm}$$, the new number of fringes $$n_2$$ is obtained from the ratio

$$\frac{n_2}{n_1}=\frac{\lambda_1}{\lambda_2}.$$

Substituting the values, we have

$$n_2 = n_1 \times \frac{\lambda_1}{\lambda_2} =16 \times \frac{700\ \text{nm}}{400\ \text{nm}} =16 \times \frac{700}{400}.$$

We simplify the fraction:

$$\frac{700}{400}=\frac{7}{4}=1.75.$$

Therefore,

$$n_2 = 16 \times 1.75 = 28.$$

So, when the wavelength is reduced to 400 nm, 28 fringes will be observed in the same segment of the screen.

Hence, the correct answer is Option D.

For a double-slit arrangement let the two slits give waves of equal individual intensity $$I_0$$. The general interference formula is first stated:

$$I \;=\; I_1 + I_2 + 2\sqrt{I_1I_2}\,\cos\phi,$$

where $$\phi$$ is the phase difference of the two waves at the observation point. Because both slits are identical, we have $$I_1 = I_2 = I_0$$, so the expression becomes

$$I \;=\; 2I_0 + 2I_0\cos\phi \;=\; 2I_0\bigl(1+\cos\phi\bigr).$$

Now we use the trigonometric identity $$1+\cos\phi = 2\cos^2\frac{\phi}{2}$$, giving

$$I \;=\; 2I_0 \times 2\cos^2\frac{\phi}{2} \;=\; 4I_0\cos^2\frac{\phi}{2}.$$

At the centre of a bright fringe (the central maximum) the two waves are in phase, so $$\phi = 0$$ and we obtain the maximum intensity

$$I_{\max} = 4I_0\cos^2 0 = 4I_0.$$

Our task is to find the intensity $$I$$ when the path difference $$\Delta$$ equals $$\dfrac{\lambda}{8}$$. Phase difference and path difference are connected by the relation

$$\phi = \dfrac{2\pi}{\lambda}\,\Delta.$$

Substituting $$\Delta = \dfrac{\lambda}{8}$$ gives

$$\phi \;=\; \dfrac{2\pi}{\lambda}\,\Bigl(\dfrac{\lambda}{8}\Bigr) = \dfrac{2\pi}{8} = \dfrac{\pi}{4}.$$

Now we insert this value of $$\phi$$ into the intensity formula:

$$I \;=\; 4I_0\cos^2\frac{\phi}{2} = 4I_0\cos^2\!\Bigl(\dfrac{\pi}{4}\times\dfrac12\Bigr) = 4I_0\cos^2\!\Bigl(\dfrac{\pi}{8}\Bigr).$$

We are interested in the ratio of this intensity to the central maximum:

$$\dfrac{I}{I_{\max}} = \dfrac{4I_0\cos^2\!\bigl(\dfrac{\pi}{8}\bigr)}{4I_0} = \cos^2\!\Bigl(\dfrac{\pi}{8}\Bigr).$$

To evaluate $$\cos\!\bigl(\dfrac{\pi}{8}\bigr)$$ we employ the half-angle identity once more:

$$\cos\!\Bigl(\dfrac{\pi}{8}\Bigr) = \sqrt{\dfrac{1+\cos\!\bigl(\dfrac{\pi}{4}\bigr)}{2}} = \sqrt{\dfrac{1+\tfrac{1}{\sqrt2}}{2}} = \sqrt{\dfrac{1+\;0.7071}{2}} = \sqrt{\dfrac{1.7071}{2}} = \sqrt{0.8536} \approx 0.9239.$$

Squaring this value we obtain

$$\cos^2\!\Bigl(\dfrac{\pi}{8}\Bigr) \;=\; (0.9239)^2 \;\approx\; 0.8536.$$

Thus,

$$\dfrac{I}{I_{\max}} \approx 0.853.$$

Hence, the correct answer is Option A.

We know that, in Young’s double slit experiment, the linear separation between two successive bright (or dark) fringes on the screen is called the fringe width $$\beta$$ and is given by the formula

$$\beta \;=\;\dfrac{\lambda\,D}{d}$$

where

$$\lambda$$ = wavelength of the monochromatic light,

$$D$$ = distance between the double-slit plane and the screen,

$$d$$ = separation between the two slits.

First, we convert every quantity to SI units:

Wavelength: $$\lambda = 589 \text{ nm} = 589 \times 10^{-9}\ \text{m}$$

Slit separation: $$d = 0.15 \text{ mm} = 0.15 \times 10^{-3}\ \text{m}$$

Screen distance: $$D = 1.5\ \text{m}$$ (already in SI units).

Now we substitute these values into the formula:

$$ \beta = \dfrac{\lambda\,D}{d} = \dfrac{(589 \times 10^{-9}\ \text{m}) \;(1.5\ \text{m})}{0.15 \times 10^{-3}\ \text{m}} $$

First we multiply the numerical values in the numerator:

$$589 \times 1.5 = 883.5$$

So,

$$ \beta = \dfrac{883.5 \times 10^{-9}\ \text{m}^{2}}{0.15 \times 10^{-3}\ \text{m}} $$

Next, we divide the powers of ten separately. We have

$$\dfrac{10^{-9}}{10^{-3}} = 10^{-9 - (-3)} = 10^{-6}$$

and for the numerical division:

$$\dfrac{883.5}{0.15} = \dfrac{883.5 \times 100}{15} = \dfrac{88350}{15} = 5890$$

Hence,

$$ \beta = 5890 \times 10^{-6}\ \text{m} = 5.890 \times 10^{-3}\ \text{m} $$

To convert metres to millimetres, we recall that $$1\ \text{m} = 1000\ \text{mm}$$, so

$$ \beta = 5.890 \times 10^{-3}\ \text{m} \times 1000\ \dfrac{\text{mm}}{\text{m}} = 5.890\ \text{mm} \approx 5.9\ \text{mm}. $$

Hence, the correct answer is Option C.

First of all, for a double-slit interference pattern the position $$y_m$$ of the $$m^{\text{th}}$$ bright fringe on the screen is given by the formula

$$y_m \;=\; \frac{m \lambda D}{d}$$

where

$$\lambda$$ is the wavelength of light,   $$D$$ is the distance between the slits and the screen,   $$d$$ is the separation between the two slits,   $$m$$ is the order number of the bright fringe (an integer 0, 1, 2, …).

We know the following numerical values:

$$d \;=\; 1\;\text{mm} \;=\; 1 \times 10^{-3}\;\text{m}$$ $$D \;=\; 100\;\text{cm} \;=\; 1\;\text{m}$$ $$\lambda \;=\; 632.8\;\text{nm} \;=\; 632.8 \times 10^{-9}\;\text{m}$$ The observed distance of the bright fringe from the central bright fringe is $$y_m \;=\; 1.27\;\text{mm} \;=\; 1.27 \times 10^{-3}\;\text{m}$$

We substitute these numbers into the formula and solve for $$m$$:

$$ m = \frac{y_m \, d}{\lambda D} = \frac{\bigl(1.27 \times 10^{-3}\bigr)\bigl(1 \times 10^{-3}\bigr)} {\bigl(632.8 \times 10^{-9}\bigr)\bigl(1\bigr)} $$

To simplify the numerator, we multiply the powers of ten:

$$1.27 \times 10^{-3} \times 1 \times 10^{-3} = 1.27 \times 10^{-6}$$

So we have

$$ m = \frac{1.27 \times 10^{-6}}{632.8 \times 10^{-9}} = \frac{1.27}{632.8} \times 10^{-6+9} = \frac{1.27}{632.8} \times 10^{3} $$

Now,

$$\frac{1.27}{632.8} \approx 0.002007$$

Hence,

$$m \approx 0.002007 \times 1000 = 2.007$$

Since $$m$$ must be an integer for a bright fringe, we take $$m = 2$$. This tells us that the observed bright band is the second-order bright fringe.

The path difference $$\Delta$$ between the waves from the two slits at the position of the $$m^{\text{th}}$$ bright fringe is

$$\Delta = m \lambda$$

Substituting $$m = 2$$ and $$\lambda = 632.8\;\text{nm}$$, we get

$$ \Delta = 2 \times 632.8\;\text{nm} = 1265.6\;\text{nm} $$

We convert nanometres to micrometres using $$1\;\mu\text{m} = 1000\;\text{nm}$$:

$$ \Delta = \frac{1265.6\;\text{nm}}{1000} = 1.2656\;\mu\text{m} \approx 1.27\;\mu\text{m} $$

Hence, the correct answer is Option A.

We have an astronomical telescope whose objective (the front mirror) has a circular aperture of diameter $$D = 5\ \text{m}$$. When two point objects on the surface of the moon are imaged, the ability of the telescope to distinguish them separately is governed by the Rayleigh criterion for a circular aperture.

According to the Rayleigh criterion, the minimum angular separation $$\theta_{\min}$$ that can just be resolved is given by the formula

$$ \theta_{\min} \;=\; 1.22\,\dfrac{\lambda}{D}. $$

Here $$\lambda$$ is the wavelength of the light used and $$D$$ is the diameter of the aperture. First, let us put every quantity into SI units:

$$ \lambda = 5500\ \text{\AA} = 5500 \times 10^{-10}\ \text{m} = 5.5 \times 10^{-7}\ \text{m}. $$

The distance between the earth and the moon, which we denote by $$L$$, is given as

$$ L = 4 \times 10^{5}\ \text{km} = 4 \times 10^{5} \times 10^{3}\ \text{m} = 4 \times 10^{8}\ \text{m}. $$

Now we substitute $$\lambda$$ and $$D$$ into the Rayleigh formula:

$$ \theta_{\min} = 1.22 \,\dfrac{5.5 \times 10^{-7}\ \text{m}}{5\ \text{m}}. $$

Simplify the numerator first:

$$ 1.22 \times 5.5 = 6.71, $$

so

$$ 1.22 \times 5.5 \times 10^{-7}\ \text{m} = 6.71 \times 10^{-7}\ \text{m}. $$

Dividing by the diameter $$D = 5 \ \text{m}$$, we get

$$ \theta_{\min} = \dfrac{6.71 \times 10^{-7}}{5} = 1.342 \times 10^{-7}\ \text{radian}. $$

The linear separation $$s$$ on the moon’s surface that corresponds to this angular separation is obtained by simple geometry:

$$ s = L \,\theta_{\min}. $$

Substituting $$L = 4 \times 10^{8}\ \text{m}$$ and $$\theta_{\min} = 1.342 \times 10^{-7}\ \text{rad}$$, we have

$$ s = \left(4 \times 10^{8}\right)\,\left(1.342 \times 10^{-7}\right)\ \text{m}. $$

Multiply the numbers explicitly:

$$ 4 \times 1.342 = 5.368, $$

and combine the powers of ten:

$$ 10^{8} \times 10^{-7} = 10^{1}. $$

Therefore

$$ s = 5.368 \times 10^{1}\ \text{m} = 53.68\ \text{m}. $$

Rounding to the nearest convenient value, the minimum resolvable separation is approximately $$54\ \text{m}$$, which is closest to $$60\ \text{m}$$ in the given options.

Hence, the correct answer is Option A.

We start with two monochromatic light waves that are initially in phase. Their wavelength in vacuum is given to be $$\lambda$$.

For any wave that moves through a medium, the phase it accumulates depends on the distance travelled and on the speed of propagation in that medium. The standard relation connecting phase $$\Phi$$, path length $$L$$ and wavelength is

$$\Phi = \frac{2\pi}{\lambda_{\text{medium}}}\,L,$$

where $$\lambda_{\text{medium}}$$ is the wavelength of the light inside the medium. Because the frequency of light does not change when it enters a medium, the wavelength in a medium of refractive index $$n$$ is shortened according to the well-known formula

$$\lambda_{\text{medium}} = \frac{\lambda}{n}.$$

Substituting this value of $$\lambda_{\text{medium}}$$ into the phase formula, we obtain

$$\Phi \;=\; \frac{2\pi}{\lambda_{\text{medium}}}\,L \;=\; \frac{2\pi}{\dfrac{\lambda}{n}}\;L \;=\; \frac{2\pi n}{\lambda}\,L \;=\; \frac{2\pi}{\lambda}\,(nL).$$

Thus a wave travelling a distance $$L$$ in a medium of refractive index $$n$$ accumulates the phase

$$\Phi = \frac{2\pi}{\lambda}\,nL.$$

Now, let us apply this to our two waves separately:

The first wave travels a distance $$L_1$$ in the medium with refractive index $$n_1$$, so its phase after emerging is

$$\Phi_1 = \frac{2\pi}{\lambda}\,n_1L_1.$$

The second wave travels a distance $$L_2$$ in the medium with refractive index $$n_2$$, so its phase after emerging is

$$\Phi_2 = \frac{2\pi}{\lambda}\,n_2L_2.$$

The phase difference $$\Delta\Phi$$ between the two waves after they have completed their respective journeys is simply the algebraic difference of these two phases. Taking the first wave minus the second, we have

$$\Delta\Phi = \Phi_1 \;-\; \Phi_2 = \frac{2\pi}{\lambda}\,n_1L_1 \;-\; \frac{2\pi}{\lambda}\,n_2L_2.$$

Factoring out the common factor $$\frac{2\pi}{\lambda}$$ gives

$$\Delta\Phi = \frac{2\pi}{\lambda}\;\bigl(n_1L_1 - n_2L_2\bigr).$$

This expression matches exactly the form given in Option C.

Hence, the correct answer is Option C.

For a single-slit Fraunhofer diffraction pattern the angular position of the dark minima is given by the well-known condition

$$a\,\sin\theta = m\,\lambda$$

where $$a$$ is the width of the slit, $$\lambda$$ is the wavelength of the incident light, $$\theta$$ is the angle measured from the central (zero-order) maximum to the minimum, and $$m=1,2,3,\ldots$$ denotes the order of the minimum.

We are told that the second minimum (that is, $$m=2$$) occurs at an angle $$\theta_2 = 60^{\circ}$$. Substituting these values into the formula gives

$$a \,\sin 60^{\circ} = 2\,\lambda.$$

First we write the numerical value of the wavelength in convenient form. The statement gives $$\lambda = 6000 \times 10^{-8}\ \text{cm}$$, and multiplying the coefficients we get

$$\lambda = 6\,000 \times 10^{-8}\ \text{cm} = 6 \times 10^{-5}\ \text{cm}.$$

Now we evaluate the sine of $$60^{\circ}$$:

$$\sin 60^{\circ} = \frac{\sqrt3}{2} \approx 0.866.$$

Putting these into the relation for $$a$$ we have

$$a = \frac{2\,\lambda}{\sin 60^{\circ}} = \frac{2 \times 6 \times 10^{-5}\ \text{cm}}{0.866} = \frac{12 \times 10^{-5}\ \text{cm}}{0.866} \approx 1.386 \times 10^{-4}\ \text{cm}.$$

The angle $$\theta_1$$ of the first diffraction minimum corresponds to $$m=1$$, so the same formula gives

$$a\,\sin\theta_1 = 1 \times \lambda.$$ Substituting the value of $$a$$ that we have just obtained, we get

$$\sin\theta_1 = \frac{\lambda}{a} = \frac{\lambda}{\dfrac{2\,\lambda}{\sin 60^{\circ}}} = \lambda \;\times\; \frac{\sin 60^{\circ}}{2\,\lambda} = \frac{\sin 60^{\circ}}{2}.$$

Because the wavelength $$\lambda$$ has cancelled out, the numerical evaluation is very easy:

$$\sin\theta_1 = \frac{0.866}{2} = 0.433.$$

We now find the angle whose sine is $$0.433$$. Taking the inverse sine (in degrees),

$$\theta_1 = \sin^{-1}(0.433) \approx 25.6^{\circ}.$$

This value lies closest to $$25^{\circ}$$ among the alternatives provided.

Hence, the correct answer is Option C.

For interference of two coherent sources in a Young’s double-slit arrangement, the resultant intensity at any point is obtained from the formula

$$I \;=\; I_0\,\bigl(1+\cos\phi\bigr),$$

where $$I_0$$ is the intensity due to each slit taken separately, and $$\phi$$ is the phase difference between the two waves at the observation point. The phase difference in terms of path difference $$\Delta x$$ is

$$\phi \;=\; \frac{2\pi}{\lambda}\,\Delta x.$$

We first use the given information for the point where the path difference is $$\lambda$$.

For $$\Delta x = \lambda$$ we have

$$\phi \;=\; \frac{2\pi}{\lambda}\,\lambda \;=\; 2\pi.$$

Substituting $$\phi = 2\pi$$ in the intensity expression, we obtain

$$I_{\lambda} \;=\; I_0\bigl(1+\cos 2\pi\bigr).$$

Since $$\cos 2\pi = 1,$$ this gives

$$I_{\lambda} \;=\; I_0(1+1) \;=\; 2I_0.$$

The problem states that this intensity equals $$K$$ units, so

$$K \;=\; 2I_0.$$

Now we move to the point where the path difference is $$\dfrac{\lambda}{6}$$.

For $$\Delta x = \dfrac{\lambda}{6},$$ the phase difference is

$$\phi \;=\; \frac{2\pi}{\lambda}\,\frac{\lambda}{6} \;=\; \frac{\pi}{3}.$$

Using the intensity formula again, we have

$$I_{\lambda/6} \;=\; I_0\bigl(1 + \cos\!\tfrac{\pi}{3}\bigr).$$

Because $$\cos\!\tfrac{\pi}{3} = \tfrac{1}{2},$$ the expression becomes

$$I_{\lambda/6} \;=\; I_0\bigl(1 + \tfrac12\bigr) \;=\; \tfrac32 I_0.$$

The question tells us that this same intensity can be written as $$\dfrac{nK}{12}$$. We therefore set

$$\tfrac32 I_0 \;=\; \frac{nK}{12}.$$

Substituting $$K = 2I_0$$ from our earlier result, we get

$$\tfrac32 I_0 \;=\; \frac{n(2I_0)}{12} \;=\; \frac{n I_0}{6}.$$

Dividing both sides by $$I_0$$ to eliminate it, we are left with

$$\tfrac32 \;=\; \frac{n}{6}.$$

Multiplying by 6, we finally obtain

$$n = 9.$$

Hence, the correct answer is Option 9.

We have a single-slit diffraction experiment in which monochromatic light of wavelength $$\lambda = 6000 \times 10^{-10}\,\text{m}$$ is incident on a slit of width $$a = 0.6 \times 10^{-4}\,\text{m}$$.

First we convert these numbers to the same power of ten so that later division is easy.

For the wavelength:

$$\lambda = 6000 \times 10^{-10}\,\text{m} = 6.0 \times 10^{-7}\,\text{m}.$$

For the slit width:

$$a = 0.6 \times 10^{-4}\,\text{m} = 6.0 \times 10^{-5}\,\text{m}.$$

The condition for the minima in a single-slit diffraction pattern is stated by the formula

$$a \sin\theta = m\lambda,$$

where $$m = \pm 1, \pm 2, \pm 3, \ldots$$ represents the order of the minima.

Because $$\sin\theta$$ can never exceed 1, the largest integer value that $$m$$ can have must satisfy

$$a \sin\theta \le a \quad\Longrightarrow\quad m\lambda \le a,$$

which gives

$$m_{\text{max}} = \frac{a}{\lambda}.$$

Substituting the numerical values, we obtain

$$m_{\text{max}} = \frac{6.0 \times 10^{-5}}{6.0 \times 10^{-7}}.$$

Now we divide the coefficients and subtract the exponents of ten:

$$\frac{6.0}{6.0} = 1,$$

and

$${10^{-5}} \div {10^{-7}} = 10^{-5 - (-7)} = 10^{2}.$$

Hence

$$m_{\text{max}} = 1 \times 10^{2} = 100.$$

This means that on one side of the central maximum we can have integers $$m = 1, 2, 3, \ldots, 100,$$ i.e. exactly $$100$$ minima.

Because the diffraction pattern is symmetric, the same number of minima appear on the other side. Therefore the total number of minima on both sides is

$$2 \times 100 = 200.$$

So, the answer is $$200$$.

We are told that 4 % of the incident light energy is reflected from the air-glass interface, and we also know that the incident electric-field amplitude is $$E_0 = 30\ \text{V m}^{-1}.$$

First, we recall the relation between intensity and amplitude. For an electromagnetic wave, the average intensity $$I$$ is proportional to the square of the peak electric-field amplitude:

$$I \propto E_0^{\,2}.$$

Because of this proportionality, the fraction of the reflected intensity is equal to the square of the fraction of the reflected amplitude. Mathematically, if $$R = 0.04$$ represents the reflected intensity fraction, then

$$\left(\dfrac{E_r}{E_0}\right)^2 = R = 0.04,$$

where $$E_r$$ is the reflected electric-field amplitude. Taking the square root on both sides gives

$$\dfrac{E_r}{E_0} = \sqrt{0.04} = 0.2.$$

Substituting the given incident amplitude $$E_0 = 30\ \text{V m}^{-1},$$ we obtain the reflected amplitude:

$$E_r = 0.2 \times 30\ \text{V m}^{-1} = 6\ \text{V m}^{-1}.$$

The transmitted (or propagated) wave is the remainder of the incident wave after the reflected portion has been subtracted. Hence, its amplitude $$E_t$$ is

$$E_t = E_0 - E_r = 30\ \text{V m}^{-1} - 6\ \text{V m}^{-1} = 24\ \text{V m}^{-1}.$$

Thus the electric-field amplitude of the wave that continues inside the glass slab is $$24\ \text{V m}^{-1}.$$

Hence, the correct answer is Option C.

Let the incident unpolarized light have intensity $$I_0$$. For unpolarized light passing through a single polarizer, the well-known result is that exactly one half of the incident intensity is transmitted. Mathematically,

$$I_1=\dfrac{I_0}{2}.$$

Here $$I_1$$ is the intensity just after the first polarizer $$P_1$$.

Now the second polarizer $$P_2$$ makes an angle of $$30^{\circ}$$ with respect to $$P_1$$ because $$P_1$$ is at $$90^{\circ}$$ from $$P_3$$ while $$P_2$$ is at $$60^{\circ}$$ from $$P_3$$, giving

$$\theta_{12}=90^{\circ}-60^{\circ}=30^{\circ}.$$

Whenever already plane-polarized light of intensity $$I$$ meets another polarizer at an angle $$\theta$$, Malus’s law states:

$$I_{\text{after}} = I_{\text{before}}\cos^2\theta.$$

Applying this law to the passage from $$P_1$$ to $$P_2$$, we obtain

$$I_2 = I_1\cos^2 30^{\circ}.$$

Because $$\cos 30^{\circ}= \dfrac{\sqrt3}{2},$$ we have

$$I_2=\dfrac{I_0}{2}\left(\dfrac{\sqrt3}{2}\right)^2=\dfrac{I_0}{2}\cdot\dfrac{3}{4}=\dfrac{3I_0}{8}.$$

Next the third polarizer $$P_3$$ is crossed with $$P_1$$, so its axis differs from that of $$P_2$$ by the given $$60^{\circ}$$. Applying Malus’s law once more with $$\theta_{23}=60^{\circ},$$

$$I_3 = I_2\cos^2 60^{\circ}.$$

Because $$\cos 60^{\circ}= \dfrac12,$$ this gives

$$I_3 = \dfrac{3I_0}{8}\left(\dfrac12\right)^2=\dfrac{3I_0}{8}\cdot\dfrac14=\dfrac{3I_0}{32}.$$

The intensity emerging from the complete set of three polarizers is therefore

$$I=\dfrac{3I_0}{32}.$$

Taking the ratio asked for in the problem,

$$\frac{I_0}{I}= \frac{I_0}{\dfrac{3I_0}{32}} = \frac{32}{3}\approx 10.67.$$

Hence, the correct answer is Option D.

For the limit of angular resolution produced by diffraction at a circular aperture we begin with the well-known Rayleigh criterion. The formula is stated first:

$$\theta_{\text{min}} = 1.22 \, \frac{\lambda}{D}$$

Here $$\theta_{\text{min}}$$ denotes the smallest angle (in radians) that the telescope can resolve, $$\lambda$$ is the wavelength of the light, and $$D$$ is the diameter of the objective lens or mirror.

We are given a wavelength $$\lambda = 500 \, \text{nm}$$ and an objective diameter $$D = 200 \, \text{cm}$$. Before substituting, both quantities must be converted to the same system of units (metres in SI).

Recall that $$1 \, \text{nm} = 10^{-9} \, \text{m}$$, so

$$\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}.$$

Also $$1 \, \text{cm} = 10^{-2} \, \text{m}$$, therefore

$$D = 200 \, \text{cm} = 200 \times 10^{-2} \, \text{m} = 2 \, \text{m}.$$

Now we substitute these values into the Rayleigh formula:

$$\theta_{\text{min}} = 1.22 \, \frac{500 \times 10^{-9} \, \text{m}}{2 \, \text{m}}.$$

First carry out the division in the numerator and denominator:

$$\frac{500 \times 10^{-9}}{2} = 250 \times 10^{-9}.$$

Next multiply by the factor $$1.22$$:

$$\theta_{\text{min}} = 1.22 \times 250 \times 10^{-9}.$$

Performing the multiplication of the numerical factors,

$$1.22 \times 250 = 305.$$

So we obtain

$$\theta_{\text{min}} = 305 \times 10^{-9} \, \text{radian}.$$

This value can also be expressed as $$3.05 \times 10^{-7} \, \text{radian}$$, but the form above matches one of the given options exactly.

Hence, the correct answer is Option A.

We first recall that in a double-slit set-up the intensity pattern observed on the screen is the product of two separate effects:

1. Single-slit diffraction from each slit of width $$a$$ gives dark (minimum-intensity) directions obeying the well-known condition

$$a\,\sin\theta \;=\; m\,\lambda,$$

where $$m = \pm 1, \pm 2, \pm 3,\dots$$. The values of $$m$$ label the successive diffraction minima. The first minimum corresponds to $$m = 1$$, the second to $$m = 2$$, and so on.

2. Interference between the two slits of centre-to-centre separation $$d$$ produces bright fringes in the directions that satisfy

$$d\,\sin\theta \;=\; n\,\lambda,$$

with $$n = 0, \pm 1, \pm 2,\dots$$. Each integer $$n$$ therefore labels one interference maximum (bright fringe).

We are asked to count how many interference maxima lie between the first and the second diffraction minima, i.e. for angles satisfying

$$\text{first minimum:}\; a\sin\theta = 1\,\lambda \quad \text{to} \quad \text{second minimum:}\; a\sin\theta = 2\,\lambda.$$

To proceed, we substitute the expression for $$\sin\theta$$ obtained from the diffraction condition into the interference condition so that the two formulae can be linked directly.

From the diffraction minima condition we have

$$\sin\theta = \dfrac{m\,\lambda}{a}.$$

Placing this value of $$\sin\theta$$ into the interference condition $$d\sin\theta = n\lambda$$ gives

$$d \left(\dfrac{m\,\lambda}{a}\right) = n\,\lambda.$$

The factor $$\lambda$$ cancels on both sides, leaving

$$n = \dfrac{d}{a}\,m.$$

This remarkably simple relation tells us that whenever $$m$$ takes an integer value at a diffraction minimum, the corresponding interference order is

$$n = \left(\dfrac{d}{a}\right)m.$$

We now put in the numerical values supplied in the question:

$$d = 19.44\;\mu \text{m}, \quad a = 4.05\;\mu \text{m}.$$

Hence

$$\dfrac{d}{a} = \dfrac{19.44}{4.05} = 4.8.$$

Let us evaluate $$n$$ at the two diffraction minima in question.

For the first diffraction minimum ($$m = 1$$) we obtain

$$n_1 = (4.8)(1) = 4.8.$$

For the second diffraction minimum ($$m = 2$$) we obtain

$$n_2 = (4.8)(2) = 9.6.$$

Between these two values of $$\theta$$, the interference maxima correspond to integer $$n$$ lying strictly between 4.8 and 9.6 (because at the minima themselves the intensity is zero, so those angular positions are excluded). All integers satisfying

$$4.8 < n < 9.6$$

are therefore counted. Listing them explicitly, we have

$$n = 5,\;6,\;7,\;8,\;9.$$

This is a total of

$$5 \text{ bright fringes}.$$

Hence, the correct answer is Option B.

For a bright (constructive) fringe in Young’s double-slit experiment we start from the condition

$$d\,\sin\theta = m\lambda,$$

where $$d$$ is the slit separation, $$\theta$$ the angular position of the fringe, $$\lambda$$ the wavelength of light and $$m$$ an integer (the order of the fringe).

We are told that the same angular position $$\theta = \dfrac{1}{40}\,{\rm rad}$$ gives bright fringes for two different wavelengths $$\lambda_1$$ and $$\lambda_2$$. Hence for the two cases we have

$$d\,\sin\theta = m_1\lambda_1 \qquad\text{and}\qquad d\,\sin\theta = m_2\lambda_2.$$

Because the left-hand side is the same in both equations, we can write

$$m_1\lambda_1 = m_2\lambda_2.$$ Now we substitute the numerical values that are common to both situations. The slit separation is

$$d = 0.1\,{\rm mm} = 0.1 \times 10^{-3}\,{\rm m} = 1.0 \times 10^{-4}\,{\rm m}.$$

To work conveniently with nanometres, we convert $$d$$ to nanometres. We recall that $$1\,{\rm m}=10^{9}\,{\rm nm},$$ therefore

$$d = 1.0 \times 10^{-4}\,{\rm m} = 1.0 \times 10^{-4}\times 10^{9}\,{\rm nm}=1.0 \times 10^{5}\,{\rm nm}.$$

The given angle is small, so $$\sin\theta \approx \theta,$$ and we evaluate

$$\sin\theta = \frac{1}{40} = 0.025.$$

Putting these numbers together, the product $$d\,\sin\theta$$ that must equal each $$m\lambda$$ is

$$d\,\sin\theta = (1.0 \times 10^{5}\,{\rm nm})\,(0.025) = 2.5 \times 10^{3}\,{\rm nm} = 2500\,{\rm nm}.$$

Thus both wavelengths must divide the value $$2500\,{\rm nm}$$ exactly so that the corresponding orders $$m_1$$ and $$m_2$$ are integers:

$$m_1 = \frac{2500}{\lambda_1},\qquad m_2 = \frac{2500}{\lambda_2}.$$

We now inspect each option to see which pair of wavelengths makes both fractions integral and still keeps the wavelengths in the visible range (380 nm - 740 nm).

Option A:  $$\lambda_1 = 400\,{\rm nm},\ \lambda_2 = 500\,{\rm nm}$$ gives $$\frac{2500}{400}=6.25\ (\text{not integer}),$$ so this fails.
Option B:  $$\lambda_1 = 380\,{\rm nm},\ \lambda_2 = 525\,{\rm nm}$$ gives $$\frac{2500}{380}=6.579,\ \frac{2500}{525}=4.762 \ (\text{neither integer}),$$ so this fails.
Option C:  $$\lambda_1 = 625\,{\rm nm},\ \lambda_2 = 500\,{\rm nm}$$ gives $$\frac{2500}{625}=4,\qquad \frac{2500}{500}=5,$$ both perfect integers, so this works.
Option D:  $$\lambda_1 = 380\,{\rm nm},\ \lambda_2 = 500\,{\rm nm}$$ gives $$\frac{2500}{380}=6.579\ (\text{not integer}),$$ so this fails.

Only Option C satisfies the integer requirement while keeping both wavelengths within the visible spectrum. For completeness we can note the corresponding fringe orders:

$$m_1 = 4\ \text{for}\ \lambda_1 = 625\,{\rm nm},\qquad m_2 = 5\ \text{for}\ \lambda_2 = 500\,{\rm nm},$$

and indeed $$m_1\lambda_1 = 4\times625 = 2500\,{\rm nm} = 5\times500 = m_2\lambda_2,$$ as required.

Hence, the correct answer is Option C.

In a Young’s double-slit experiment two coherent waves originating from the two slits reach a point on the screen with a certain path difference. Let this path difference be denoted by $$\Delta x$$. For the present point we are told that

$$\Delta x=\dfrac{\lambda}{8},$$

where $$\lambda$$ is the wavelength of the monochromatic light being used.

Whenever two waves of the same amplitude interfere, the intensity at any point on the screen is obtained from the interference formula

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi.$$

Because both slits are identical, the individual intensities are the same, i.e. $$I_1 = I_2 = I_s$$. Substituting $$I_1 = I_2 = I_s$$ in the above expression we get

$$I = I_s + I_s + 2\sqrt{I_s I_s}\cos\phi$$

$$\;\; = 2I_s + 2I_s\cos\phi$$

$$\;\; = 2I_s(1+\cos\phi).$$

The phase difference $$\phi$$ between the two waves is related to the path difference $$\Delta x$$ through the relation

$$\phi = \dfrac{2\pi}{\lambda}\,\Delta x.$$

Substituting $$\Delta x = \dfrac{\lambda}{8}$$, we obtain

$$\phi = \dfrac{2\pi}{\lambda}\left(\dfrac{\lambda}{8}\right) = \dfrac{2\pi}{8} = \dfrac{\pi}{4}.$$

Now, the cosine of this phase difference is

$$\cos\phi = \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \approx 0.707.$$

Putting this value back into the intensity expression, the intensity at the given point becomes

$$I = 2I_s\left(1 + \dfrac{\sqrt{2}}{2}\right).$$

The intensity at the centre of the central bright fringe, commonly called the maximum intensity $$I_{\text{max}}$$, occurs when the two waves meet in perfect phase (that is, $$\phi = 0$$). Then the formula gives

$$I_{\text{max}} = 2I_s(1 + \cos 0) = 2I_s(1 + 1) = 4I_s.$$

We now require the ratio of the intensity at the given point to the maximum intensity. Taking the ratio, we have

$$\dfrac{I}{I_{\text{max}}} = \dfrac{2I_s\left(1 + \dfrac{\sqrt{2}}{2}\right)}{4I_s} = \dfrac{1 + \dfrac{\sqrt{2}}{2}}{2}.$$

Simplifying the numerator first, write $$1 = \dfrac{2}{2}$$, so

$$1 + \dfrac{\sqrt{2}}{2} = \dfrac{2 + \sqrt{2}}{2}.$$

Substituting this back into the ratio,

$$\dfrac{I}{I_{\text{max}}} = \dfrac{\dfrac{2 + \sqrt{2}}{2}}{2} = \dfrac{2 + \sqrt{2}}{4}.$$

Because $$\sqrt{2} \approx 1.414$$, the numerator becomes

$$2 + 1.414 = 3.414.$$

Dividing by 4 gives

$$\dfrac{3.414}{4} \approx 0.8535.$$

This numerical value is closest to 0.85 among the choices provided.

Hence, the correct answer is Option B.

In a Young’s double-slit experiment each slit behaves like an independent source of light. The light energy that finally reaches the screen from a slit is proportional to the geometric width of that slit, because both slits are illuminated uniformly by the same extended source.

We are told that the ratio of the widths of the two slits is 4 : 1. Therefore the individual intensities of light emerging from the two slits are also in the same ratio, i.e.

$$I_1 : I_2 = 4 : 1.$$

For any coherent source, the amplitude of the electric field is the square root of the intensity. Hence

$$a_1 = \sqrt{I_1}, \qquad a_2 = \sqrt{I_2}.$$

Taking the ratio, we have

$$\frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{4}{1}} = 2.$$

Thus the individual amplitudes may be written, for convenience, as

$$a_1 = 2A, \quad a_2 = A,$$

where $$A$$ is some common amplitude factor.

The standard interference formula for the resultant intensity due to two coherent waves of amplitudes $$a_1$$ and $$a_2$$ is stated as

$$I = a_1^{\,2} + a_2^{\,2} + 2\,a_1 a_2 \cos\phi,$$

where $$\phi$$ is the phase difference between the two waves at the observation point.

For a bright fringe (maximum) the condition is $$\cos\phi = +1$$, giving

$$I_{\text{max}} = a_1^{\,2} + a_2^{\,2} + 2\,a_1 a_2 = (a_1 + a_2)^2.$$

For a dark fringe (minimum) the condition is $$\cos\phi = -1$$, giving

$$I_{\text{min}} = a_1^{\,2} + a_2^{\,2} - 2\,a_1 a_2 = (a_1 - a_2)^2.$$

Substituting $$a_1 = 2A$$ and $$a_2 = A$$ in these expressions, we get

$$I_{\text{max}} = (2A + A)^2 = (3A)^2 = 9A^{\,2},$$

$$I_{\text{min}} = (2A - A)^2 = (A)^2 = A^{\,2}.$$

Therefore the ratio of the intensity of a maximum to that of the adjacent minimum is

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9A^{\,2}}{A^{\,2}} = 9 : 1.$$

Hence, the correct answer is Option 2.

In a Young’s double-slit experiment the condition for bright (constructive) interference is stated first:

$$d \sin\theta = m\lambda$$

Here, $$d$$ is the slit separation, $$\theta$$ is the angular position of the bright fringe with respect to the central axis, $$\lambda$$ is the wavelength of light, and $$m$$ is an integer (… , -3, -2, -1, 0, 1, 2, 3, …).

We are told that

$$d = 0.320\ \text{mm} = 0.320 \times 10^{-3}\ \text{m} = 3.20 \times 10^{-4}\ \text{m},$$

$$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5.00 \times 10^{-7}\ \text{m},$$

and that the observation is restricted to the angular range

$$-30^{\circ} \le \theta \le 30^{\circ}.$$

For any bright fringe to lie inside this range, the following inequality must hold:

$$|\,\sin\theta\,| \le \sin 30^{\circ} = \tfrac12.$$

Using the bright-fringe formula, we rewrite this requirement in terms of $$m$$:

$$\left|\frac{m\lambda}{d}\right| \le \tfrac12.$$

Isolating $$m$$ gives

$$|m| \le \frac{d}{\lambda}\,\frac12.$$

We now compute the numerical factor $$\dfrac{d}{\lambda}$$ first:

$$\frac{d}{\lambda} = \frac{3.20 \times 10^{-4}}{5.00 \times 10^{-7}} = \frac{3.20}{5.00} \times 10^{\,(-4+7)} = 0.64 \times 10^{3} = 640.$$

Multiplying by $$\tfrac12$$ (that is, 0.5) we obtain

$$\frac{d}{\lambda}\,\frac12 = 640 \times 0.5 = 320.$$

Hence the largest integer value (in magnitude) that $$m$$ can take is

$$m_{\text{max}} = 320.$$

Therefore, allowable orders run from $$m = -320$$ up through $$m = +320,$$ inclusive. The total count of integers in this set is calculated as follows:

$$N = (320 - (-320)) + 1 = 640 + 1 = 641.$$

This number $$N$$ is precisely the total number of bright fringes observed within the stated angular limits.

Hence, the correct answer is Option B.

We are told that the numerical aperture of the objective lens is $$\text{NA}=1.25$$ and that light of wavelength $$\lambda=5000\ \text{\AA}$$ is used.

First, the wavelength must be expressed in metres. We recall that $$1\ \text{\AA}=10^{-10}\ \text{m}.$$ Hence

$$ \lambda = 5000\ \text{\AA}=5000 \times 10^{-10}\ \text{m}=5 \times 10^{-7}\ \text{m}. $$

For a microscope objective, the Rayleigh criterion gives the minimum resolvable distance (limit of resolution) as

$$ d = \frac{0.61\,\lambda}{\text{NA}}. $$

Substituting the known values, we have

$$ d = \frac{0.61 \times 5 \times 10^{-7}\ \text{m}}{1.25}. $$

First multiply the numerator:

$$ 0.61 \times 5 = 3.05, $$

so

$$ d = \frac{3.05 \times 10^{-7}\ \text{m}}{1.25}. $$

Now divide by 1.25:

$$ \frac{3.05}{1.25}=2.44, $$

hence

$$ d = 2.44 \times 10^{-7}\ \text{m}. $$

To convert this result into micrometres, we remember that $$1\ \mu\text{m}=10^{-6}\ \text{m}.$$ Therefore

$$ d = 2.44 \times 10^{-7}\ \text{m} = 2.44 \times 10^{-7}\ \text{m} \times \frac{1\ \mu\text{m}}{10^{-6}\ \text{m}} = 0.244\ \mu\text{m}. $$

Rounded to two significant figures, this is $$0.24\ \mu\text{m}.$$

Hence, the correct answer is Option B.

We are dealing with the diffraction-limited resolving power of a telescope whose objective is a circular aperture. For such an aperture, the smallest angular separation $$\theta_{\min}$$ that can be resolved is given by the Rayleigh criterion

$$\theta_{\min}=1.22\,\frac{\lambda}{D}$$

where $$\lambda$$ is the wavelength of the light and $$D$$ is the diameter of the objective.

First we convert all quantities to SI units. The diameter is given as 250 cm:

$$D = 250\;\text{cm} = 250 \times 10^{-2}\;\text{m} = 2.5\;\text{m}$$

The wavelength is 600 nm:

$$\lambda = 600\;\text{nm} = 600 \times 10^{-9}\;\text{m}$$

We simplify the wavelength value for easier calculation:

$$600 \times 10^{-9}\;\text{m} = 6.0 \times 10^{2} \times 10^{-9}\;\text{m} = 6.0 \times 10^{-7}\;\text{m}$$

Now we substitute $$\lambda = 6.0 \times 10^{-7}\;\text{m}$$ and $$D = 2.5\;\text{m}$$ into the Rayleigh formula:

$$\theta_{\min}=1.22 \times \frac{6.0 \times 10^{-7}\;\text{m}}{2.5\;\text{m}}$$

We calculate the numerator first:

$$1.22 \times 6.0 \times 10^{-7} = 7.32 \times 10^{-7}$$

Now we divide by 2.5:

$$\theta_{\min} = \frac{7.32 \times 10^{-7}}{2.5}$$

To divide, we handle the coefficients and powers of ten separately:

$$\frac{7.32}{2.5} = 2.928$$

The power of ten remains $$10^{-7}$$, so

$$\theta_{\min} \approx 2.928 \times 10^{-7}\;\text{rad}$$

Rounding to two significant figures to match the precision of the options, we write

$$\theta_{\min} \approx 3.0 \times 10^{-7}\;\text{rad}$$

This value matches Option B.

Hence, the correct answer is Option B.

In a Young’s double-slit experiment, the condition for constructive interference at any point on the screen is that the path difference between the two interfering waves must be an integral multiple of the wavelength. Expressed mathematically, the path difference $$\Delta$$ required for a bright fringe is

$$\Delta = m\lambda \quad (m = 0,1,2,\dots)$$

Normally, at the centre of the screen the geometrical path difference between the two slits is zero, so the central point corresponds to $$m = 0$$ and is therefore bright.

Now we introduce a thin transparent film of thickness $$t$$ and refractive index $$\mu$$ just in front of one of the slits. Whenever a wave travels a distance $$t$$ inside a medium of refractive index $$\mu$$, its optical path length becomes $$\mu t$$. In the absence of the film the same geometrical distance would count as simply $$t$$ (because it would be travelled in air or vacuum, whose refractive index is $$1$$). Therefore, the additional optical path introduced by the film is

$$\text{Extra path difference} = \mu t - t = (\mu - 1)t.$$

This extra path difference shifts the entire interference pattern. If the shift of the central maximum equals one complete fringe width, it means that the point which was originally the central maximum now satisfies the bright-fringe condition for the next integral order. In other words, the optical path difference has effectively increased by exactly one wavelength $$\lambda$$:

$$ (\mu - 1)t = \lambda. $$

We now solve this simple algebraic equation for the thickness $$t$$. Dividing both sides by $$(\mu - 1)$$ we get

$$ t = \frac{\lambda}{\mu - 1}. $$

This is precisely the required thickness that shifts the central bright fringe by one fringe width when the film is inserted in front of one slit.

Hence, the correct answer is Option D.

We are told that the two interfering beams are coherent and that their amplitudes are in the ratio

$$\frac{a_1}{a_2}=\frac{1}{3}.$$

To work comfortably, we assign actual symbols that satisfy this ratio. Let the common constant of proportionality be $$k.$$ Then we can write

$$a_1 = k \quad\text{and}\quad a_2 = 3k.$$

Interference formulas relate intensities to these amplitudes. First, we recall the basic result:

Intensity is proportional to the square of amplitude, so

$$I_1 \propto a_1^{\,2}, \qquad I_2 \propto a_2^{\,2}.$$

Now, in a two-source interference pattern, the expressions for maximum and minimum intensities are obtained by adding the wave amplitudes in phase (for a bright fringe) and subtracting them out of phase (for a dark fringe). Explicitly, the standard formulas are

$$I_{\text{max}} = (a_1 + a_2)^{2},$$

$$I_{\text{min}} = (a_1 - a_2)^{2}.$$

We substitute the concrete values $$a_1 = k$$ and $$a_2 = 3k.$$ Carrying out each calculation step by step:

For the maximum intensity,

$$I_{\text{max}} = (a_1 + a_2)^{2} = (\,k + 3k\,)^{2} = (4k)^{2} = 16k^{2}.$$

For the minimum intensity,

$$I_{\text{min}} = (a_1 - a_2)^{2} = (\,k - 3k\,)^{2} = (-2k)^{2} = 4k^{2}.$$

The ratio of these two intensities is then

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{16k^{2}}{4k^{2}} = \frac{16}{4} = 4.$$

Notice that the factor $$k^{2}$$ cancels, as expected, so the result is independent of our initial scaling choice.

Hence, the required ratio of maximum to minimum intensities of the fringes is $$4.$$

Hence, the correct answer is Option A.

We start with the well-known expressions for the maximum and minimum intensities obtained when two coherent light waves of individual intensities $$I_1$$ and $$I_2$$ interfere.

The formulae are stated as:

$$I_{\text{max}}=(\sqrt{I_1}+\sqrt{I_2})^{2}$$

$$I_{\text{min}}=(\sqrt{I_1}-\sqrt{I_2})^{2}$$

The question tells us that after interference the ratio $$I_{\text{max}}:I_{\text{min}}$$ equals $$16:1$$. Translating this into an equation, we have

$$\frac{I_{\text{max}}}{I_{\text{min}}}=16$$

Substituting the expressions for $$I_{\text{max}}$$ and $$I_{\text{min}}$$,

$$\frac{(\sqrt{I_1}+\sqrt{I_2})^{2}}{(\sqrt{I_1}-\sqrt{I_2})^{2}}=16$$

To simplify, we let the ratio of the amplitudes (square-root intensities) be

$$a=\frac{\sqrt{I_1}}{\sqrt{I_2}}$$

Rewriting the equation in terms of $$a$$ gives

$$\frac{(a+1)^{2}}{(a-1)^{2}}=16$$

Cross-multiplying,

$$(a+1)^{2}=16(a-1)^{2}$$

Now we expand both binomials:

$$a^{2}+2a+1 = 16(a^{2}-2a+1)$$

$$a^{2}+2a+1 = 16a^{2}-32a+16$$

Bringing every term to the right side yields zero on the left:

$$0 = 16a^{2}-32a+16 - (a^{2}+2a+1)$$

$$0 = 15a^{2}-34a+15$$

The quadratic $$15a^{2}-34a+15=0$$ can be factorised as

$$(5a-3)(3a-5)=0$$

So, the possible values of $$a$$ are

$$a=\frac{3}{5} \quad \text{or} \quad a=\frac{5}{3}$$

Because amplitudes are positive and we conventionally take $$a\ge 1$$ by choosing the larger intensity in the numerator, we select

$$a=\frac{5}{3}$$

The ratio of the original intensities is the square of the amplitude ratio:

$$\frac{I_1}{I_2}=a^{2}=\left(\frac{5}{3}\right)^{2}=\frac{25}{9}$$

Thus the two waves have intensities in the ratio $$25:9$$.

Hence, the correct answer is Option A.

We start with unpolarized light of intensity $$I$$ incident on polarizer A. The basic result for an unpolarized beam passing through a single polarizer is Malus’ law in its first consequence:

Unpolarized → first polarizer gives $$I_{\text{after A}} \;=\;\dfrac{I}{2}.$$

Now this light meets polarizer B. Let the angle between the transmission axes of A and B be $$\phi$$. Malus’ law in general form states:

For light of intensity $$I_0$$ incident on a polarizer whose axis makes an angle $$\alpha$$ with the light’s existing plane of polarization, the transmitted intensity is $$I = I_0\cos^{2}\alpha.$$

Applying the law to polarizer B we have

$$I_{\text{after B}} = \left(\dfrac{I}{2}\right)\cos^{2}\phi.$$

The problem tells us that the emergent intensity after A and B alone is exactly $$\dfrac{I}{2}$$. Equating, we obtain

$$\left(\dfrac{I}{2}\right)\cos^{2}\phi = \dfrac{I}{2}\;\;\Longrightarrow\;\;\cos^{2}\phi = 1,$$

so $$\phi = 0^{\circ}.$$ Hence polarizers A and B are parallel.

Next, a third polarizer C is inserted between A and B, making an angle $$\theta$$ with A. Because A and B are parallel, the angle between C and B is also $$\theta$$.

Step by step through the three polarizers:

1. After A: $$I_{1}=\dfrac{I}{2}.$$

2. After C: apply Malus’ law with angle $$\theta$$, giving $$I_{2}=I_{1}\cos^{2}\theta =\left(\dfrac{I}{2}\right)\cos^{2}\theta.$$

3. After B: again use Malus’ law with the same angle $$\theta$$ (since C and B differ by $$\theta$$), obtaining $$I_{3}=I_{2}\cos^{2}\theta =\left(\dfrac{I}{2}\right)\cos^{2}\theta\cos^{2}\theta =\left(\dfrac{I}{2}\right)\cos^{4}\theta.$$

According to the question this final intensity equals $$\dfrac{I}{3}$$. Therefore

$$\left(\dfrac{I}{2}\right)\cos^{4}\theta = \dfrac{I}{3} \;\;\Longrightarrow\;\; \cos^{4}\theta = \dfrac{2}{3}.$$

Taking the fourth root,

$$\cos\theta = \left(\dfrac{2}{3}\right)^{\frac14}.$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.

We begin with unpolarised light of initial intensity $$I$$ incident on the first ideal polariser A. For unpolarised light, the standard result is that:

$$I_{\text{after A}}=\dfrac{I}{2}$$

This happens because an ideal polariser transmits exactly one‐half of the incident unpolarised intensity.

Now the light emerging from A is plane-polarised. Let the transmission axis of the second polariser B make an angle $$\theta_{AB}$$ with that of A. Malus’s Law states:

$$I_{\text{after B}}=I_{\text{after A}}\cos^{2}\theta_{AB}$$

According to the question, the observed intensity after B (with only A and B present) is still $$\dfrac{I}{2}$$. Substituting $$I_{\text{after A}}=\dfrac{I}{2}$$ gives

$$\dfrac{I}{2}\cos^{2}\theta_{AB}=\dfrac{I}{2}$$

Dividing both sides by $$\dfrac{I}{2}$$ yields

$$\cos^{2}\theta_{AB}=1$$

Hence $$\theta_{AB}=0^{\circ}$$. So the transmission axes of A and B are parallel.

Next, a third identical polariser C is inserted between A and B. Let the angle between the axes of A and C be $$\theta$$. Because B is parallel to A, the angle between C and B is also $$\theta$$.

Step-wise calculation of the intensity now proceeds as follows:

1. After A (unchanged): $$I_{1}=\dfrac{I}{2}$$

2. After C: apply Malus’s Law once more, now with angle $$\theta$$ between A and C:

$$I_{2}=I_{1}\cos^{2}\theta=\dfrac{I}{2}\cos^{2}\theta$$

3. After B: the light from C meets B at the same angle $$\theta$$, so again using Malus’s Law,

$$I_{3}=I_{2}\cos^{2}\theta=\left(\dfrac{I}{2}\cos^{2}\theta\right)\cos^{2}\theta=\dfrac{I}{2}\cos^{4}\theta$$

The experiment tells us that this final intensity is $$\dfrac{I}{8}$$. Therefore,

$$\dfrac{I}{2}\cos^{4}\theta=\dfrac{I}{8}$$

Cancelling the common factor $$\dfrac{I}{2}$$ from both sides gives

$$\cos^{4}\theta=\dfrac{1}{4}$$

Taking the square root of both sides,

$$\cos^{2}\theta=\dfrac{1}{2}$$

Again taking the square root, and noting that the cosine of a small positive angle is positive,

$$\cos\theta=\dfrac{1}{\sqrt{2}}$$

This corresponds to

$$\theta=45^{\circ}$$

Hence, the correct answer is Option D.

We are told that the angular width of the central maximum in the single-slit diffraction pattern is $$60^\circ$$. By definition, the angular width of the central maximum equals the angle between the first minima on either side of the central bright band. Hence if we denote the angle of the first minimum (on one side) by $$\theta_1$$, we have

$$2\theta_1 = 60^\circ \Longrightarrow \theta_1 = 30^\circ.$$

For a single slit of width $$a$$ illuminated by monochromatic light of wavelength $$\lambda$$, the condition for the first diffraction minimum is stated by the formula

$$a \sin \theta_1 = m\lambda, \quad m = 1.$$

Substituting $$m = 1$$, $$a = 1\;\mu\text{m} = 1 \times 10^{-6}\,\text{m}$$ and $$\theta_1 = 30^\circ$$, we get

$$\lambda = a \sin \theta_1 = 1 \times 10^{-6}\,\text{m}\; \sin 30^\circ = 1 \times 10^{-6}\,\text{m}\; \times \frac12 = 0.5 \times 10^{-6}\,\text{m} = 5 \times 10^{-7}\,\text{m}.$$

Thus the wavelength of the monochromatic light is $$\lambda = 5 \times 10^{-7}\,\text{m} \, (500\;\text{nm}).$$

Now a second identical slit is made, producing Young’s double-slit interference. The distance between the slits is denoted by $$d$$ and the screen is placed at a distance $$D = 50\;\text{cm} = 0.5\;\text{m}$$. In Young’s experiment the fringe width (distance between successive bright or dark fringes) is given by the formula

$$\beta = \frac{\lambda D}{d}.$$

The observed fringe width is $$\beta = 1\;\text{cm} = 0.01\;\text{m}$$. Substituting the known values into the formula, we have

$$0.01 = \frac{(5 \times 10^{-7}) \times 0.5}{d}.$$

First multiply the numerator:

$$(5 \times 10^{-7}) \times 0.5 = 2.5 \times 10^{-7}.$$

Now solve for $$d$$:

$$d = \frac{2.5 \times 10^{-7}}{0.01} = 2.5 \times 10^{-7} \times \frac{1}{10^{-2}} = 2.5 \times 10^{-7} \times 10^{2} = 2.5 \times 10^{-5}\,\text{m}.$$

Convert metres to micrometres (since $$1\;\mu\text{m} = 10^{-6}\,\text{m}$$):

$$d = 2.5 \times 10^{-5}\,\text{m} = \frac{2.5 \times 10^{-5}}{10^{-6}}\;\mu\text{m} = 25\;\mu\text{m}.$$

Hence, the correct answer is Option B.

For diffraction of light by a single slit, the positions of the minima are obtained from the condition

$$b \sin\theta_m = m\lambda,$$

where $$b$$ is the slit width, $$\lambda$$ is the wavelength, $$\theta_m$$ is the angle that the $$m^{\text{th}}$$ minimum makes with the central axis, and $$m = 1,2,3,\ldots$$ labels the order of the minimum.

The screen is placed at a distance $$D$$ from the slit. When the angles are small, we use the small-angle approximations

$$\tan\theta_m \approx \sin\theta_m \approx \theta_m \quad\text{(in radians)},$$

so the transverse distance $$y_m$$ of the $$m^{\text{th}}$$ minimum from the central maximum on the screen is

$$y_m = D\tan\theta_m \;\approx\; D\sin\theta_m.$$

Substituting $$\sin\theta_m = m\lambda/b$$ from the diffraction condition, we get an explicit formula for the position of the minima on the screen:

$$y_m \approx D\left(\frac{m\lambda}{b}\right) = \frac{D\lambda}{b}\,m.$$

This shows that, under the small-angle approximation, the distances of successive minima are directly proportional to the order $$m$$.

We are told that

$$y_2 = 3\ \text{cm}, \qquad y_4 = 6\ \text{cm}.$$

To confirm consistency, notice that

$$\frac{y_4}{y_2} = \frac{6\ \text{cm}}{3\ \text{cm}} = 2 = \frac{4}{2},$$

which matches the proportionality predicted by the formula, so the data are self-consistent.

From the expression $$y_m = (D\lambda/b)\,m,$$ solve for the constant factor $$D\lambda/b$$ using the second-order minimum (any order would work):

$$\frac{D\lambda}{b} = \frac{y_2}{2} = \frac{3\ \text{cm}}{2} = 1.5\ \text{cm}.$$

Now, the first-order minima (which bound the central maximum) correspond to $$m = 1$$. Their distances from the central axis are therefore

$$y_1 = \frac{D\lambda}{b}\, (1) = 1.5\ \text{cm}.$$

The central bright fringe extends from the first minimum on one side to the first minimum on the other side, so its full width is twice $$y_1$$:

$$\text{Width of central maximum} = 2y_1 = 2 \times 1.5\ \text{cm} = 3.0\ \text{cm}.$$

Hence, the correct answer is Option D.

We have a single-slit diffraction problem. For a slit of width $$a$$ that is illuminated by monochromatic light of wavelength $$\lambda$$, the condition for the dark (minimum-intensity) bands on either side of the central bright band is first stated as

$$a \sin\theta = m\lambda,$$

where $$m = 1,2,3,\ldots$$ denotes the order of the dark band and $$\theta$$ is the angle the corresponding ray makes with the original direction of the beam.

Because the screen is much farther from the slit than the separation of the fringes, the angles are very small, so we use the small-angle approximations $$\sin\theta \approx \tan\theta \approx \theta.$$ If the distance from the slit to the screen is $$D$$ and the distance of the $$m^{\text{th}}$$ dark band from the central bright band measured along the screen is $$y_m,$$ then $$\tan\theta = y_m/D.$$ Substituting $$\tan\theta$$ for $$\sin\theta$$ in the condition gives

$$a\,\frac{y_m}{D} = m\lambda.$$

Solving for $$y_m$$ we get the working formula

$$y_m = \frac{m\lambda D}{a}.$$

Now we insert the numerical data step by step. The width of the slit is given as $$0.1\ \text{mm}$$, so

$$a = 0.1\ \text{mm} = 0.1 \times 10^{-3}\ \text{m} = 1.0 \times 10^{-4}\ \text{m}.$$

The wavelength of light is $$6000\ \unicode{x212B}$$; recalling that $$1\ \unicode{x212B} = 10^{-10}\ \text{m},$$ we write

$$\lambda = 6000 \times 10^{-10}\ \text{m} = 6.0 \times 10^{-7}\ \text{m}.$$

The screen distance is given as $$D = 0.5\ \text{m}.$$ We need the third dark band, so $$m = 3.$$ Substituting all these values into the expression for $$y_m$$ gives

$$\begin{aligned} y_3 &= \frac{m\lambda D}{a} \\ &= \frac{3 \,\bigl(6.0 \times 10^{-7}\ \text{m}\bigr)\,(0.5\ \text{m})}{1.0 \times 10^{-4}\ \text{m}}. \end{aligned}$$

First multiply the wavelength by the screen distance:

$$6.0 \times 10^{-7}\ \text{m} \times 0.5 = 3.0 \times 10^{-7}\ \text{m}.$$

Next multiply by the order $$m = 3$$:

$$3 \times 3.0 \times 10^{-7}\ \text{m} = 9.0 \times 10^{-7}\ \text{m}.$$

Now divide by the slit width:

$$\frac{9.0 \times 10^{-7}\ \text{m}}{1.0 \times 10^{-4}\ \text{m}} = 9.0 \times 10^{-3}\ \text{m}.$$

Finally convert metres to millimetres, remembering that $$1\ \text{m} = 1000\ \text{mm}:$$

$$9.0 \times 10^{-3}\ \text{m} = 9.0 \times 10^{-3} \times 1000\ \text{mm} = 9.0\ \text{mm}.$$

So, the third dark band lies $$9\ \text{mm}$$ away from the central bright band.

Hence, the correct answer is Option A.

In a Young’s double slit experiment the distance between the slits is given as $$d = 0.5\;\text{mm} = 0.5 \times 10^{-3}\;\text{m}$$ and the distance of the screen from the slits is $$D = 150\;\text{cm} = 1.5\;\text{m}.$$ The two wavelengths in the incident light are $$\lambda_1 = 650\;\text{nm} = 650 \times 10^{-9}\;\text{m}$$ and $$\lambda_2 = 520\;\text{nm} = 520 \times 10^{-9}\;\text{m}.$$

For any wavelength, the position $$y$$ of the $$m^{\text{th}}$$ bright (constructive) fringe from the central maximum is obtained from the standard fringe-width relation

$$ y = m \,\frac{\lambda D}{d}. $$

A point on the screen will be simultaneously bright for both wavelengths if their bright fringes coincide there. If the orders of the two coincident bright fringes are $$m_1$$ (for $$\lambda_1$$) and $$m_2$$ (for $$\lambda_2$$), we must have the same geometrical position $$y$$ for both:

$$ m_1 \,\frac{\lambda_1 D}{d} \;=\; m_2 \,\frac{\lambda_2 D}{d}. $$

The common factors $$D$$ and $$d$$ cancel, leaving

$$ m_1 \,\lambda_1 \;=\; m_2 \,\lambda_2. $$

Substituting the two wavelengths,

$$ m_1 \,(650 \times 10^{-9}) \;=\; m_2 \,(520 \times 10^{-9}). $$

Dividing both sides by $$130 \times 10^{-9}$$ simplifies the numerical coefficients:

$$ m_1 \times 5 \;=\; m_2 \times 4. $$

So the ratio of the orders must satisfy

$$ \frac{m_1}{m_2} = \frac{4}{5}. $$

The smallest positive integers with this ratio are therefore $$m_1 = 4$$ and $$m_2 = 5.$$ Taking any higher multiple would move the coincidence farther out, so this choice gives the least (nearest) distance we seek.

Now we calculate that distance using $$m_1 = 4$$ with the wavelength $$\lambda_1 = 650 \times 10^{-9}\;\text{m}:$$

$$ y_{\text{least}} = m_1 \,\frac{\lambda_1 D}{d} = 4 \times \frac{650 \times 10^{-9}\,\text{m} \times 1.5\;\text{m}}{0.5 \times 10^{-3}\;\text{m}}. $$

First multiply in the numerator:

$$ 4 \times 650 = 2600, \qquad 2600 \times 10^{-9}\,\text{m} = 2.6 \times 10^{-6}\,\text{m}. $$

Then include the factor $$1.5$$ from $$D$$:

$$ 2.6 \times 10^{-6}\,\text{m} \times 1.5 = 3.9 \times 10^{-6}\,\text{m}. $$

Now divide by the denominator $$0.5 \times 10^{-3}\,\text{m}:$$

$$ y_{\text{least}} = \frac{3.9 \times 10^{-6}}{0.5 \times 10^{-3}} = \frac{3.9}{0.5} \times 10^{-6 + 3} = 7.8 \times 10^{-3}\,\text{m}. $$

Finally convert metres to millimetres (1 mm = $$10^{-3}$$ m):

$$ y_{\text{least}} = 7.8 \;\text{mm}. $$

Hence, the correct answer is Option C.

We begin by analysing the two independent reasons for which the bright spot on the screen of a pin-hole camera acquires a finite size.

First reason (pure geometry). Rays that pass straight through the circular hole of radius $$a$$ reach the screen (placed at a distance $$L$$ behind the hole) without bending. Consequently the geometrical radius of the spot is exactly the same as the radius of the hole. Hence

$$\text{geometrical spread } = a.$$

Second reason (diffraction). Light emerging from the circular aperture behaves like light coming from a circular source; it diffracts and forms an Airy pattern. The linear radius of the central maximum on a distant screen is obtained from the simple Fraunhofer condition for the first minimum

$$\sin\theta \approx \theta = \frac{\lambda}{a},$$

so that for small angles the linear size on a screen a distance $$L$$ away is

$$\text{diffraction spread } = L\,\theta = L\left(\frac{\lambda}{a}\right)=\frac{\lambda L}{a}.$$

The two effects act independently along the same straight line, so the total radius $$b$$ of the spot is taken as the algebraic sum

$$b = a + \frac{\lambda L}{a}.$$

Our task is to find the value of $$a$$ that minimises $$b$$. For this we differentiate $$b$$ with respect to $$a$$ and equate the derivative to zero.

First write the explicit expression again:

$$b(a)=a+\frac{\lambda L}{a}.$$

Now differentiate:

$$\frac{db}{da}=\frac{d}{da}\left(a\right)+\frac{d}{da}\left(\frac{\lambda L}{a}\right)=1-\frac{\lambda L}{a^{2}}.$$

The minimum occurs when

$$\frac{db}{da}=0\quad\Longrightarrow\quad 1-\frac{\lambda L}{a^{2}}=0.$$

Solving the above equation step by step, we obtain

$$\frac{\lambda L}{a^{2}}=1,$$

$$a^{2}=\lambda L,$$

$$\boxed{a=\sqrt{\lambda L}}.$$

Substituting this optimal value of $$a$$ back into the formula for $$b$$ gives the minimum possible radius $$b_{\text{min}}$$:

$$\begin{aligned} b_{\text{min}} &=a+\frac{\lambda L}{a}\\[4pt] &=\sqrt{\lambda L}+\frac{\lambda L}{\sqrt{\lambda L}}\\[4pt] &=\sqrt{\lambda L}+\sqrt{\lambda L}\\[4pt] &=2\sqrt{\lambda L}\\[4pt] &=\sqrt{4\lambda L}. \end{aligned}$$

Thus the smallest spot is obtained when

$$a = \sqrt{\lambda L}\quad\text{and}\quad b_{\text{min}} = \sqrt{4\lambda L}.$$

These two results coincide exactly with the statements in Option A.

Hence, the correct answer is Option A.

We have monochromatic light of wavelength $$\lambda = 600 \text{ nm}=600\times10^{-9}\,\text{m}$$ falling on a pair of slits separated by a distance $$d$$. The interference pattern is observed on a screen placed at a distance $$L=1\ \text{m}$$, and a person keeps his eye very close to the slits to look at that pattern.

For a double-slit arrangement, the angular position of the $$m^{\text{th}}$$ bright fringe is given, for small angles, by the well-known formula

$$\theta_m \approx \frac{m\lambda}{d}\,.$$

Hence the angular separation between two successive bright fringes is obtained by writing

$$\Delta\theta=\theta_{m+1}-\theta_m =\frac{(m+1)\lambda}{d}-\frac{m\lambda}{d} =\frac{\lambda}{d}\,.$$

The eye can resolve two distinct bright lines only if the angle between them is not smaller than the eye’s minimum resolvable angle. The question tells us that the angular resolution of the eye is

$$\theta_{\min}=\frac{1}{60}^{\circ}\,.$$

First we change this to radians because our interference formula is in radians. We use the relation $$1^{\circ}=\frac{\pi}{180}\ \text{rad}$$, so

$$\theta_{\min}= \frac{1}{60}\times\frac{\pi}{180} =\frac{\pi}{10800}\ \text{rad}\,.$$ Numerically,

$$\theta_{\min}= \frac{3.1416}{10800}\approx2.91\times10^{-4}\ \text{rad}\,.$$

As the slit separation $$d$$ is gradually increased, the angular fringe spacing $$\Delta\theta=\frac{\lambda}{d}$$ steadily decreases. The fringes will cease to be distinguishable exactly when the separation between adjacent bright fringes becomes equal to the eye’s limiting angle, that is when

$$\Delta\theta=\theta_{\min}\; \Longrightarrow\; \frac{\lambda}{d_0}= \theta_{\min}\,.$$ Therefore

$$d_0=\frac{\lambda}{\theta_{\min}} =\frac{600\times10^{-9}} {\dfrac{\pi}{10800}} =600\times10^{-9}\times\frac{10800}{\pi}\,.$$ Carrying out the multiplications,

$$600\times10800=6\,480\,000, \quad 6\,480\,000\times10^{-9}=6.48\times10^{-3}\,,$$ so

$$d_0=\frac{6.48\times10^{-3}}{3.1416} \approx2.06\times10^{-3}\,\text{m} =2.06\ \text{mm}\,.$$

This value is closest to $$2\ \text{mm}$$ among the given options.

Hence, the correct answer is Option C.

The two stars will appear as two distinct objects to the eye only if the telescope can just resolve them. For a telescope with a circular aperture, the condition for just-resolution is given by Rayleigh’s criterion, which states:

$$\theta_{\text{min}} \;=\;1.22 \,\frac{\lambda}{D}$$

Here $$\theta_{\text{min}}$$ is the smallest angular separation (in radians) that the telescope can resolve, $$\lambda$$ is the wavelength of the light being observed, and $$D$$ is the diameter of the objective.

We have the numerical values

$$\lambda = 600 \text{ nm} = 600 \times 10^{-9}\,\text{m},$$

$$D = 30 \text{ cm} = 0.30\,\text{m}.$$

Substituting these into Rayleigh’s formula, we obtain

$$ \theta_{\text{min}} =1.22 \,\frac{600 \times 10^{-9}}{0.30} =1.22 \times \left( \frac{600}{0.30}\right)\times 10^{-9} =1.22 \times 2000 \times 10^{-9} =2440 \times 10^{-9} =2.44 \times 10^{-6}\,\text{rad}. $$

The linear distance $$s$$ between the two stars that corresponds to this angular separation is related by the simple geometry formula

$$s = \theta_{\text{min}}\; L,$$

where $$L$$ is the distance of the stars from the earth.

Each star is 10 light-years away, so

$$ L = 10 \times (1\;\text{light year}) = 10 \times 9.46 \times 10^{15}\,\text{m} = 9.46 \times 10^{16}\,\text{m}. $$

Now we multiply:

$$ s = (2.44 \times 10^{-6}) \times (9.46 \times 10^{16}) = (2.44 \times 9.46) \times 10^{\,(-6 + 16)} = 23.0824 \times 10^{10}\,\text{m}. $$

Shifting the decimal place to get a single significant figure before the power of ten

$$ s = 2.30824 \times 10^{11}\,\text{m}. $$

To express this in kilometres, we divide by $$1000$$:

$$ s = \frac{2.30824 \times 10^{11}}{10^{3}} = 2.30824 \times 10^{8}\,\text{km}. $$

This is of the order of $$10^{8}\,\text{km}.$$

Hence, the correct answer is Option A.

The uncertainty in the y-position ($$\Delta y$$) is approximately equal to the width of the slit, $$\Delta y \approx d$$

According to Heisenberg’s Uncertainty Principle, $$\Delta y \cdot \Delta p_y \ge \frac{h}{4\pi}$$. This implies that by restricting the position, an uncertainty in the y-momentum ($$\Delta p_y$$) is inevitably introduced.

$$d \sin \theta = \lambda$$, where $$\lambda$$ is the de Broglie wavelength ($$\lambda = \frac{h}{p}$$)

The y-component of momentum for an electron deflected at an angle $$\theta$$ is $$p_y = p \sin \theta$$

$$p_y = p \left( \frac{\lambda}{d} \right)$$

$$p_y = \frac{h}{d} \implies p_y d = h$$

The relationship $$p_y d = h$$ corresponds to the electrons reaching the first minimum of the diffraction pattern. However, the question asks for the majority of electrons. 

The majority of electrons fall within the central maximum of the diffraction pattern. For these electrons, the angle of deviation is less than the angle of the first minimum. Consequently, their y-momentum magnitude $$|p_y|$$ is less than $$\frac{h}{d}$$.

Therefore, $$|p_y| < \frac{h}{d}$$

$$|p_y| d < h$$

We begin with the standard expression for the intensity distribution on the screen in a Young’s double slit experiment. If the two individual slits, kept in phase, each produce an intensity $$I_0,$$ then at a point situated a distance $$y$$ from the central line on the screen the resultant intensity is given by the interference formula

$$I \;=\; 4I_0\cos^2\!\theta,$$

where the phase angle $$\theta$$ is related to the geometry by

$$\theta \;=\;\dfrac{\pi\,d\,y}{\lambda D}.$$

Here $$d$$ is the slit separation, $$D$$ the distance of the screen from the slits, and $$\lambda$$ the wavelength of light. The factor $$4I_0$$ represents the maximum possible intensity when the two waves interfere constructively in perfect phase.

For the central bright fringe, the path difference is zero, so $$\theta = 0$$ and we have

$$I_{\text{max}} \;=\; 4I_0\cos^2 0 \;=\; 4I_0.$$

We are asked to locate the point (on either side of the central maximum) where the intensity falls to half of this maximum value, that is

$$I \;=\;\dfrac{I_{\text{max}}}{2} \;=\;\dfrac{4I_0}{2} \;=\; 2I_0.$$

Substituting $$I = 4I_0\cos^2\!\theta$$ into the above condition gives

$$4I_0\cos^2\!\theta \;=\; 2I_0.$$

Dividing both sides by $$2I_0$$ we get

$$2\cos^2\!\theta \;=\; 1 \;\Longrightarrow\; \cos^2\!\theta \;=\;\dfrac{1}{2}.$$

Taking the square root (and remembering that the cosine function is even) yields

$$\cos\!\theta \;=\;\dfrac{1}{\sqrt{2}}.$$

Since $$\cos 45^{\circ} = 1/\sqrt{2},$$ the smallest positive value of $$\theta$$ satisfying the above is

$$\theta \;=\;\dfrac{\pi}{4}.$$

Now we insert this value back into the geometrical relation for $$\theta$$:

$$\dfrac{\pi\,d\,y}{\lambda D} \;=\;\dfrac{\pi}{4}.$$

We cancel $$\pi$$ from both sides to obtain

$$\dfrac{d\,y}{\lambda D} \;=\;\dfrac{1}{4}.$$

Multiplying through by $$\dfrac{\lambda D}{d}$$ gives the required distance $$y$$ of the half-intensity point from the central maximum:

$$y \;=\;\dfrac{\lambda D}{4d}.$$

Next we recall the definition of the fringe width $$\beta,$$ which is the separation between two successive bright (or dark) fringes. For Young’s experiment it is

$$\beta \;=\;\dfrac{\lambda D}{d}.$$

Substituting this value for $$\dfrac{\lambda D}{d}$$ into the expression for $$y$$ we get

$$y \;=\;\dfrac{\beta}{4}.$$

This distance $$y = \beta/4$$ is measured from the central maximum to the position where the intensity has dropped to one-half of the maximum on either side. Therefore the required distance is $$\dfrac{\beta}{4}.$$

Hence, the correct answer is Option A.

We begin by recalling the Huygens-Fresnel principle. It tells us that every point of a wave-front is the source of secondary spherical wavelets, and the new wave-front at a later instant is the common tangent (envelope) to all those wavelets. In a medium where the speed of light changes from point to point, the radii of the successive secondary wavelets are different at different points, so the envelope is no longer a straight line and the ray (which is always drawn perpendicular to the wave-front) bends.

Let us translate the verbal description of the atmosphere given in the question into a refractive-index function:

Near the ground the refractive index is minimum and as we go to a height $$y$$ above the ground the refractive index $$n$$ increases. Symbolically, we may write

$$\frac{dn}{dy}>0.$$

The speed of light in any medium is related to the refractive index by the simple formula

$$v=\frac{c}{n},$$

where $$c$$ is the speed of light in vacuum. So if $$n$$ increases with $$y$$, the speed $$v$$ decreases with $$y$$, i.e.

$$\frac{dv}{dy}<0.$$

Now imagine the initial wave-front to be vertical because the beam is launched perfectly horizontal. Consider two neighbouring points on that wave-front: point $$A$$ at a small height $$y$$ and point $$B$$ infinitesimally lower, at height $$y-\Delta y$$. Because $$n(y)>n(y-\Delta y)$$, the speed at $$A$$, which is $$v(y)=c/n(y),$$ is smaller than the speed at $$B$$,

$$v(y)<v(y-\Delta y).$$

In an infinitesimal time interval $$\Delta t$$ the secondary wavelet from $$B$$ therefore advances a greater horizontal distance than the one from $$A$$. Geometrically the top portion of the wave-front lags behind the bottom portion, so the wave-front tilts such that its upper end stays behind. Because the ray is everywhere perpendicular to the wave-front, the ray tilts so that it starts pointing slightly upward.

We can make this more algebraic by invoking the differential form of Snell’s law for a stratified medium. Snell’s law in its usual form for two media is

$$n_1\sin\theta_1 = n_2\sin\theta_2.$$

For a medium whose refractive index varies continuously with height, the product $$n\sin\theta$$ remains constant along the ray. Hence, writing the constant as $$K$$, we have for every position $$y$$,

$$n(y)\sin\theta(y)=K.$$

Differentiating with respect to $$y$$,

$$\frac{d}{dy}\left[n(y)\sin\theta(y)\right]=0.$$

Using the product rule,

$$\sin\theta\,\frac{dn}{dy}+n\cos\theta\,\frac{d\theta}{dy}=0.$$

Solving for $$\dfrac{d\theta}{dy}$$ we get

$$\frac{d\theta}{dy}=-\frac{\sin\theta}{n\cos\theta}\,\frac{dn}{dy}.$$ $$\displaystyle\Longrightarrow\quad\frac{d\theta}{dy}=-\tan\theta\;\frac{1}{n}\,\frac{dn}{dy}.$$

We already established that $$\dfrac{dn}{dy}>0$$; $$n$$ itself is positive, and for a near-horizontal ray the initial $$\theta$$ (measured from the normal) is less than $$90^{\circ}$$, so $$\tan\theta>0$$. Therefore

$$\frac{d\theta}{dy}<0.$$

A negative derivative means that $$\theta$$ decreases as we move upward, i.e. the ray turns toward the higher layer (upward). So the continuously varying atmosphere behaves like a prism that bends the initially horizontal beam upward.

Hence, the Huygens’ principle together with the given variation of refractive index leads us to conclude that the light beam bends upward.

Hence, the correct answer is Option A.

For the eye, the minimum angular separation that can be distinguished is governed by the Rayleigh criterion for a circular aperture. We first state the formula:

$$\theta_{\min}=1.22\,\frac{\lambda}{D}$$

Here $$\theta_{\min}$$ is the least resolvable angle (in radians), $$\lambda$$ is the wavelength of light and $$D$$ is the diameter of the aperture, which in this case is the pupil.

We are given the radius of the pupil as $$r=0.25\ \text{cm}$$. Converting this into metres, we write

$$r=0.25\ \text{cm}=0.25\times10^{-2}\ \text{m}=2.5\times10^{-3}\ \text{m}.$$

Hence the diameter of the pupil is twice the radius:

$$D = 2r = 2\left(2.5\times10^{-3}\ \text{m}\right)=5.0\times10^{-3}\ \text{m}.$$

The wavelength of the light used is $$\lambda = 500\ \text{nm}$$. Converting to metres,

$$\lambda = 500\ \text{nm}=500\times10^{-9}\ \text{m}=5.0\times10^{-7}\ \text{m}.$$

Now we substitute these values into the Rayleigh formula:

$$\theta_{\min}=1.22\,\frac{\lambda}{D}=1.22\,\frac{5.0\times10^{-7}\ \text{m}}{5.0\times10^{-3}\ \text{m}}.$$

Dividing the powers of ten first, we have

$$\frac{5.0\times10^{-7}}{5.0\times10^{-3}}=\frac{5.0}{5.0}\times10^{-7-(-3)}=1\times10^{-4}=10^{-4}.$$

Hence

$$\theta_{\min}=1.22\times10^{-4}\ \text{radian}.$$

The problem finally asks for the linear or spatial separation of two point objects that can just be resolved when they are at the comfortable viewing distance of the eye, given here as $$d=25\ \text{cm}$$. Converting the distance into metres, we get

$$d=25\ \text{cm}=25\times10^{-2}\ \text{m}=0.25\ \text{m}.$$

The small-angle approximation relates the minimum separation $$s_{\min}$$ on the object to the minimum angle by

$$s_{\min} = d\,\theta_{\min}.$$

Substituting $$d=0.25\ \text{m}$$ and $$\theta_{\min}=1.22\times10^{-4}\ \text{rad}$$, we find

$$s_{\min}=0.25\ \text{m}\times1.22\times10^{-4}=0.305\times10^{-4}\ \text{m}.$$

Multiplying,

$$0.305\times10^{-4}\ \text{m}=3.05\times10^{-5}\ \text{m}.$$

To express this in micrometres (1 μm = 10⁻⁶ m):

$$3.05\times10^{-5}\ \text{m}=3.05\times10^{-5}\ \text{m}\times\frac{10^{6}\ \mu\text{m}}{1\ \text{m}}=30.5\ \mu\text{m}.$$

This is approximately $$30\ \mu\text{m}.$$ Comparing with the given options, this corresponds to option C.

Hence, the correct answer is Option C.

When unpolarized light is incident on any surface, we always begin by remembering that it can be thought of as an equal mixture of two mutually perpendicular linear polarisations. Hence, for an incident intensity $$I_0$$, we may write

$$I_{\parallel}^{\;(inc)}=\dfrac{I_0}{2},\qquad I_{\perp}^{\;(inc)}=\dfrac{I_0}{2}$$

where the $$\parallel$$ component has its electric field vector in the plane of incidence and the $$\perp$$ component has its electric field vector perpendicular to that plane.

Now, we recall Brewster’s law. It states that for a ray travelling from a medium of refractive index $$n_1$$ to one of refractive index $$n_2$$, there exists an angle of incidence $$\theta_B$$, called Brewster’s angle, for which the reflected component of the $$\parallel$$ polarisation vanishes. In symbols, Fresnel’s reflection coefficient for the parallel component is

$$r_{\parallel}=0\quad\text{when}\quad\theta_i=\theta_B.$$

So, at Brewster’s angle,

$$I_{\parallel}^{\;(refl)} = 0.$$

However, for the perpendicular component, the corresponding reflection coefficient $$r_{\perp}$$ is not zero. Its magnitude is less than one, so a certain fraction of the $$\perp$$ component is reflected while the rest is transmitted. Hence

$$I_{\perp}^{\;(refl)} = R_{\perp}\left(\dfrac{I_0}{2}\right),\qquad 0 < R_{\perp} < 1,$$

where $$R_{\perp}=|r_{\perp}|^{\,2}$$ is the reflectance for the perpendicular polarisation. Combining the two reflected intensities, we have

$$I_{\text{reflected}} = I_{\parallel}^{\;(refl)} + I_{\perp}^{\;(refl)} = 0 + R_{\perp}\left(\dfrac{I_0}{2}\right) = \dfrac{R_{\perp}I_0}{2}.$$

Because $$0<R_{\perp}<1,$$ the numerical value of $$\dfrac{R_{\perp}I_0}{2}$$ is strictly less than $$\dfrac{I_0}{2}.$$ Very importantly, the entire reflected beam now contains only the $$\perp$$ polarisation, i.e. only one linear polarisation. Therefore the reflected light is completely plane-polarised.

Putting these two conclusions together:

• The reflected beam is completely (100%) polarised.
• Its intensity is a fraction $$R_{\perp}/2$$ of $$I_0$$, which is always less than $$I_0/2$$.

So the statement that matches these facts is: “reflected light is completely polarized with intensity less than $$\dfrac{I_0}{2}$$.” This is exactly Option D.

Hence, the correct answer is Option D.

In Young's double-slit experiment, we have two identical slits, and the distance between the centers of these slits is denoted by $$d$$. The width of each slit is denoted by $$a$$. According to the problem, the distance between the slits is 6.1 times larger than the slit width, so we can write:

We need to find the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern. The central maximum of the single-slit diffraction pattern is the region between the first minima on either side of the central peak. For a single slit of width $$a$$, the first minimum occurs at an angle $$\theta$$ where:

Here, $$\lambda$$ is the wavelength of light. Therefore, the angular positions of the first minima are at $$\sin \theta = \pm \frac{\lambda}{a}$$. The central maximum extends from $$\sin \theta = -\frac{\lambda}{a}$$ to $$\sin \theta = \frac{\lambda}{a}$$.

In the double-slit experiment, the interference maxima occur at angles where the path difference is an integer multiple of the wavelength. The condition for the $$m$$-th interference maximum is:

where $$m$$ is an integer (the order of the maximum). To find which interference maxima lie within the central maximum of the diffraction pattern, we require that $$\sin \theta$$ satisfies:

Substituting $$\sin \theta = \frac{m \lambda}{d}$$ from the interference condition, we get:

Since $$\lambda > 0$$, we can divide all parts of the inequality by $$\lambda$$:

Multiplying through by $$d$$:

Given that $$d = 6.1a$$, we substitute $$\frac{d}{a} = 6.1$$:

Since $$m$$ must be an integer, the possible values of $$m$$ are:

This gives 13 values: from $$-6$$ to $$6$$ inclusive.

However, the central interference maximum at $$m = 0$$ is the brightest and is considered separately in some contexts. The problem asks for the number of intensity maxima within the central maximum of the single-slit diffraction pattern, and in many interpretations, this excludes the central interference maximum ($$m = 0$$) but includes all other maxima inside the central diffraction envelope. Therefore, we exclude $$m = 0$$ and count the remaining maxima.

The values excluding $$m = 0$$ are:

This gives 12 maxima: six on the left side ($$m = -6$$ to $$m = -1$$) and six on the right side ($$m = 1$$ to $$m = 6$$).

It is important to verify that these maxima are visible and not suppressed by diffraction minima. The diffraction minima occur at $$a \sin \theta = n \lambda$$ for $$n = \pm 1, \pm 2, \pm 3, \ldots$$. The interference maxima at $$d \sin \theta = m \lambda$$ coincide with diffraction minima when $$\frac{d}{a} = \frac{m}{n}$$, which is $$\frac{6.1}{1} = \frac{61}{10}$$. For $$m$$ and $$n$$ integers, this would require $$m$$ to be a multiple of 61 and $$n$$ a multiple of 10. Within the range $$|m| \leq 6$$, the only possible $$m$$ is 0 (when $$n = 0$$), but $$n = 0$$ corresponds to the central maximum, not a minimum. Thus, no interference maximum in this range coincides with a diffraction minimum, and all 12 maxima are visible within the central diffraction envelope.

Therefore, the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern, excluding the central interference maximum, is 12.

Hence, the correct answer is Option D.

The direct path traveled by the first ray is $$x_1 = SP = 2l$$

For the second ray reflected from surface $$R$$, let the point of reflection be $$O$$. $$SO = OP = \frac{l}{\cos 30^\circ} = \frac{2l}{\sqrt{3}}$$

The total path length of the reflected ray is $$x_2 = SO + OP = \frac{4l}{\sqrt{3}}$$

The geometric path difference between the two rays is $$\Delta x_{geom} = x_2 - x_1 = \frac{4l}{\sqrt{3}} - 2l = 2l \left( \frac{2 - \sqrt{3}}{\sqrt{3}} \right)$$

Reflection from the perfect reflecting surface $$R$$ (denser medium) introduces a phase change of $$\pi$$, which is equivalent to an additional path difference of $$\frac{\lambda}{2}$$. The net path difference is $$\Delta x = \Delta x_{geom} + \frac{\lambda}{2}$$

$$\Delta x = 2l \left( \frac{2 - \sqrt{3}}{\sqrt{3}} \right) + \frac{\lambda}{2}$$

For constructive interference (maxima) at point $$P$$: $$\Delta x = n\lambda$$

$$2l \left( \frac{2 - \sqrt{3}}{\sqrt{3}} \right) + \frac{\lambda}{2} = n\lambda$$

$$2l \left( \frac{2 - \sqrt{3}}{\sqrt{3}} \right) = \left(n - \frac{1}{2}\right)\lambda = \frac{(2n - 1)\lambda}{2}$$

$$l = \frac{(2n - 1)\lambda \sqrt{3}}{4(2 - \sqrt{3})}$$

The standard formula for fringe width is $$\beta = \frac{\lambda D}{d}$$.

From Case 1: The initial configuration has a screen distance $$D$$ and slit separation $$d$$. $$\beta = \frac{\lambda D}{d}$$

From Case 3: The screen distance is doubled to $$2D$$, while the slit separation remains $$d$$. The new fringe width is $$\beta' = \frac{\lambda (2D)}{d} = 2\beta$$

From Case 2: When a transparent mica sheet of thickness $$t$$ and refractive index $$\mu$$ is introduced in front of one of the slits, the entire fringe pattern shifts. The distance $$y$$ that the central fringe shifts is given by the formula: 

$$y = \frac{D}{d}(\mu - 1)t$$

We are given the condition: $$y = \beta'$$

$$\frac{D}{d}(\mu - 1)t = \frac{\lambda (2D)}{d}$$

$$(\mu - 1)t = 2\lambda$$

$$(1.6 - 1) \times (1.8 \times 10^{-6}) = 2\lambda$$

$$0.6 \times 1.8 \times 10^{-6} = 2\lambda$$

$$1.08 \times 10^{-6} = 2\lambda$$

$$\lambda = \frac{1.08 \times 10^{-6}}{2} = 0.54 \times 10^{-6}\text{ m}$$

$$\lambda = 540 \times 10^{-9}\text{ m} = 540\text{ nm}$$

In a single slit diffraction experiment, the position of minima and maxima are determined by specific conditions. For the first minimum of red light, the condition is given by:

$$ a \sin \theta = n \lambda $$

where $$ a $$ is the slit width, $$ \theta $$ is the angle from the central maximum, $$ \lambda $$ is the wavelength, and $$ n $$ is the order of the minimum. For the first minimum, $$ n = 1 $$. Given the wavelength of red light $$ \lambda_r = 6600 \text{Å} $$, we have:

$$ a \sin \theta = 1 \cdot 6600 = 6600 \text{Å} \quad \text{(Equation 1)} $$

For the first maximum of another light with wavelength $$ \lambda $$, the condition is approximately:

$$ a \sin \theta = \left( m + \frac{1}{2} \right) \lambda $$

where $$ m $$ is the order of the maximum. For the first maximum, $$ m = 1 $$, so:

$$ a \sin \theta = \left( 1 + \frac{1}{2} \right) \lambda = \frac{3}{2} \lambda \quad \text{(Equation 2)} $$

The problem states that the first minimum for red light coincides with the first maximum of the other light. This means the angular position $$ \theta $$ is the same for both, so $$ \sin \theta $$ is identical. Therefore, the expressions for $$ a \sin \theta $$ from Equation 1 and Equation 2 must be equal:

$$ 6600 = \frac{3}{2} \lambda $$

Solving for $$ \lambda $$:

$$ \lambda = 6600 \times \frac{2}{3} $$

Performing the multiplication:

$$ \lambda = \frac{6600 \times 2}{3} = \frac{13200}{3} = 4400 \text{Å} $$

Thus, the wavelength of the other light for which the first maximum coincides with the first minimum of red light is 4400 Å.

Comparing with the options:

A. 3300 Å

B. 4400 Å

C. 5500 Å

D. 6600 Å

Hence, the correct answer is Option B.

We recall Malus’s law, which states that when plane-polarized light of incident intensity $$I_0$$ meets a polaroid whose transmission axis makes an angle $$\theta$$ with the light’s plane of polarization, the emerging intensity is

$$I = I_0 \cos^2\theta.$$

Let the transmission axis of the polaroid be initially aligned with the plane of polarization of beam A. Thus, for beam A the angle is $$\theta_A = 0^\circ$$, while for beam B, whose polarization is perpendicular to that of A, the angle is $$\theta_B = 90^\circ.$$

Applying Malus’s law to the initial position, we have

$$I_{A\,(\text{through})} = I_A \cos^2 0^\circ = I_A \times 1 = I_A,$$

$$I_{B\,(\text{through})} = I_B \cos^2 90^\circ = I_B \times 0 = 0.$$

This matches the given condition: beam A is brightest and beam B is extinguished.

Now the polaroid is rotated through $$30^\circ$$. The planes of polarization of the two beams stay fixed in space; only the axis of the polaroid turns. Consequently,

• For beam A, the new angle with the axis is $$\theta_A' = 30^\circ.$$
• For beam B, originally at $$90^\circ$$ from A, the new angle is $$\theta_B' = 90^\circ - 30^\circ = 60^\circ.$$

Using Malus’s law again, the transmitted intensities become

$$I_{A}' = I_A \cos^2 30^\circ,$$

$$I_{B}' = I_B \cos^2 60^\circ.$$

According to the statement, the two beams now appear equally bright, so

$$I_{A}' = I_{B}'.$$

Substituting the expressions,

$$I_A \cos^2 30^\circ = I_B \cos^2 60^\circ.$$

We insert the numerical values $$\cos 30^\circ = \frac{\sqrt3}{2}$$ and $$\cos 60^\circ = \frac12$$:

$$I_A\left(\frac{\sqrt3}{2}\right)^2 = I_B\left(\frac12\right)^2,$$

$$I_A \left(\frac{3}{4}\right) = I_B \left(\frac{1}{4}\right).$$

Dividing both sides by $$\dfrac14$$ gives

$$3I_A = I_B.$$

Rearranging, we find

$$\frac{I_A}{I_B} = \frac13.$$

Hence, the correct answer is Option D.

Two monochromatic light beams with intensities $$I_1 = 16$$ units and $$I_2 = 9$$ units interfere. We need to find the ratio of the intensity of the bright parts to the intensity of the dark parts in the resultant interference pattern.

For interference of two coherent monochromatic light beams, the intensity at any point is given by:

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$$

where $$\phi$$ is the phase difference between the two beams.

The maximum intensity (bright parts) occurs when $$\cos \phi = 1$$, so:

$$I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2}$$

The minimum intensity (dark parts) occurs when $$\cos \phi = -1$$, so:

$$I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2}$$

We are asked for the ratio $$ \frac{I_{\text{max}}}{I_{\text{min}}} $$.

First, calculate $$\sqrt{I_1 I_2}$$:

$$I_1 = 16, \quad I_2 = 9$$

$$\sqrt{I_1 I_2} = \sqrt{16 \times 9} = \sqrt{144} = 12$$

Now substitute into the formula for $$I_{\text{max}}$$:

$$I_{\text{max}} = 16 + 9 + 2 \times 12 = 25 + 24 = 49$$

Next, substitute into the formula for $$I_{\text{min}}$$:

$$I_{\text{min}} = 16 + 9 - 2 \times 12 = 25 - 24 = 1$$

Therefore, the ratio is:

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{49}{1} = 49$$

Now, comparing with the options:

A. $$\frac{16}{9}$$

B. $$\frac{4}{3}$$

C. $$\frac{7}{1}$$

D. $$\frac{49}{1}$$

The ratio $$\frac{49}{1}$$ matches option D.

Hence, the correct answer is Option D.

In Young's double slit experiment, without any glass plate, at a point equidistant from both slits, the path difference is zero. This results in constructive interference, and the intensity is maximum. Let the intensity from each slit be $$ I_0 $$. The resultant intensity is given by:

$$ I_{\text{before}} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(0^\circ) = 2I_0 + 2I_0 \times 1 = 4I_0 $$

Now, a glass plate of thickness $$ t = \frac{2500}{3} \lambda $$ and refractive index $$ \mu = 1.5 $$ is inserted in front of one slit. This introduces an additional path difference for the light passing through that slit. The additional path difference $$ \Delta x $$ is calculated as:

$$ \Delta x = (\mu - 1) t = (1.5 - 1) \times \frac{2500}{3} \lambda = 0.5 \times \frac{2500}{3} \lambda = \frac{1}{2} \times \frac{2500}{3} \lambda = \frac{2500}{6} \lambda = \frac{1250}{3} \lambda $$

This path difference causes a phase difference $$ \phi $$, given by:

$$ \phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{1250}{3} \lambda = \frac{2\pi \times 1250}{3} = \frac{2500\pi}{3} $$

Since phase difference is periodic with $$ 2\pi $$, we reduce $$ \frac{2500\pi}{3} $$ modulo $$ 2\pi $$:

$$ \frac{2500\pi}{3} \div 2\pi = \frac{2500\pi}{3} \times \frac{1}{2\pi} = \frac{2500}{6} = \frac{1250}{3} $$

Now, $$ \frac{1250}{3} = 416 + \frac{2}{3} $$ (since $$ 416 \times 3 = 1248 $$, remainder $$ 1250 - 1248 = 2 $$). The fractional part is $$ \frac{2}{3} $$, so:

$$ \phi = \frac{2}{3} \times 2\pi = \frac{4\pi}{3} $$

After inserting the glass plate, the amplitudes from both slits remain the same (assuming no absorption), so the intensities are still $$ I_0 $$ for each slit. The resultant intensity is:

$$ I_{\text{after}} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{4\pi}{3}\right) = 2I_0 + 2I_0 \cos\left(\frac{4\pi}{3}\right) $$

Since $$ \cos\left(\frac{4\pi}{3}\right) = \cos(240^\circ) = -\frac{1}{2} $$:

$$ I_{\text{after}} = 2I_0 + 2I_0 \times \left(-\frac{1}{2}\right) = 2I_0 - I_0 = I_0 $$

The ratio of intensities before and after inserting the glass plate is:

$$ \frac{I_{\text{before}}}{I_{\text{after}}} = \frac{4I_0}{I_0} = 4 : 1 $$

Hence, the correct answer is Option C.

In the double-slit interference experiment with two wavelengths, λ₁ = 500 nm and λ₂ = 600 nm, we need to find the path difference Δ where the maximum of one pattern coincides with the minimum of the other, causing indistinct fringes.

For a maximum (bright fringe) at wavelength λ₁, the path difference must satisfy Δ = nλ₁, where n is an integer. For a minimum (dark fringe) at wavelength λ₂, the path difference must satisfy Δ = (m + ½)λ₂, where m is an integer. Setting these equal gives:

$$n \lambda_1 = \left(m + \frac{1}{2}\right) \lambda_2$$

Substituting λ₁ = 500 nm and λ₂ = 600 nm:

$$n \times 500 = \left(m + \frac{1}{2}\right) \times 600$$

Simplify the equation:

$$500n = 600m + 300$$

Divide both sides by 100:

$$5n = 6m + 3$$

Rearrange to:

$$5n - 6m = 3$$

Solve for integer solutions. Express n in terms of m:

$$n = \frac{6m + 3}{5}$$

So, (6m + 3) must be divisible by 5. Thus, 6m + 3 ≡ 0 mod 5. Simplify modulo 5:

$$6 \equiv 1 \pmod{5}, \quad 3 \equiv 3 \pmod{5} \implies 1 \cdot m + 3 \equiv 0 \pmod{5} \implies m \equiv 2 \pmod{5}$$

Therefore, m = 5k + 2 for integer k ≥ 0. Substitute back:

$$n = \frac{6(5k + 2) + 3}{5} = \frac{30k + 12 + 3}{5} = \frac{30k + 15}{5} = 6k + 3$$

The path difference Δ is:

$$\Delta = n \lambda_1 = (6k + 3) \times 500$$

For k = 0 (smallest positive Δ):

$$\Delta = (6 \times 0 + 3) \times 500 = 3 \times 500 = 1500 \text{ nm}$$

Verify with λ₂:

$$\Delta = \left(m + \frac{1}{2}\right) \lambda_2 = \left(5 \times 0 + 2 + \frac{1}{2}\right) \times 600 = \left(2.5\right) \times 600 = 1500 \text{ nm}$$

At Δ = 1500 nm:

• For λ₁ = 500 nm: Δ = 1500 = 3 × 500 → maximum (n=3).
• For λ₂ = 600 nm: Δ = 1500 = 2.5 × 600 = (2 + ½) × 600 → minimum (m=2).

Thus, a maximum of λ₁ coincides with a minimum of λ₂, satisfying the condition. Checking other k values gives larger Δ (e.g., k=1 yields 4500 nm, not in options). Considering the reverse case (maximum of λ₂ and minimum of λ₁) leads to no integer solutions, as shown by:

$$n \lambda_2 = \left(m + \frac{1}{2}\right) \lambda_1 \implies 600n = 500\left(m + \frac{1}{2}\right) \implies 12n = 10m + 5 \implies 12n - 10m = 5$$

The left side is even, right side odd, so no integer solutions exist. Therefore, the only valid solution in the options is Δ = 1500 nm.

Hence, the correct answer is Option D (1500 nm).

We begin with an unpolarised light beam whose initial intensity is denoted by $$I_0$$.

First, the beam passes through polaroid A. For unpolarised light, the standard result is that a single ideal polaroid transmits exactly one-half of the incident intensity. We can state this as:

Formula: “When unpolarised light passes through one perfect polaroid, the transmitted intensity becomes half of the incident intensity.”

Using this, we write $$I_1 = \frac{I_0}{2},$$ where $$I_1$$ is the intensity of the light emerging from polaroid A.

Now this partially polarised light encounters polaroid B. The transmission through a second polaroid is governed by Malus’ Law.

Formula: “If polarised light of intensity $$I$$ is incident on a polaroid whose transmission axis makes an angle $$\theta$$ with the plane of polarisation of that light, the emergent intensity $$I'$$ is given by $$I' = I \cos^2\theta.$$”

Here, the transmission axis of B is at $$\theta = 45^\circ$$ with respect to that of A, and the intensity incident on B is $$I_1 = \dfrac{I_0}{2}.$$ Applying Malus’ Law gives

$$ I_2 = I_1 \cos^2 45^\circ = \frac{I_0}{2} \times \cos^2 45^\circ. $$

We know the numerical value $$\cos 45^\circ = \frac{1}{\sqrt2}.$$ Substituting this value step by step, we obtain

$$ I_2 = \frac{I_0}{2} \times \left(\frac{1}{\sqrt2}\right)^2 = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}. $$

Thus the intensity of the light emerging from polaroid B is $$\dfrac{I_0}{4}.$$

Hence, the correct answer is Option A.

Ray AB is produced by the first partial reflection at point A.

Reflection Coefficient ($$R$$): $$25\%$$ or $$0.25$$. Intensity ($$I_1$$): $$I_1 = R \times I = \mathbf{0.25 I}$$.

Ray A'B' follows a path of refraction, reflection, and another refraction.

Transmission Coefficient ($$T$$): $$1 - 0.25 = 0.75$$.

At point A (Refraction): Intensity enters the slab = $$T \times I = 0.75 I$$.

At point C (Reflection): Intensity inside = $$R \times (0.75 I) = 0.25 \times 0.75 I = 0.1875 I$$.

At point A' (Refraction): Intensity exiting as ray A'B' = $$T \times (0.1875 I) = 0.75 \times 0.1875 I = \mathbf{0.140625 I}$$.

Intensity ($$I$$) is proportional to the square of the Amplitude ($$A$$). Therefore, $$A \propto \sqrt{I}$$.

$$\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{0.25 I}{0.140625 I}} = \sqrt{\frac{1/4}{9/64}} = \sqrt{\frac{16}{9}} = \mathbf{\frac{4}{3}}$$.

$$I_{max} \propto (A_1 + A_2)^2 \quad \text{and} \quad I_{min} \propto (A_1 - A_2)^2$$

$$\frac{I_{max}}{I_{min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$$

$$\frac{I_{max}}{I_{min}} = \left( \frac{4 + 3}{4 - 3} \right)^2 = \left( \frac{7}{1} \right)^2 = \mathbf{49 : 1}$$