JEE Wave Optics Questions

The relation for the position of the $$m^{\text{th}}$$ single-slit diffraction maximum is

$$\frac{bd}{D}=m\lambda \;\Longrightarrow\; b=\frac{m\lambda D}{d}$$

Given data:
$$m=3,\; \lambda = 600\,\text{nm}=600\times 10^{-9}\,\text{m},\; D=1\,\text{m},\; d=5\,\text{mm}=0.005\,\text{m}$$

Substituting,

$$b=\frac{3\,(600\times 10^{-9})\,(1)}{0.005}=3.6\times 10^{-4}\,\text{m}=360\,\mu\text{m}$$

The least counts (LC) of the measuring scales are
LC for $$D$$ : $$1\,\text{cm}=0.01\,\text{m}\;\Longrightarrow\; \Delta D = 0.01\,\text{m}$$
LC for $$d$$ : $$1\,\text{mm}=0.001\,\text{m}\;\Longrightarrow\; \Delta d = 0.001\,\text{m}$$

For a quantity obtained as a product/quotient, the maximum fractional error adds algebraically:

$$\frac{\Delta b}{b}= \frac{\Delta D}{D}+\frac{\Delta d}{d}$$

Compute each term:
$$\frac{\Delta D}{D}= \frac{0.01}{1}=0.01$$
$$\frac{\Delta d}{d}= \frac{0.001}{0.005}=0.20$$

Hence
$$\frac{\Delta b}{b}=0.01+0.20 = 0.21$$

Absolute error in $$b$$:

$$\Delta b = b \left(0.21\right)=\left(3.6\times 10^{-4}\right)(0.21)=7.56\times 10^{-5}\,\text{m}$$

Convert to micrometres ($$1\,\text{m}=10^{6}\,\mu\text{m}$$):

$$\Delta b = 7.56\times 10^{-5}\times 10^{6}=75.6\,\mu\text{m}$$

Therefore, the absolute error in the estimated slit width lies in the range $$75\text{-}79\,\mu\text{m}$$.

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

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

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

Using Snell's law, the angle of refraction of the incident beam in medium 2 is $$\theta$$.

Evaluating optical path difference before the slits between $$S_1$$ and $$S_2$$:

$$\Delta x_1 = n_2 d \cos\theta$$

Evaluating optical path difference after the slits to the far detector at angle $$\theta$$:

$$\Delta x_2 = -n_2 d \cos\theta$$

$$\Delta x_{\text{total}} = \Delta x_1 + \Delta x_2 = n_2 d \cos\theta - n_2 d \cos\theta = 0$$

$$\Delta \phi = \frac{2\pi}{\lambda_0} \Delta x_{\text{total}} = 0$$

Since $$\Delta \phi = 0$$, the phase difference is completely independent of $$d$$ and the rays interfere constructively.

Answer: Option (A), Option (B)

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.

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.

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.

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: $$y = \beta'$$

$$\frac{D}{d}(\mu - 1)t = \frac{\lambda (2D)}{d}$$

$$(\mu - 1)t = 2\lambda$$

$$(1.6 - 1) \times (1.8 \times 10^{-6}) = 2\lambda$$

$$0.6 \times 1.8 \times 10^{-6} = 2\lambda$$

$$1.08 \times 10^{-6} = 2\lambda$$

$$\lambda = \frac{1.08 \times 10^{-6}}{2} = 0.54 \times 10^{-6}\text{ m}$$

$$\lambda = 540 \times 10^{-9}\text{ m} = 540\text{ nm}$$

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.

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.

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, $$A \propto \sqrt{I}$$.

$$\frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{0.25 I}{0.140625 I}} = \sqrt{\frac{1/4}{9/64}} = \sqrt{\frac{16}{9}} = \mathbf{\frac{4}{3}}$$.

$$I_{max} \propto (A_1 + A_2)^2 \quad \text{and} \quad I_{min} \propto (A_1 - A_2)^2$$

$$\frac{I_{max}}{I_{min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$$

$$\frac{I_{max}}{I_{min}} = \left( \frac{4 + 3}{4 - 3} \right)^2 = \left( \frac{7}{1} \right)^2 = \mathbf{49 : 1}$$

When unpolarized light of intensity $$I_0$$ passes through a polarizer, its intensity reduces to half: $$I_1 = \frac{I_0}{2}$$

$$I_f = I_1 \cos^2\theta$$

$$\frac{3}{8}I_0 = \left(\frac{I_0}{2}\right) \cos^2\theta$$

$$\frac{3}{8} = \frac{1}{2} \cos^2\theta$$

$$\theta = 30^\circ$$

Use Rayleigh criterion for telescope:

$$\theta=1.22\frac{\lambda}{D}$$

given:

$$θ=3.0\times10^{-7},λ=500nm=5\times10^{-7}m$$

$$D=\frac{1.22\lambda}{\theta}=\frac{1.22\times5\times10^{-7}}{3\times10^{-7}}$$

$$=\frac{6.1}{3}\approx2.03m$$

convert to cm:

2.03 m=203 cm

Let the two slits be $$S_1$$ and $$S_2$$, separated by $$d = 2\;{\rm mm}$$.
The glass combination has a total (constant) physical thickness $$t = 12\;\mu{\rm m}$$ at every point, but the fractional thicknesses of the two glasses vary linearly along the length of the wedge, as shown in the figure.

Denote the thickness of glass A (refractive index $$n_A = 1.7$$) in front of a point on the slit plate by $$x$$.
Consequently, the thickness of glass B (refractive index $$n_B = 1.5$$) at the same point is $$t-x$$, because the total thickness remains $$t$$.

The slanted interface of the wedges makes the thickness of glass A change from $$0$$ to $$t$$ over the horizontal length $$l = 1\;{\rm mm}$$.
Therefore, the rate of change (slope) of thickness of glass A along the slit plate is

$$\frac{t}{l}= \frac{12\;\mu{\rm m}}{1\;{\rm mm}} = 12\;\frac{\mu{\rm m}}{{\rm mm}}$$

The two slits are $$d = 2\;{\rm mm}$$ apart. Hence the difference in the thickness of glass A present in front of the two slits is obtained by simple proportion (similar triangles):

$$\Delta x = \frac{t}{l}\times \frac{l}{2} = t\;\frac{l}{d} = 12\;\mu{\rm m}\;\times\;\frac{1\;{\rm mm}}{2\;{\rm mm}} = 6\;\mu{\rm m}$$

This means glass A is $$6\;\mu{\rm m}$$ thicker in front of one slit than in front of the other. Because the total thickness is fixed, glass B will automatically be $$6\;\mu{\rm m}$$ thinner at that slit:

$$t_{A1}-t_{A2}= +6\;\mu{\rm m}, \qquad t_{B1}-t_{B2}= -6\;\mu{\rm m}$$

The optical path length contributed by a layer of refractive index $$n$$ and physical thickness $$x$$ is $$(n-1)x$$ (extra path in excess of the same distance in air).
Hence the net path difference introduced between the two slits is

$$\delta_0 =(n_A-1)\,(t_{A1}-t_{A2})+(n_B-1)\,(t_{B1}-t_{B2}) =(n_A-n_B)\,\Delta x$$

Substituting the numerical values:

$$\delta_0 =(1.7-1.5)\times 6\;\mu{\rm m} =0.2\times 6\;\mu{\rm m} =1.2\;\mu{\rm m}$$

In Young’s double-slit arrangement, if one slit receives an additional optical path $$\delta_0$$, the point on the screen where the path difference due to geometry cancels this extra path is the new central maximum. For small angles (screen kept at distance $$D$$), the geometric path difference at a point $$y$$ from the original centre $$O$$ is $$\frac{d\,y}{D}$$. Setting the total path difference to zero gives

$$\frac{d\,y_0}{D} +\delta_0 = 0 \;\;\Longrightarrow\;\; |y_0| = \frac{D}{d}\,\delta_0$$

With $$D = 2\;{\rm m}=2000\;{\rm mm}$$ and $$d = 2\;{\rm mm}$$, we have

$$|y_0| = \frac{2000\;{\rm mm}}{2\;{\rm mm}} \times 1.2\;\mu{\rm m} = 1000 \times 1.2\times 10^{-3}\;{\rm mm} = 1.2\;{\rm mm}$$

Thus the central maximum shifts by $$1.2\;{\rm mm}$$ (towards the slit that has the smaller thickness of glass A).

Final answer: 1.2 mm

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 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 $$\mathbf{6\ \mu\text{m}}$$.

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 $$15$$.

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.

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 $$8$$.

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.

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.

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.

Two light waves of wavelengths $$\lambda_1 = 800$$ nm and $$\lambda_2 = 600$$ nm are used in Young's double slit experiment. Screen is at $$D = 7$$ m and slit separation $$d = 0.35$$ mm.

$$ \beta_1 = \frac{\lambda_1 D}{d} = \frac{800 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = \frac{5600 \times 10^{-9}}{3.5 \times 10^{-4}} = 16 \times 10^{-3} \text{ m} = 16 \text{ mm} $$

$$ \beta_2 = \frac{\lambda_2 D}{d} = \frac{600 \times 10^{-9} \times 7}{0.35 \times 10^{-3}} = 12 \times 10^{-3} \text{ m} = 12 \text{ mm} $$

Bright fringes coincide when:

$$ n_1 \beta_1 = n_2 \beta_2 $$

$$ 16n_1 = 12n_2 \Rightarrow \frac{n_1}{n_2} = \frac{12}{16} = \frac{3}{4} $$

The smallest positive integers satisfying this ratio are $$n_1 = 3$$ and $$n_2 = 4$$.

$$ y = n_1 \beta_1 = 3 \times 16 = 48 \text{ mm} $$

Verification: $$n_2 \beta_2 = 4 \times 12 = 48$$ mm $$\checkmark$$

The shortest distance from the central bright maximum to the point of coincidence is 48 mm.

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°.

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 $$x = 2$$.

Fringe width definition: $$\beta = \frac{\lambda D}{d}$$

Total linear width of a section containing $$n$$ fringes: $$\Delta x = n\beta = \frac{n\lambda D}{d}$$

Since the section width $$\Delta x$$, slit distance $$d$$, and screen distance $$D$$ remain constant:

$$n_1 \lambda_1 = n_2 \lambda_2$$

$$15 \times 500 = 10 \times \lambda_2$$

$$\lambda_2 = \frac{7500}{10} = 750\text{ nm}$$

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.

Step-by-Step Derivation

1. Shift due to a Transparent Sheet

When a transparent sheet of thickness $$t$$ and refractive index $$\mu$$ is introduced in front of one of the slits, it introduces an additional optical path difference ($$\Delta x$$) given by:

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

Due to this path difference, the entire fringe pattern shifts on the screen. The linear shift of the central maximum ($$\Delta y$$) is given by the formula:

$$\Delta y = \frac{D}{a}(\mu - 1)t$$

Where:

• $$D$$ = distance between the slits and the screen
• $$a$$ = distance between the two slits

2. Relating the Shift to Fringe Width

The width of a single fringe ($$\beta$$) in Young's double slit experiment is defined as:

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

According to the problem statement, the central maximum shifts by a distance equal to $$n$$ fringe widths:

$$\Delta y = n\beta$$

3. Equating and Finding Thickness ($$t$$)

Substitute the expressions for $$\Delta y$$ and $$\beta$$ into the equation:

$$\frac{D}{a}(\mu - 1)t = n \left(\frac{D\lambda}{a}\right)$$

We can cancel out the common terms $$\frac{D}{a}$$ from both sides of the equation:

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

Solving for the thickness $$t$$:

$$t = \frac{n\lambda}{\mu - 1}$$

Correct Answer

The correct option is (2) $$\frac{n\lambda}{\mu - 1}$$.

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.

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.

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.

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.

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.

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 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.

We are given $$n$$ identical waves, each of intensity $$I_0$$. We need to find the ratio of the maximum intensities for two cases: (i) coherent interference and (ii) incoherent interference.

First, recall that intensity is proportional to the square of the amplitude. For a single wave, intensity $$I_0$$ corresponds to an amplitude of $$\sqrt{I_0}$$. Let the amplitude of each wave be $$a$$, so $$I_0 = a^2$$.

Now, consider case (i): coherent interference. In coherent interference, the waves have a constant phase difference. The maximum intensity occurs when all waves are in phase, meaning their phase difference is zero. In this case, the amplitudes add up linearly.

The resultant amplitude for $$n$$ waves in phase is:

$$A_{\text{coherent}} = a + a + \cdots + a \quad (n \text{ times}) = n a$$

Since $$a = \sqrt{I_0}$$, we substitute:

$$A_{\text{coherent}} = n \sqrt{I_0}$$

The maximum intensity for coherent interference is proportional to the square of the resultant amplitude:

$$I_{\text{max, coherent}} = \left(A_{\text{coherent}}\right)^2 = \left(n \sqrt{I_0}\right)^2 = n^2 I_0$$

So, $$I_{\text{max, coherent}} = n^2 I_0$$.

Now, consider case (ii): incoherent interference. In incoherent interference, the phase differences between the waves are random and vary rapidly with time. Therefore, the waves do not maintain a constant phase relationship, and the cross terms in the interference average to zero. As a result, the total intensity is simply the sum of the individual intensities.

For $$n$$ waves, each of intensity $$I_0$$, the total intensity is:

$$I_{\text{incoherent}} = I_0 + I_0 + \cdots + I_0 \quad (n \text{ times}) = n I_0$$

Since the intensity is constant and does not vary with position (no interference pattern), this is also the maximum intensity for incoherent interference. So, $$I_{\text{max, incoherent}} = n I_0$$.

The ratio of the maximum intensities for coherent to incoherent interference is:

$$\text{Ratio} = \frac{I_{\text{max, coherent}}}{I_{\text{max, incoherent}}} = \frac{n^2 I_0}{n I_0} = \frac{n^2}{n} = n$$

Hence, the ratio is $$n$$.

Comparing with the options:

A. $$n^2$$

B. $$\frac{1}{n}$$

C. $$\frac{1}{n^2}$$

D. $$n$$

The correct option is D.

Hence, the correct answer is Option D.

Fringes are formed at the locus of points where the path difference ($$\Delta x$$) between waves from $$S_1$$ and $$S_2$$ is constant.

$$\Delta x = |S_1Q - S_2Q| = \text{constant}$$

The common center of these circular fringes is the point $$P$$, where the line joining the sources $$S_1$$ and $$S_2$$ meets the screen.

Every point on the screen that is at an equal radial distance from the central point $$P$$ will have the same path difference, leading to a pattern of concentric circular fringes.

In Young's double slit experiment, we need to evaluate Statement-1 and Statement-2.

First, recall the formula for fringe width, denoted by $$\beta$$. The fringe width is given by:

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

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

Statement-2 claims that the fringe width depends directly on the wavelength of light. From the formula, if $$D$$ and $$d$$ remain constant, then $$\beta$$ is proportional to $$\lambda$$, meaning $$\beta \propto \lambda$$. Therefore, Statement-2 is true.

Now, Statement-1 claims that the number of fringes observed in the field of view is small with longer wavelength and large with shorter wavelength. The field of view refers to a fixed region on the screen with a constant width, say $$W$$. The number of fringes, denoted by $$N$$, is approximately the total width of the field of view divided by the fringe width. So:

$$N = \frac{W}{\beta}$$

Substituting the expression for $$\beta$$:

$$N = \frac{W}{\frac{\lambda D}{d}} = \frac{W d}{\lambda D}$$

Assuming $$W$$, $$d$$, and $$D$$ are constant for the setup, then $$N \propto \frac{1}{\lambda}$$. This means that as the wavelength $$\lambda$$ increases, the number of fringes $$N$$ decreases, and as $$\lambda$$ decreases, $$N$$ increases. Therefore, with longer wavelength, the number of fringes is small, and with shorter wavelength, it is large. Hence, Statement-1 is true.

Now, we must determine if Statement-2 correctly explains Statement-1. Statement-2 states that fringe width is directly proportional to wavelength, which is correct. The number of fringes decreases with increasing wavelength because the fringe width increases, leading to fewer fringes fitting within the fixed field of view. However, for this explanation to hold, we must assume that the field of view width $$W$$ is constant and does not depend on wavelength. Statement-2 does not mention the field of view or its constancy. Therefore, while Statement-2 is true and related, it is not by itself a complete explanation for Statement-1 without the additional implicit assumption of a fixed field of view.

Hence, both statements are true, but Statement-2 is not the correct explanation for Statement-1.

So, the answer is Option C.

For coherent sources of equal amplitude $$A$$, the resulting amplitude at the midpoint of the screen (where path difference is zero, so phase difference $$\phi = 0$$) is: $$$A_{1} = A + A = 2A$$$ Since intensity is proportional to the square of amplitude ($$I \propto A^2$$): $$$I_1 \propto (2A)^2 = 4A^2$$$ For incoherent sources of equal amplitude $$A$$, the intensities simply add up without interference: $$$I_2 = I_0 + I_0 \propto A^2 + A^2 = 2A^2$$$ Taking the ratio of the intensities in the two cases: $$$\frac{I_1}{I_2} = \frac{4A^2}{2A^2} = 2:1$$$ The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is 2:1 (or 2).

For a single slit diffraction, the position of minima is given by:

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

Given:

First minimum, $$n = 1$$

Angle of diffraction, $$\theta = 30^\circ$$

Wavelength of light, $$\lambda = 5000\text{ \AA} = 5000 \times 10^{-10}\text{ m} = 5 \times 10^{-7}\text{ m}$$

Substituting the values into the formula:

$$a \sin 30^\circ = 1 \times 5 \times 10^{-7}$$

$$a \times \frac{1}{2} = 5 \times 10^{-7}$$

$$a = 2 \times 5 \times 10^{-7}\text{ m}$$

$$a = 10^{-6}\text{ m} = 1\text{ }\mu\text{m}$$

Wavelengths: $$\lambda_1 = 650$$ nm and $$\lambda_2 = 550$$ nm, slit separation: $$d = 2$$ mm $$= 2 \times 10^{-3}$$ m, and screen distance: $$D = 1.2$$ m.

The bright fringe positions for wavelength $$\lambda$$ in Young's double slit experiment are given by $$y_n = \frac{n \lambda D}{d}$$.

For bright fringes of both wavelengths to coincide at the same point, we require $$n_1 \lambda_1 = n_2 \lambda_2$$.

Since $$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$$, the smallest positive integers satisfying this are $$n_1 = 11$$ and $$n_2 = 13$$.

The distance of their least coincidence from the central maximum is $$y = \frac{n_1 \lambda_1 D}{d} = \frac{11 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$$.

$$y = \frac{11 \times 650 \times 1.2}{2} \times 10^{-9+3}$$

$$y = \frac{8580}{2} \times 10^{-6} = 4290 \times 10^{-6} = 429 \times 10^{-5} \text{ m}$$

The answer is 429.

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.

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 $$24$$.

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.

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 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 $$\boxed{7}$$.

The shift in the position of the central maxima ($$\Delta y$$) is given by:

$$\Delta y = \frac{D}{d}(\mu - 1)t$$

The fringe width $$\beta_0$$ is defined as $$\beta_0 = \frac{\lambda D}{d}$$

$$x \beta_0 = \frac{D}{d}(\mu - 1)t$$

$$x \left( \frac{\lambda D}{d} \right) = \frac{D}{d}(\mu - 1)t$$

$$x = \frac{(\mu - 1)t}{\lambda}$$

$$x = \frac{(1.2 - 1) \times 10 \times 10^{-6}}{500 \times 10^{-9}}$$

$$x = \frac{20}{5} = 4$$

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.

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.

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 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 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.

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.

$$S_1S_2 = d, \quad D = 2d$$

$$S_1P = D = 2d$$

$$S_2P = \sqrt{D^2 + d^2} = \sqrt{(2d)^2 + d^2} = \sqrt{5d^2} = \sqrt{5}d$$

$$\Delta x = S_2P - S_1P = \sqrt{5}d - 2d = (\sqrt{5} - 2)d$$

Condition for the first minima: $$\Delta x = \frac{\lambda}{2}$$

$$(\sqrt{5} - 2)d = \frac{\lambda}{2} \implies d = \frac{\lambda}{2(\sqrt{5} - 2)}$$

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.

For a single-slit of width $$a$$, the first diffraction minimum occurs at $$a \sin\theta = \lambda$$. For small angles in diffraction experiments $$\sin\theta \approx \theta$$ (in radians).

Given: wavelength $$\lambda = 628\ \text{nm} = 628 \times 10^{-9}\ \text{m}$$ slit width $$a = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m} = 2 \times 10^{-4}\ \text{m}$$

First minimum angle: $$\theta = \frac{\lambda}{a} = \frac{628 \times 10^{-9}}{2 \times 10^{-4}}$$ $$\theta = 314 \times 10^{-7} = 3.14 \times 10^{-3}\ \text{rad}$$

The central maximum extends from $$-\theta$$ to $$+\theta$$, so its angular width is $$2\theta = 2 \times 3.14 \times 10^{-3} = 6.28 \times 10^{-3}\ \text{rad}$$

Convert radians to degrees using $$1\ \text{rad} = \frac{180}{\pi} \approx 57.3^{\circ}$$: $$\text{width in degrees} = 6.28 \times 10^{-3} \times 57.3 \approx 0.36^{\circ}$$

Write this as $$0.36^{\circ} = 36 \times 10^{-2}\ ^{\circ}$$, hence $$\alpha = 36$$.

Final Answer: 36

The mica sheet is placed in front of one of the slits, so all rays from that slit travel an extra optical path.
Extra optical path introduced by a sheet of thickness $$t$$ and refractive index $$n$$ is

$$\Delta = (n-1)\,t$$

In Young’s double slit experiment, a path difference of one wavelength $$\lambda$$ shifts the fringe pattern by one fringe width (i.e. the central maximum moves to the position of the next bright band).
Hence, a shift through $$m$$ bright fringes requires a path difference

$$\Delta = m\,\lambda \quad -(1)$$

Given data:
• Refractive index of mica: $$n = 1.56$$
• Wavelength of light: $$\lambda = 450\;\text{nm} = 450 \times 10^{-9}\,\text{m}$$
• Central fringe is observed at the position of the 7-th bright fringe: $$m = 7$$

Equate the two expressions for $$\Delta$$:

$$(n-1)\,t = m\,\lambda$$

Solve for $$t$$:

$$t = \frac{m\,\lambda}{n-1}$$

Substitute the numbers:

$$t = \frac{7 \times 450 \times 10^{-9}}{1.56 - 1}$$

$$t = \frac{3150 \times 10^{-9}}{0.56}$$

$$t = 5625 \times 10^{-9}\;\text{m}$$

Hence, the thickness of the mica sheet is $$\alpha \times 10^{-9}\,\text{m}$$ with

$$\alpha = 5625$$

Answer (numerical): 5625

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.

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.

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

Intensity after passing orthogonal polarizers $$P_1$$ and $$P_2$$:

$$I_1 = I_2 = \frac{I_0}{2}$$

Beams from $$s_1$$ and $$s_2$$ passing through $$P_3$$ at $$45^\circ$$ (using Malus's Law):

$$I_{s1} = I_1 \cos^2(45^\circ) \implies I_{s1} = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

$$I_{s2} = I_2 \cos^2(45^\circ) \implies I_{s2} = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

Since both beams pass through $$P_3$$, they have parallel transmission axes and can interfere.

$$\Delta x = \frac{\lambda}{3} \implies \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{3}$$

Resultant intensity:

$$I_{\text{net}} = I_{s1} + I_{s2} + 2\sqrt{I_{s1}I_{s2}}\cos\phi$$

$$I_{\text{net}} = \frac{I_0}{4} + \frac{I_0}{4} + 2\left(\frac{I_0}{4}\right)\cos\left(\frac{2\pi}{3}\right) \implies I_{\text{net}} = \frac{I_0}{2} + \frac{I_0}{2}\left(-\frac{1}{2}\right) = \frac{I_0}{4}$$

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).

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),

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

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.

In Young's double slit experiment, the amplitude ratio of light from the two slits is 2:1. We need to find the ratio of maximum to minimum intensity.

To begin,

The intensity of a wave is proportional to the square of its amplitude:

$$ I \propto a^2 $$

If the amplitudes from the two slits are $$a_1$$ and $$a_2$$, then the individual intensities are $$I_1 \propto a_1^2$$ and $$I_2 \propto a_2^2$$.

Next,

In an interference pattern, the resultant amplitude depends on the phase difference $$\phi$$ between the two waves:

$$ a_{\text{resultant}} = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos\phi} $$

Maximum intensity occurs when $$\cos\phi = 1$$ (constructive interference):

$$ I_{max} \propto (a_1 + a_2)^2 $$

Minimum intensity occurs when $$\cos\phi = -1$$ (destructive interference):

$$ I_{min} \propto (a_1 - a_2)^2 $$

From this,

$$ \frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} $$

Given $$a_1 : a_2 = 2 : 1$$, let $$a_1 = 2$$ and $$a_2 = 1$$:

$$ \frac{I_{max}}{I_{min}} = \frac{(2 + 1)^2}{(2 - 1)^2} = \frac{9}{1} = 9 $$

The ratio of maximum to minimum intensity is 9 : 1.

The correct answer is Option 4: 9 : 1.

Solution :

For first minima in Young’s double slit experiment :

$$\text{Path difference} = \frac{\lambda}{2}$$

At point P,

$$A_1P = D$$

$$A_2P = \sqrt{D^2+a^2}$$

Therefore,

$$\sqrt{D^2+a^2} - D = \frac{\lambda}{2}$$

Given :

$$\lambda = 800\text{ nm} = 8 \times 10^{-7}\text{ m}$$

$$D = 5\text{ cm} = 5 \times 10^{-2}\text{ m}$$

Substituting :

$$\sqrt{D^2+a^2} = D + \frac{\lambda}{2}$$

Squaring both sides :

$$D^2+a^2 = D^2 + D\lambda + \frac{\lambda^2}{4}$$

Since $$\ \ lambda^2$$ is very small, neglecting it :

$$a^2 = D\lambda$$

$$= (5 \times 10^{-2})(8 \times 10^{-7})$$

$$= 4 \times 10^{-8}$$

$$a = 2 \times 10^{-4}\text{ m}$$

$$= 0.2\text{ mm}$$

Final Answer :

$$0.2\text{ mm}$$

Solution :

According to Malus law :

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

Given initial intensity :

$$I_0 = 256\text{ W m}^{-2}$$

After passing through first polaroid $$P_1$$ :

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

$$= \frac{256}{2}$$

$$= 128\text{ W m}^{-2}$$

Angle between axes of $$P_1$$ and $$P_2$$ :

$$\theta = 60^\circ$$

After passing through $$P_2$$ :

$$I_2 = I_1\cos^2 60^\circ$$

$$= 128 \times \left(\frac{1}{2}\right)^2$$

$$= 32\text{ W m}^{-2}$$

Angle between axes of $$P_2$$ and $$P_3$$ :

$$90^\circ - 60^\circ = 30^\circ$$

After passing through $$P_3$$ :

$$I_3 = I_2\cos^2 30^\circ$$

$$= 32 \times \left(\frac{\sqrt3}{2}\right)^2$$

$$= 32 \times \frac{3}{4}$$

$$= 24\text{ W m}^{-2}$$

Final Answer :

$$24\text{ W m}^{-2}$$

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 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.

Let the intensity of the incident unpolarised light be $$I_0$$. When unpolarised light passes through a polariser, its intensity is reduced to half.

Case 1: Only the original two polarisers (axes at $$30^{\circ}$$ and $$90^{\circ}$$)
After the first polariser (axis $$30^{\circ}$$):
$$I_1 = \frac{I_0}{2}$$

The second polariser’s axis is at $$90^{\circ}$$, i.e. it is at an angle $$\theta = 90^{\circ}-30^{\circ}=60^{\circ}$$ to the first axis.
By Malus’ law, transmitted intensity is
$$I_{\text{without}} = I_1 \cos^{2} \theta = \frac{I_0}{2}\cos^{2}60^{\circ} = \frac{I_0}{2}\left(\frac{1}{2}\right)^{2} = \frac{I_0}{8}$$

Case 2: A third polariser (axis $$60^{\circ}$$) inserted between the first two
Order of axes: $$30^{\circ} \rightarrow 60^{\circ} \rightarrow 90^{\circ}$$.

Step-1 First polariser (axis $$30^{\circ}$$):
$$I_1 = \frac{I_0}{2}$$

Step-2 Second polariser (axis $$60^{\circ}$$) is at $$30^{\circ}$$ to the first.
$$I_2 = I_1 \cos^{2}30^{\circ} = \frac{I_0}{2}\left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{I_0}{2}\left(\frac{3}{4}\right) = \frac{3I_0}{8}$$

Step-3 Third polariser (axis $$90^{\circ}$$) is at $$30^{\circ}$$ to the second.
$$I_{\text{with}} = I_2 \cos^{2}30^{\circ} = \frac{3I_0}{8}\left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3I_0}{8}\left(\frac{3}{4}\right) = \frac{9I_0}{32}$$

Required ratio
$$\frac{I_{\text{with}}}{I_{\text{without}}} = \frac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \frac{9}{32}\times\frac{8}{1} = \frac{9}{4}$$

Hence, the intensity becomes $$\dfrac{9}{4}$$ times larger when the third polariser at $$60^{\circ}$$ is inserted.

Option C which is: $$\frac{9}{4}$$

We need to find the ratio $$D_1/D_2$$ given that two Young’s double-slit setups produce fringes of equal width.

The fringe width in Young’s double-slit experiment is given by

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

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

Since the fringe widths are equal, $$\beta_1 = \beta_2$$, which implies

$$ \frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2} $$

Substituting the given ratios $$d_1 : d_2 = 2 : 1$$ and $$\lambda_1 : \lambda_2 = 1 : 2$$, we rearrange for $$D_1/D_2$$:

$$ \frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2} $$

Evaluating this gives

$$ \frac{D_1}{D_2} = \frac{2}{1} \times \frac{2}{1} = 4 $$

The answer is 4.

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.

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.

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 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.

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.

An unpolarised beam can be resolved into two mutually perpendicular components: one whose electric field lies in the plane of incidence (p-polarisation) and one whose electric field is perpendicular to the plane of incidence (s-polarisation).

Let the interface be the $$xy$$-plane and let the incident ray lie in the $$xz$$-plane. Hence the $$xz$$-plane is the plane of incidence; the direction $$\hat{y}$$ is perpendicular to this plane.

For reflection at an angle of incidence equal to the Brewster angle $$\theta_B$$ we have $$\tan\theta_B = \frac{n_2}{n_1}$$, and the Fresnel reflection coefficient for the p-component becomes zero: $$r_p(\theta_B)=0$$. Therefore the parallel (p) component is completely transmitted and the reflected light contains only the s-component.

Consequently the electric field of the reflected wave must be perpendicular to the plane of incidence, i.e. directed along $$\hat{y}$$. No component along $$\hat{x}$$ (parallel to the plane) or $$\hat{z}$$ (normal to the interface) can survive in the reflected beam.

Next, consider the phase of the reflected wave. For an incident wave whose wave vector lies in the $$xz$$-plane, reflection changes the sign of the normal component $$k_z$$ but leaves the tangential component $$k_x$$ unchanged. Hence the phase of the reflected wave can be written as $$k_x x - k_z z - \omega t \;.$$

Among the given expressions, the only one that

• contains the correct phase $$kx - kz - \omega t$$ for a wave propagating in the $$xz$$-plane after reflection, and
• allows the electric field to lie entirely perpendicular to the plane of incidence (choose $$E_x = 0,\,E_y \neq 0$$),

is

$$\bigl(E_x\hat{i}+E_y\hat{j}\bigr)\sin(kx-kz-\omega t).$$

Thus the reflected wave is represented by

Option A which is: $$(E_x \hat{i} + E_y \hat{j})\sin(kx - kz - \omega t).$$

Let the individual (monochromatic) sources in Young’s experiment have the same intensity $$I_0$$.
For any point on the screen, the resultant intensity is

$$I = 4I_0 \cos^{2}\!\left(\frac{\delta}{2}\right)$$

where $$\delta$$ is the phase difference between the two interfering waves at that point and $$4I_0$$ is the maximum intensity $$I_{\text{max}}$$.

The problem states that the observed intensity is $$\tfrac{3}{4}$$ of the maximum, so

$$I = \frac{3}{4}\,I_{\text{max}} = \frac{3}{4}\,(4I_0) = 3I_0$$

and therefore

$$3I_0 = 4I_0 \cos^{2}\!\left(\frac{\delta}{2}\right)$$

Dividing by $$4I_0$$ gives

$$\cos^{2}\!\left(\frac{\delta}{2}\right) = \frac{3}{4}$$

Taking the positive square root (intensity depends on $$\cos^{2}$$, so the sign of the cosine itself is immaterial):

$$\cos\!\left(\frac{\delta}{2}\right) = \frac{\sqrt{3}}{2}$$

The standard angle that satisfies this is

$$\frac{\delta}{2} = \frac{\pi}{6} \quad\Longrightarrow\quad \delta = \frac{\pi}{3}$$

Phase difference $$\delta$$ and path difference $$\Delta$$ are connected by

$$\delta = \frac{2\pi}{\lambda}\,\Delta$$

Hence

$$\frac{\pi}{3} = \frac{2\pi}{\lambda}\,\Delta \; \Longrightarrow \; \Delta = \frac{\lambda}{6}$$

The statement of the question writes the same path difference as $$\Delta = \frac{\lambda}{x}$$. Equating the two forms:

$$\frac{\lambda}{x} = \frac{\lambda}{6} \;\Longrightarrow\; x = 6$$

Therefore the required value is 6.

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 $$6$$.

The minimum angular resolution of an optical telescope is governed by the Rayleigh criterion:

$$\theta_{\text{min}} = 1.22 \frac{\lambda}{D}$$

where
  • $$\theta_{\text{min}}$$ is the smallest resolvable angle (in radians),
  • $$\lambda$$ is the wavelength of light used, and
  • $$D$$ is the diameter of the objective aperture.

Given data:
$$\theta_{\text{min}} = 5 \times 10^{-7}\ {\text{rad}}$$
$$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5 \times 10^{-7}\ \text{m}$$

Rearrange the Rayleigh formula to find $$D$$:

$$D = 1.22 \frac{\lambda}{\theta_{\text{min}}}$$

Substitute the numerical values:

$$D = 1.22 \times \frac{5 \times 10^{-7}\ \text{m}}{5 \times 10^{-7}\ \text{rad}}$$

The factor $$5 \times 10^{-7}$$ cancels out in numerator and denominator, giving

$$D = 1.22\ \text{m}$$

Convert metres to centimetres:

$$1.22\ \text{m} = 1.22 \times 100\ \text{cm} = 122\ \text{cm}$$

Thus the required diameter of the telescope objective is $$R = 122\ \text{cm}$$.

Option B which is: 122

The point is described as being "exactly opposite to one of the slits." Since the two slits are separated by a distance $$d$$ symmetrically about the central axis, the distance from the central maximum to a point directly opposite one of the slits is: $$y = \frac{d}{2}$$

$$\Delta x = \frac{y \cdot d}{D} = \frac{\left(\frac{d}{2}\right) \cdot d}{10d} = \frac{d^2}{20d} = \frac{d}{20}$$

$$\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$$

$$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}$$

$$I = I_0 \cos^2\left(\frac{\pi / 2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$$

$$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$$

$$I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$

$$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4}$$

$$\frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$$

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

$$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \cdot \Delta x$$

$$\Delta x = \frac{6000 \text{ \AA}}{3} = 2000 \text{ \AA}$$

fringe width β = λD / d

in a medium of refractive index n, wavelength becomes λ/n

so new fringe width:

$$β'=(λ/n)D/d=β/n$$

given:

$$β=2.4μm$$
$$n=1.2$$

$$β′=\frac{2.4}{1.2}=2μm$$

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

$$a \sin\theta_m = m\lambda, \qquad m = 1,2,3,\dots$$

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

When the screen is at a large distance $$D$$, the linear distance $$y_m$$ of the $$m^{\text{th}}$$ minimum from the central axis is obtained using the small-angle approximation $$\sin\theta_m \approx \tan\theta_m \approx \theta_m$$:

$$y_m = D \tan\theta_m \approx D \theta_m = D \sin\theta_m = D \left(\frac{m\lambda}{a}\right) = \frac{m\lambda D}{a}$$

Thus,
for the first minimum $$\bigl(m = 1\bigr): \; y_1 = \frac{\lambda D}{a}$$
for the third minimum $$\bigl(m = 3\bigr): \; y_3 = \frac{3\lambda D}{a}$$

The required linear separation between the first and third minima is

$$y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2\lambda D}{a}$$

Hence the correct choice is:
Option C which is: $$\frac{2D\lambda}{a}$$

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

$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$

$$\phi_A = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$$

$$I_A = 4I_0 \cos^2\left(\frac{2\pi / 3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$$

$$I_A = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \left(\frac{1}{4}\right) = I_0$$

$$\phi_B = \frac{2\pi}{\lambda} \left(\frac{\lambda}{6}\right) = \frac{\pi}{3}$$

$$I_B = 4I_0 \cos^2\left(\frac{\pi / 3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{6}\right)$$

$$I_B = 4I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \left(\frac{3}{4}\right) = 3I_0$$

$$\frac{I_A}{I_B} = \frac{I_0}{3I_0} = \frac{1}{3}$$