JEE Ray Optics Questions

A concave mirror produces an image with object-image distance = 20 cm and magnification = $$-3$$.

We start by noting that for a concave mirror, using the standard convention: object distance $$u$$ is negative (object on the same side as incoming light), and for a real image, $$v$$ is also negative. The magnification is given by $$m = -\frac{v}{u} = -3$$, which leads to $$v = 3u$$, and since $$m = -3$$ (negative), the image is real and inverted.

Next, the distance between object and image is $$|v - u| = 20$$ cm. Substituting $$v = 3u$$ gives $$|3u - u| = |2u| = 20$$, so $$|u| = 10$$ cm. Therefore $$u = -10$$ cm (negative by convention) and $$v = 3(-10) = -30$$ cm.

Using the mirror formula, $$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-10} = -\frac{1}{30} - \frac{1}{10} = -\frac{1}{30} - \frac{3}{30} = -\frac{4}{30} = -\frac{2}{15} $$. This gives $$ f = -\frac{15}{2} = -7.5 \text{ cm} $$.

The radius of curvature is $$R = 2f = 2 \times (-7.5) = -15$$ cm. The magnitude is $$|R| = 15$$ cm.

The correct answer is Option C: 15 cm.

$$\tan \alpha = \sqrt{3} \implies \alpha = 60^\circ$$

$$\sqrt{3} \sin \beta = 1 \times \sin \alpha$$

$$\sqrt{3} \sin \beta = \sin 60^\circ = \frac{\sqrt{3}}{2} \implies \sin \beta = \frac{1}{2} \implies \beta = 30^\circ$$

Using the sine rule for the triangle formed inside the geometry:

$$\frac{R}{2 \sin 30^\circ} = \frac{x}{\sin 120^\circ}$$

$$\frac{R}{\sin 120^\circ} = \frac{R}{\frac{\sqrt{3}}{2} \times \sin \theta} \implies \sin \theta = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$$

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

Let the spherical surface have its vertex at point $$V$$ on the optic axis, with the centre of curvature $$C$$ lying inside the glass medium at a distance $$R$$ from $$V$$.
Choose the Cartesian sign convention: light travels from left (air) to right (glass).
Hence, distances measured to the right of $$V$$ are taken as positive.

Given data
  • Refractive index in air, $$n_1 = 1$$
  • Refractive index in glass, $$n_2 = 1.5$$
  • Radius of curvature, $$R$$, is positive (centre is on the image side).

The object $$O$$ is in air (left side), so its distance from the vertex is $$u = -PO$$ (negative).
The real image $$I$$ is in glass (right side), so its distance from the vertex is $$v = PI$$ (positive).
It is given that $$PO = PI$$, therefore the magnitudes of the object and image distances are equal:
  $$|u| = v \quad\Longrightarrow\quad v = -u$$ $$-(1)$$

The refraction at a spherical surface is governed by the formula
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ $$-(2)$$

Substitute $$n_1 = 1$$, $$n_2 = 1.5$$ and use $$(1)$$, i.e. $$u = -v$$:

$$\frac{1.5}{v} - \frac{1}{-v} = \frac{1.5 - 1}{R}$$

Simplify the left-hand side:
$$\frac{1.5}{v} + \frac{1}{v} = \frac{2.5}{v}$$

Thus
$$\frac{2.5}{v} = \frac{0.5}{R}$$

Solving for $$v$$:
$$v = \frac{2.5}{0.5}\,R = 5R$$

The distance asked in the problem is $$PO$$, and $$PO = v = 5R$$.

Therefore, the correct option is Option A: $$5R$$.

For the original equiconvex lens of focal length $$f$$:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = \frac{2(\mu - 1)}{R}$$

For part $$L_1$$ (cut along horizontal plane AB):

The curvature of the surfaces remains unchanged. $$f_1 = f$$

For part $$L_3$$ (cut along vertical plane XY):

One surface becomes plane ($$R_2 = \infty$$), making it a planoconvex lens.

$$\frac{1}{f_3} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{\mu - 1}{R} = \frac{1}{2f} \implies f_3 = 2f$$

Ratio of focal lengths of $$L_1$$ and $$L_3$$: $$\frac{f_1}{f_3} = \frac{f}{2f} = \frac{1}{2}$$

$$\tan\alpha = \frac{\text{Vertical Height}}{\text{Horizontal Length}} = \frac{dh}{du} = \frac{2\text{ cm}}{1\text{ cm}} = 2$$

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v_A} - \frac{1}{-30} = \frac{1}{20}$$

$$\frac{1}{v_A} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60} \implies v_A = +60\text{ cm}$$

$$m_T = \frac{v_A}{u_A} = \frac{60}{-30} = -2$$

$$dh' = m_T \cdot dh = (-2) \times 2\text{ cm} = -4\text{ cm}$$ (Vertical dimension of image)

$$dv = m_L \cdot du = -m_T^2 \cdot du$$

$$dv = -(-2)^2 \times 1\text{ cm} = -4\text{ cm}$$ (Horizontal dimension of image)

$$\tan\theta = \frac{\text{Vertical dimension of image}}{\text{Horizontal dimension of image}} = \frac{dh'}{dv}$$

$$\tan\theta = \frac{-4\text{ cm}}{4\text{ cm}} = -1$$

$$\theta = \tan^{-1}(-1) = -45^\circ$$

$$P = (\mu - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = \frac{2(\mu - 1)}{R} = 4\text{ D}$$

$$P_{\text{horizontal half}} = P = 4\text{ D}$$

Cutting perpendicular to the principal axis (plane CD):

$$R_1 = R, \quad R_2 = \infty \implies P_{\text{vertical half}} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{\mu - 1}{R} = \frac{P}{2}$$

For a quarter part (divided by both planes AB and CD):

$$P' = \frac{P}{2} \implies P' = \frac{4\text{ D}}{2} = 2\text{ D}$$

We need to find the angle of prism $$P_2$$ such that the combination of $$P_1$$ and $$P_2$$ produces dispersion without deviation.

- Prism $$P_1$$: angle $$A_1 = 4°$$, refractive index $$n_1 = 1.54$$

- Prism $$P_2$$: angle $$A_2 = ?$$, refractive index $$n_2 = 1.72$$

For a thin prism, the deviation is: $$\delta = (n - 1)A$$

For dispersion without deviation, the two prisms must be arranged to cancel each other's deviation. The prisms are placed with opposite orientations:

$$\delta_1 = \delta_2$$

$$(n_1 - 1)A_1 = (n_2 - 1)A_2$$

$$A_2 = \frac{(n_1 - 1)}{(n_2 - 1)} \times A_1 = \frac{(1.54 - 1)}{(1.72 - 1)} \times 4° = \frac{0.54}{0.72} \times 4°$$

$$= \frac{3}{4} \times 4° = 3°$$

The angle of prism $$P_2$$ is 3 degrees.

The correct answer is Option 1: 3.

A thin convex lens with refractive index $$\mu_2$$ is immersed in a liquid of refractive index $$\mu_1$$ ($$\mu_1 < \mu_2$$), and its second surface is silvered, effectively forming a combination of a lens and a mirror. We seek the object distance for which the image coincides with the object position.

In this silvered lens system light passes through the lens, reflects off the silvered surface, and passes through the lens again, so the system acts as an equivalent mirror whose power is given by $$P_{eq} = 2P_{lens} + P_{mirror}$$.

For the thin lens in the surrounding medium, the lens maker’s formula yields

$$\frac{1}{f_{lens}} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{|R_1|} + \frac{1}{|R_2|}\right) = \frac{(\mu_2 - \mu_1)}{\mu_1}\cdot\frac{|R_1| + |R_2|}{|R_1|\cdot |R_2|}.$$

The silvered surface behaves as a concave mirror of radius $$|R_2|$$, so

$$\frac{1}{f_{mirror}} = \frac{2}{|R_2|}.$$

Therefore, the equivalent focal length satisfies

$$\frac{1}{f_{eq}} = \frac{2}{f_{lens}} + \frac{1}{f_{mirror}} = \frac{2\mu_2(|R_1| + |R_2|) - 2\mu_1|R_2|}{\mu_1|R_1|\cdot |R_2|}$$

For the image to coincide with the object, the object must be placed at the center of curvature of the equivalent mirror, i.e., at $$R_{eq} = 2f_{eq}$$. Hence the required object distance is

$$R_{eq} = 2f_{eq} = \frac{2\mu_1|R_1|\cdot |R_2|}{2[\mu_2(|R_1| + |R_2|) - \mu_1|R_2|]} = \frac{\mu_1|R_1|\cdot |R_2|}{\mu_2(|R_1| + |R_2|) - \mu_1|R_2|},$$

which matches Option A.

$$\frac{1}{F} = \frac{1}{f_{\text{liquid}}} + \frac{1}{f_{\text{glass}}}$$ (Net focal length)

$$\frac{1}{f} = (\mu_{\text{lens}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ (individual focal length)

$$\frac{1}{f_{\text{glass}}} = (1.5 - 1) \left( \frac{1}{-30} - \frac{1}{-20} \right)$$

$$\frac{1}{f_{\text{glass}}} = 0.5 \left( -\frac{1}{30} + \frac{1}{20} \right) = 0.5 \left( \frac{-2 + 3}{60} \right) = 0.5 \times \frac{1}{60} = \frac{1}{120}\text{ cm}^{-1}$$

$$\frac{1}{f_{\text{liquid}}} = (1.3 - 1) \left( \frac{1}{\infty} - \frac{1}{-30} \right)$$

$$\frac{1}{f_{\text{liquid}}} = 0.3 \left( 0 + \frac{1}{30} \right) = \frac{0.3}{30} = \frac{1}{100}\text{ cm}^{-1}$$

$$\frac{1}{F} = \frac{1}{100} + \frac{1}{120}$$

$$F = \frac{600}{11}\text{ cm}$$

For a thin symmetrical convex lens each surface has radius of curvature $$R$$ and refractive index $$\mu$$ with respect to air.

Step 1 : Focal length of the glass lens
Lens-maker’s formula for a thin lens in air is
$$\frac{1}{f_{\text{lens}}} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
For a biconvex lens $$R_1 = +R$$ (centre to the right) and $$R_2 = -R$$ (centre to the left).
Hence
$$\frac{1}{f_{\text{lens}}} = (\mu-1)\left(\frac{1}{R}-\frac{-1}{R}\right) = (\mu-1)\left(\frac{2}{R}\right)$$
Therefore
$$f_{\text{lens}} = \frac{R}{2(\mu-1)} \quad -(1)$$

Step 2 : Power of the lens
Optical power is $$P = \dfrac{1}{f}$$ (in dioptres when $$f$$ is in metres).
From $$(1)$$
$$P_{\text{lens}} = \frac{1}{f_{\text{lens}}} = \frac{2(\mu-1)}{R} \quad -(2)$$

Step 3 : Power of the silvered (concave) surface
After silvering the second surface acts as a concave mirror of radius $$R$$.
Focal length of a spherical mirror is $$f_{\text{mirror}} = \dfrac{R}{2}$$,
so its power is
$$P_{\text{mirror}} = \frac{1}{f_{\text{mirror}}} = \frac{2}{R} \quad -(3)$$

Step 4 : Equivalent focal length of the silvered lens
Light passes through the lens, reflects, and then passes through the lens again, so the lens power is used twice. The total power of the system is
$$P_{\text{eq}} = 2P_{\text{lens}} + P_{\text{mirror}}$$
Substituting from $$(2)$$ and $$(3)$$:
$$P_{\text{eq}} = 2\left(\frac{2(\mu-1)}{R}\right) + \frac{2}{R} = \frac{4(\mu-1) + 2}{R} = \frac{2(2\mu-1)}{R}$$
Hence the equivalent focal length $$F$$ of the silvered lens is
$$F = \frac{1}{P_{\text{eq}}} = \frac{R}{2(2\mu-1)} \quad -(4)$$

Step 5 : Condition for object and image to coincide
The silvered lens behaves like a concave mirror of focal length $$F$$. For a concave mirror, an object placed at its centre of curvature (distance $$2F$$) produces an image at the same position.
Therefore the required object distance from the lens is
$$u = 2F = 2\left(\frac{R}{2(2\mu-1)}\right) = \frac{R}{2\mu-1}$$

Answer : The object must be placed at a distance $$\displaystyle \frac{R}{2\mu-1}\;$$ from the lens (Option D).

A concave mirror of focal length f in air is dipped in a liquid. What is its focal length in the liquid?

Key concept: The focal length of a mirror depends only on the radius of curvature ($$f = R/2$$), which is a geometric property. Unlike a lens, a mirror's focal length does not depend on the refractive index of the surrounding medium.

Therefore, the focal length in the liquid remains f.

The correct answer is Option 2: f.

For grazing emergence along the surface BE to reach point E:

$$\text{Angle of refraction at surface near B}, \quad r = 90^\circ$$

From the geometry of the hemispherical vessel of radius $$R$$:

$$OC = R, \quad CB = R \implies \angle OBC = 45^\circ$$

$$\text{Angle of incidence at point B}, \quad i = \angle OBC = 45^\circ$$

Applying Snell's law at interface at B:

$$\mu \sin i = 1 \cdot \sin r$$

$$\mu \sin(45^\circ) = 1 \cdot \sin(90^\circ)$$

$$\mu \left(\frac{1}{\sqrt{2}}\right) = 1 \implies \mu = \sqrt{2}$$

The focal length of a thin lens in a surrounding medium is obtained from the Lens-Maker’s formula:

$$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad -(1)$$

For a given lens the radii $$R_1$$ and $$R_2$$ do not change; only the refractive-index term varies when the medium is changed.

Case 1: Lens in air
Here $$\mu_{\text{medium}} = 1$$, $$\mu_{\text{lens}} = 1.6$$ and the focal length is given as $$f_1 = 12\ \text{cm}$$.

Substituting in $$(1)$$:

$$\frac{1}{12} = \left(\frac{1.6}{1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\Rightarrow \frac{1}{12} = (1.6 - 1)K$$ where $$K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$.

Hence $$K = \frac{1}{12 \times 0.6} = \frac{1}{7.2}\quad -(2)$$

Case 2: Lens in water
Now $$\mu_{\text{medium}} = 1.28$$, while $$\mu_{\text{lens}} = 1.6$$ remains the same. Let the new focal length be $$f_2$$.

Using $$(1)$$ again:

$$\frac{1}{f_2} = \left(\frac{1.6}{1.28} - 1\right)K$$ First evaluate the refractive-index factor: $$\frac{1.6}{1.28} = 1.25 \; \Rightarrow \; 1.25 - 1 = 0.25$$.

Therefore $$\frac{1}{f_2} = 0.25 \times K$$ Insert $$K$$ from $$(2)$$: $$\frac{1}{f_2} = 0.25 \times \frac{1}{7.2} = \frac{0.25}{7.2}$$ $$\frac{1}{f_2} = 0.034722\dots$$ Hence $$f_2 = \frac{1}{0.034722\dots} = 28.8\ \text{cm}$$.

Convert to millimetres: $$28.8\ \text{cm} = 28.8 \times 10 = 288\ \text{mm}$$.

Thus, the focal length of the lens in water is $$288\ \text{mm}$$.

Option B is correct.

For a spherical mirror the transverse magnification is defined as
$$m \;=\; \frac{h_i}{h_o}\;=\;-\frac{v}{u} \quad -(1)$$
where $$u$$ is the object distance and $$v$$ is the image distance, both measured from the pole using the Cartesian sign convention.

The given magnitude of magnification is $$\frac{1}{4}$$.
Because the value is positive and less than $$1$$, the image is erect and virtual. Hence the mirror must be convex, so we take
$$m = +\frac{1}{4} \quad\Longrightarrow\quad \frac{1}{4}= -\frac{v}{u} \quad -(2)$$

From $$(2)$$,
$$v = -\frac{u}{4} \quad -(3)$$

With the sign convention, the object lies in front of the mirror, so $$u$$ is negative. Equation $$(3)$$ therefore makes $$v$$ positive, placing the virtual image behind the mirror, as expected for a convex mirror.

The distance between the object and its image is given to be $$40\ \text{cm}$$. Along the principal axis these two points are on opposite sides of the pole, so their separation equals the sum of their magnitudes:
$$|u| + |v| = 40 \quad -(4)$$

Put $$|v| = \frac{|u|}{4}$$ from $$(3)$$ into $$(4)$$:
$$|u| + \frac{|u|}{4} = 40 \;\Longrightarrow\; \frac{5|u|}{4} = 40 \;\Longrightarrow\; |u| = 32\ \text{cm}$$

Thus,
$$u = -32\ \text{cm}, \qquad v = +8\ \text{cm}$$

Apply the mirror formula
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \quad -(5)$$

Substituting $$v = 8\ \text{cm}$$ and $$u = -32\ \text{cm}$$ into $$(5)$$ gives
$$\frac{1}{f} = \frac{1}{8} - \frac{1}{32} = \frac{4 - 1}{32} = \frac{3}{32}$$

Therefore
$$f = \frac{32}{3}\ \text{cm} \approx 10.7\ \text{cm}$$

The focal length is positive, confirming that the mirror is indeed convex. Hence the correct option is Option C (10.7 cm).

Two liquids A ($$\mu_A = 1.3$$) and B ($$\mu_B = 1.4$$) form a convex meniscus towards A. An object at 13 cm in A forms an image with magnification $$m = -2$$.

We start by applying the refraction formula at a spherical surface: $$\frac{\mu_B}{v} - \frac{\mu_A}{u} = \frac{\mu_B - \mu_A}{R}$$. Since the meniscus is convex towards A, the center of curvature is on the A side, and taking the direction of light from A to B gives $$u = -13$$ cm by sign convention.

Next, the magnification condition for refraction at a single surface is $$m = \frac{\mu_A \cdot v}{\mu_B \cdot u}$$. Substituting $$m = -2$$ yields $$-2 = \frac{1.3 \times v}{1.4 \times (-13)}$$ and hence $$-2 = \frac{1.3v}{-18.2}$$.

This gives $$v = \frac{-2 \times (-18.2)}{1.3} = \frac{36.4}{1.3} = 28$$ cm.

Now, finding R from the refraction formula, $$\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R}$$ leads to $$0.05 + 0.1 = \frac{0.1}{R}$$ and therefore $$0.15 = \frac{0.1}{R}$$.

Therefore, $$R = \frac{0.1}{0.15} = \frac{2}{3}$$ cm.

Hence, the correct answer is Option D) $$\frac{2}{3}$$ cm.

Initial power is $$P_1 = 2.5$$ D, which gives $$f_1 = 1/2.5 = 0.4$$ m.

When the power increases to $$P_2 = 2.6$$ D, the focal length becomes $$f_2 = 1/2.6$$ m.

The relative decrease is $$\frac{f_1-f_2}{f_1} = 1 - \frac{f_2}{f_1} = 1 - \frac{P_1}{P_2} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} \approx 0.0385 \approx 0.04$$.

Therefore the correct answer is Option 2: 0.04.

The frequency of monochromatic light never changes when it passes from one medium to another.
Given frequency $$f = 5 \times 10^{14}\, \text{Hz}$$ remains the same in every medium.

Step 1 : Wavelength in air
Speed of light in air is practically the same as in vacuum, $$c = 3 \times 10^{8}\,\text{m s}^{-1}$$.
Using $$\lambda = \dfrac{v}{f}$$, the wavelength in air is
$$\lambda_{\text{air}} = \frac{c}{f} = \frac{3 \times 10^{8}}{5 \times 10^{14}} = 6 \times 10^{-7}\,\text{m} = 600\,\text{nm}.$$

Step 2 : Speed in the second medium
Refractive index $$\mu$$ of the medium is defined by $$\mu = \frac{c}{v}$$.
For $$\mu = 2$$,
$$v = \frac{c}{\mu} = \frac{3 \times 10^{8}}{2} = 1.5 \times 10^{8}\,\text{m s}^{-1}.$$

Step 3 : Wavelength in the second medium
Again using $$\lambda = \dfrac{v}{f}$$,
$$\lambda_{\text{medium}} = \frac{v}{f} = \frac{1.5 \times 10^{8}}{5 \times 10^{14}} = 3 \times 10^{-7}\,\text{m} = 300\,\text{nm}.$$

Thus the wavelength of the light inside the medium of refractive index $$2$$ is $$300\,\text{nm}$$.

Hence, the correct choice is Option A.

The mirror formula and the lateral magnification formula for a spherical mirror are

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\qquad -(1)$$
$$m=-\frac{v}{u}\qquad\qquad\!\!\!\! -(2)$$

The object is kept in front of the mirror, so by the Cartesian sign convention
  • object distance $$u$$ is always negative,
  • for a convex mirror the focal length $$f$$ is positive and the image distance $$v$$ is positive (virtual, behind the mirror).
Since the given magnification $$\dfrac{1}{2}$$ is positive (erect image), the mirror must be convex.

Given data for the first position:
  $$u_1=-40\text{ cm},\qquad m_1=\frac{1}{2}$$

From $$-(2)$$ : $$-\frac{v_1}{u_1}=m_1\Longrightarrow v_1=-m_1u_1=-\frac{1}{2}\times(-40)=+20\text{ cm}$$

Substituting $$u_1$$ and $$v_1$$ in $$-(1)$$ to find the focal length:

$$\frac{1}{f}=\frac{1}{20}+\frac{1}{-40}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}$$
$$\Rightarrow\; f=+40\text{ cm}$$

For the new position the required magnification is
  $$m_2=\frac{1}{3}\quad(\text{still positive, erect image})$$

Using $$-(2)$$ again:
$$-\frac{v_2}{u_2}=m_2\Longrightarrow v_2=-m_2u_2=-\frac{1}{3}u_2$$ $$-(3)$$

Insert $$v_2$$ from $$-(3)$$ into the mirror formula $$-(1)$$:

$$\frac{1}{f}=\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{-\dfrac{u_2}{3}}+\frac{1}{u_2}=-\frac{3}{u_2}+\frac{1}{u_2}=-\frac{2}{u_2}$$

With $$f=+40\text{ cm}$$, we get

$$\frac{1}{40}=-\frac{2}{u_2}\;\;\Longrightarrow\;\;u_2=-80\text{ cm}$$

Comparison of object positions:
Initial distance $$|u_1|=40\text{ cm}$$
Final distance $$|u_2|=80\text{ cm}$$

The object must therefore be shifted farther from the mirror by

$$|u_2|-|u_1|=80-40=40\text{ cm}$$

Case 1: Moving the object 40 cm away from the mirror produces the required magnification $$\dfrac{1}{3}$$.

Hence, the correct option is Option A (40 cm away from the mirror).

For the plano-convex lens immersed in liquid:

$$\mu_g = 1.5, \quad \mu_l = 1.2, \quad R_1 = R, \quad R_2 = \infty$$

$$P_l = \frac{1}{f_l} = \left(\frac{\mu_g}{\mu_l} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \implies P_l = \left(\frac{1.5}{1.2} - 1\right)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{0.25}{R} = \frac{1}{4R}$$

For the plane silvered surface acting as a plane mirror:

$$R_m = \infty \implies P_m = -\frac{1}{f_m} = -\frac{1}{\infty} = 0$$

Equivalent power of silvered system behaving as a concave mirror:

$$F_{\text{eq}} = -0.2\text{ m}$$

$$P_{\text{eq}} = -\frac{1}{F_{\text{eq}}} = 2P_l + P_m \implies -\frac{1}{-0.2} = 2\left(\frac{1}{4R}\right) + 0 \implies 5 = \frac{1}{2R} \implies R = \frac{1}{10} = 0.10\text{ m}$$

Refractive index $$\mu = \sqrt{3}$$ and angle of minimum deviation $$\delta_m$$ equals the angle of the prism A.

Use the prism formula at minimum deviation:

$$\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Given $$\delta_m = A$$:

$$\sqrt{3} = \frac{\sin\left(\frac{A+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin(A/2)}$$

Apply double angle formula:

$$\sqrt{3} = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$$

$$\cos(A/2) = \frac{\sqrt{3}}{2}$$

$$A/2 = 30°$$

$$A = 60°$$

The correct answer is Option 1: 60°.

The equivalent focal length of two thin lenses separated by a distance is found with the lens-combination formula
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1\,f_2}$$
where
$$f_1, f_2$$ are the focal lengths of the individual lenses (signs taken from the Cartesian convention) and $$d$$ is the separation between their optical centres.

All three quantities are first expressed in the same unit. Working in centimetres:
$$f_1 = 30\; \text{cm},\; f_2 = 10\; \text{cm},\; d = 10\; \text{cm}$$

Substituting into the formula gives
$$\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{30 \times 10}$$

Simplify each term:
$$\frac{1}{30} = 0.0333\ldots,\quad \frac{1}{10} = 0.1,\quad \frac{10}{30 \times 10} = \frac{10}{300} = 0.0333\ldots$$

Add and subtract term by term:
$$\frac{1}{F} = 0.0333\ldots + 0.1 - 0.0333\ldots = 0.1$$

Thus
$$F = \frac{1}{0.1} = 10\; \text{cm}$$

Convert the focal length to metres for power calculation:
$$F = 10\; \text{cm} = 0.10\; \text{m}$$

Lens power is defined as
$$P = \frac{1}{F\;(\text{in metres})}$$
so
$$P = \frac{1}{0.10} = 10\; \text{dioptre}$$

Therefore, the power of the lens combination is $$10\; \text{D}$$, which corresponds to Option D.

The refractive index of the lens (glass) is $$n_g = 1.5$$ and the refractive index of the surrounding medium is

• for air: $$n_{\text{air}} = 1$$
• for water: $$n_w = 1.33$$

For a thin lens placed in a medium of refractive index $$n_m$$ the lens-maker’s formula is

$$\frac{1}{f_m} = \left(\frac{n_g}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$-(1)$$

Here $$R_1, R_2$$ are the radii of curvature of the two spherical surfaces. The term $$\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ depends only on the shape of the lens and therefore remains the same when the surrounding medium is changed.

Step 1: Focal length in air
Putting $$n_m = n_{\text{air}} = 1$$ in $$(1)$$,

$$\frac{1}{f_{\text{air}}} = (n_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$-(2)$$

Given $$f_{\text{air}} = 24 \text{ cm}$$, so

$$\frac{1}{24} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\Rightarrow$$ $$\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{24 \times 0.5} = \frac{1}{12}$$ $$-(3)$$

Step 2: Focal length in water
Now place the lens in water, i.e. $$n_m = n_w = 1.33$$. Using $$(1)$$ again,

$$\frac{1}{f_w} = \left(\frac{n_g}{n_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$-(4)$$

Substitute $$n_g = 1.5$$, $$n_w = 1.33$$ and the value from $$(3)$$:

$$\frac{1}{f_w} = \left(\frac{1.5}{1.33} - 1\right)\left(\frac{1}{12}\right)$$

Compute the bracket:

$$\frac{1.5}{1.33} = \frac{150}{133} \approx 1.1278$$
$$\frac{1.5}{1.33} - 1 \approx 0.1278$$

Hence

$$\frac{1}{f_w} \approx 0.1278 \times \frac{1}{12} = \frac{0.1278}{12} \approx 0.01065$$

Therefore

$$f_w \approx \frac{1}{0.01065} \approx 93.9 \text{ cm}$$

The nearest value among the options is $$96 \text{ cm}$$.

Final answer: in water the focal length of the lens becomes approximately $$96 \text{ cm}$$ - Option B.

$$\text{Object distance} = -u \quad (u > 0)$$

$$\text{Focal length} = +f$$

$$\frac{1}{v} - \frac{1}{-u} = \frac{1}{f} \implies \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Isolating image distance $$v$$: $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} \implies \frac{1}{v} = \frac{u - f}{uf} \implies v = \frac{uf}{u - f}$$

Given condition $$|u| > f \implies u > f$$:

$$\text{As } u \to f^{+}, \quad v \to +\infty$$

$$\text{As } u \to +\infty, \quad v \to f^{+}$$

For two thin lenses kept in contact, the equivalent focal length $$F$$ is given by the formula
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$

The convex lens has $$f_1 = +30\ \text{cm}$$ and the concave lens has $$f_2 = -20\ \text{cm}$$ (negative because it is concave).

Substituting,
$$\frac{1}{F} = \frac{1}{30} + \frac{1}{-20} = \frac{2}{60} - \frac{3}{60} = -\frac{1}{60}$$
Hence,
$$F = -60\ \text{cm}$$

The combination therefore behaves like a single concave lens of focal length $$60\ \text{cm}$$.

Using the lens formula for a thin lens,
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{F} \quad -(1)$$
where
• $$u$$ = object distance (negative when the object is on the left of the lens)
• $$v$$ = image distance (positive to the right, negative to the left)
• $$F$$ = focal length of the equivalent lens

The object is placed $$20\ \text{cm}$$ to the left, so $$u = -20\ \text{cm}$$. The focal length is $$F = -60\ \text{cm}$$. Substituting in $$(1)$$:

$$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-60}$$
$$\frac{1}{v} + \frac{1}{20} = -\frac{1}{60}$$

Move the second term to the right side:
$$\frac{1}{v} = -\frac{1}{60} - \frac{1}{20}$$

Convert to a common denominator of $$60$$:
$$\frac{1}{v} = -\frac{1}{60} - \frac{3}{60} = -\frac{4}{60} = -\frac{1}{15}$$

Therefore,
$$v = -15\ \text{cm}$$

The negative sign shows the image forms on the same side as the object (to the left of the lens system). The distance from the lens is the magnitude:

Image distance = $$15\ \text{cm}$$.

Hence, the correct option is Option D (15 cm).

The photograph shows an area of $$400\;{\rm km^{2}}$$ on the ground. If this area is square, its side is the square-root of its area.

Side of landscape, $$L = \sqrt{400\;{\rm km^{2}}} = 20\;{\rm km}$$.

The film is a square of side $$l = 2\;{\rm cm} = 0.02\;{\rm m}$$.

Linear magnification of a lens is $$m = \frac{\text{image size}}{\text{object size}} = \frac{l}{L}$$

Substituting the sizes, $$m = \frac{0.02\;{\rm m}}{20\,000\;{\rm m}} = 1 \times 10^{-6}$$.

Magnification is also $$m = \frac{v}{u}$$, where   $$u$$ = object distance = height of drone $$= 18\;{\rm km}=18\,000\;{\rm m}$$,   $$v$$ = image distance (distance of film from lens).

Therefore $$v = m\,u = 1\times10^{-6}\times 18\,000 = 0.018\;{\rm m}=1.8\;{\rm cm}$$.

For a very distant object ($$u \gg f$$), the lens formula $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ gives $$v \approx f$$ because $$\frac{1}{u}$$ is negligible compared with $$\frac{1}{v}$$.

Hence focal length, $$f \approx v = 1.8\;{\rm cm}$$.

Answer: $$1.8\;{\rm cm}$$  (Option A)

For the first planoconvex lens in air:

$$\mu_g = 1.5, \quad \mu_{\text{air}} = 1, \quad R_1 = 2\text{ cm}, \quad R_2 = \infty$$

$$\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{2} - \frac{1}{\infty}\right) \implies \frac{1}{f_1} = 0.5 \times \frac{1}{2} = \frac{1}{4}\text{ cm}^{-1} \implies f_1 = 4\text{ cm}$$

For the second planoconvex lens in liquid:

$$\mu_g = 1.5, \quad \mu_l = 1.2, \quad R_3 = 3\text{ cm}, \quad R_4 = \infty$$

$$\frac{1}{f_2} = \left(\frac{1.5}{1.2} - 1\right)\left(\frac{1}{3} - \frac{1}{\infty}\right) \implies \frac{1}{f_2} = (1.25 - 1) \times \frac{1}{3} = 0.25 \times \frac{1}{3} = \frac{1}{12}\text{ cm}^{-1} \implies f_2 = 12\text{ cm}$$

Ratio of focal lengths:

$$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$$

The assertion states: “Refractive index of glass is higher than that of air.”
The refractive index $$n$$ of a medium is defined by $$n = \frac{c}{v}$$, where $$c$$ is the speed of light in vacuum and $$v$$ is the speed of light in the medium. For air, $$v$$ is almost equal to $$c$$, so $$n_{\text{air}}\approx 1.0003$$. For ordinary crown glass, $$v\approx 2\times 10^{8}\,\text{m s}^{-1}$$, so $$n_{\text{glass}}\approx 1.5$$. Hence $$n_{\text{glass}}\gt n_{\text{air}}$$, making the assertion true.

The reason claims: “Optical density of a medium is directly proportionate to its mass density which results in a proportionate refractive index.”
Optical density describes how much a medium slows light, i.e. its refractive index. Mass density is the ratio of mass to volume. There is no fixed proportionality between mass density and refractive index. For example, turpentine has a higher refractive index than water even though its mass density is lower. Therefore the reason is false.

Conclusion: The assertion (A) is correct, but the reason (R) is not correct.
Hence the appropriate choice is Option C.

Radius of curvature of each mirror is $$R = 12\ \text{cm}$$. For a spherical mirror the focal length is given by $$f = \frac{R}{2}$$.

Therefore $$f = 6\ \text{cm}$$ in magnitude.
  • For the concave mirror, the focus lies on the object side, so $$f = -6\ \text{cm}$$.
  • For the convex mirror, the focus lies behind the mirror, so $$f = +6\ \text{cm}$$.

The objects are placed at the same distance in front of each mirror: $$u = -18\ \text{cm}$$ (negative by the new Cartesian sign convention).

Mirror formula: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$.

Substituting $$f = -6$$ and $$u = -18$$:
$$\frac{1}{-6} = \frac{1}{v} + \frac{1}{-18}$$

$$\frac{1}{v} = -\frac{1}{6} + \frac{1}{18} = -\frac{3}{18} + \frac{1}{18} = -\frac{2}{18} = -\frac{1}{9}$$

$$v = -9\ \text{cm}$$ (image is real and formed in front of the mirror).

Linear magnification: $$m_c = -\frac{v}{u} = -\frac{-9}{-18} = -\frac{1}{2}$$

Magnitude of image size for concave mirror: $$|m_c| = \frac{1}{2}$$.

Using $$f = +6$$ and $$u = -18$$ in the mirror formula:
$$\frac{1}{6} = \frac{1}{v} + \frac{1}{-18}$$

$$\frac{1}{v} = \frac{1}{6} + \frac{1}{18} = \frac{3}{18} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}$$

$$v = \frac{9}{2}\ \text{cm} = 4.5\ \text{cm}$$ (image is virtual, behind the mirror).

Linear magnification: $$m_v = -\frac{v}{u} = -\frac{4.5}{-18} = \frac{4.5}{18} = \frac{1}{4}$$

Magnitude of image size for convex mirror: $$|m_v| = \frac{1}{4}$$.

Ratio of the sizes (convex : concave) is
$$\frac{|m_v|}{|m_c|} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$

Hence the required ratio is $$\mathbf{\frac{1}{2}}$$, which corresponds to Option A.

$$\cos r = \frac{h}{L} \implies L = \frac{h}{\cos r}$$

$$\sin(i - r) = \frac{d}{L} \implies d = L \sin(i - r)$$

$$d = \frac{h \sin(i - r)}{\cos r}$$

For refraction at surface B:

Light travels from air ($$\mu_1 = 1$$) to glass ($$\mu_2 = 1.5$$).

$$u = -\frac{R}{2}, \quad R_{\text{curvature}} = -R$$

$$\frac{1.5}{V_B} - \frac{1}{-R/2} = \frac{1.5 - 1}{-R}$$

$$\frac{1.5}{V_B} + \frac{2}{R} = -\frac{0.5}{R}$$

$$\frac{1.5}{V_B} = -\frac{2.5}{R} \implies V_B = -0.6R$$

For refraction at surface A:

Light travels from air ($$\mu_1 = 1$$) to glass ($$\mu_2 = 1.5$$).

$$u = -\left(R + \frac{R}{2}\right) = -\frac{3R}{2}, \quad R_{\text{curvature}} = -R$$

$$\frac{1.5}{V_A} - \frac{1}{-3R/2} = \frac{1.5 - 1}{-R}$$

$$\frac{1.5}{V_A} + \frac{2}{3R} = -\frac{0.5}{R}$$

$$\frac{1.5}{V_A} = -\frac{7}{6R} \implies V_A = -\frac{9}{7}R \approx -1.286R$$

Distance between the two images measured from a common reference point:

$$\text{Separation} = \text{Total Distance } (AB) - |V_A| - |V_B|$$

$$\text{Separation} = 2R - \frac{9}{7}R - 0.6R$$

$$\text{Separation} = 2R - 1.286R - 0.6R = 0.114R$$

The focal length (and therefore the power) of a thin lens is given by the Lens-Maker’s formula

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where $$\mu$$ is the refractive index of the lens material, $$R_1$$ is the radius of curvature of the first surface (positive for a convex surface facing the incident light) and $$R_2$$ is the radius of curvature of the second surface (negative for the outgoing convex surface of a bi-convex lens).

Case 1: Given bi-convex lens with equal radii
$$R_1 = +\frac16 \text{ cm}, \; R_2 = -\frac16 \text{ cm}$$ (sign taken according to the convention).

Curvature term for this lens:
$$\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{\frac16} - \frac{1}{-\frac16} = 6 - (-6) = 12 \; \text{cm}^{-1}$$

Therefore the power of the original lens is

$$P = \frac{1}{f} = (\mu - 1)\times 12 \; \text{cm}^{-1}$$ $$-(1)$$

Case 2: New convex lens with unequal radii $$R_1$$ and $$R_2$$ must satisfy the same curvature term so that its power remains unchanged:

$$\frac{1}{R_1} - \frac{1}{R_2} = 12 \; \text{cm}^{-1}$$ $$-(2)$$

Now test the four given pairs (remember the second radius must be taken negative for the second convex surface):

Option A: $$R_1 = \frac13 \text{ cm}, R_2 = -\frac13 \text{ cm}$$
$$\frac{1}{R_1} - \frac{1}{R_2} = 3 - (-3) = 6 \neq 12$$

Option B: $$R_1 = \frac15 \text{ cm}, R_2 = -\frac17 \text{ cm}$$
$$\frac{1}{R_1} - \frac{1}{R_2} = 5 - (-7) = 12$$  ✓  satisfies $$(2)$$

Option C: $$R_1 = \frac13 \text{ cm}, R_2 = -\frac17 \text{ cm}$$
$$3 - (-7) = 10 \neq 12$$

Option D: $$R_1 = \frac16 \text{ cm}, R_2 = -\frac19 \text{ cm}$$
$$6 - (-9) = 15 \neq 12$$

Only Option B preserves the value of $$\frac{1}{R_1} - \frac{1}{R_2}$$ and hence the power.

Therefore the required radii of curvature are $$\frac15 \text{ cm}$$ and $$\frac17 \text{ cm}$$, i.e. Option B.

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

$$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{+0.4}$$

$$\frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4}$$

$$\frac{1.5}{v} + 5 = 1.25$$

$$v = -\frac{2}{5} = -0.4\text{ m}$$

Since the calculated value of $$v$$ is negative, the image is formed on the same side as the incoming light. This means the image is located 0.4 m to the left of the spherical surface (forming a virtual image).

Taking downward direction as positive:

$$P_1 = \left(\frac{4}{3} - 1\right)\left(\frac{1}{\infty} - \frac{1}{-|R_1|}\right) \implies P_1 = \frac{1}{3|R_1|}$$

$$P_2 = \left(\frac{3}{2} - 1\right)\left(\frac{1}{-|R_1|} - \frac{1}{-|R_2|}\right) \implies P_2 = \frac{1}{2}\left(\frac{1}{|R_2|} - \frac{1}{|R_1|}\right)$$

$$P_3 = \left(\frac{4}{3} - 1\right)\left(\frac{1}{-|R_2|} - \frac{1}{\infty}\right) \implies P_3 = -\frac{1}{3|R_2|}$$

Equivalent power:

$$P_{\text{eq}} = P_1 + P_2 + P_3 \implies P_{\text{eq}} = \frac{1}{3|R_1|} + \frac{1}{2|R_2|} - \frac{1}{2|R_1|} - \frac{1}{3|R_2|}$$

$$P_{\text{eq}} = -\frac{1}{6|R_1|} + \frac{1}{6|R_2|} \implies P_{\text{eq}} = -\frac{1}{6}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)$$

The thin lens is placed in air, so we use the lens-maker’s formula

$$\frac{1}{f} \;=\; (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where
$$f$$ = focal length of the lens,
$$n$$ = refractive index of lens material with respect to air,
$$R_1$$ = radius of curvature of the first surface (the surface that light meets first),
$$R_2$$ = radius of curvature of the second surface.

Sign convention (Cartesian):
• Distances measured to the right of the surface are positive.
• For a double-convex lens, the first surface bulges toward the left (center of curvature on the right), so $$R_1 \gt 0$$.
• The second surface bulges toward the right (center of curvature on the left), so $$R_2 \lt 0$$.

Given magnitudes:
$$|R_1| = 10 \text{ cm},\; |R_2| = 15 \text{ cm},\; f = 12 \text{ cm}$$

Applying the signs:
$$R_1 = +10 \text{ cm},\; R_2 = -15 \text{ cm}$$

Substitute into the lens-maker’s formula:

$$\frac{1}{12} = (n-1)\left(\frac{1}{10} - \frac{1}{-15}\right)$$

Simplify the bracket first:

$$\frac{1}{10} - \left(-\frac{1}{15}\right) = \frac{1}{10} + \frac{1}{15}$$

Take the LCM (30):

$$\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$$

Thus

$$\frac{1}{12} = (n-1)\left(\frac{1}{6}\right)$$

Cross-multiply:

$$(n-1) = \frac{1}{12} \times 6 = \frac{6}{12} = 0.5$$

Add 1 on both sides:

$$n = 1 + 0.5 = 1.5$$

Therefore, the refractive index of the lens material is $$1.5$$.

Option C is correct.

The object is placed on the principal axis $$30 \text{ cm}$$ in front of a convex mirror of focal length $$f = +30 \text{ cm}$$ (positive for a convex mirror in the new Cartesian sign convention).

Using the mirror formula $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$, with object distance $$u = -30 \text{ cm}$$, we get

$$\frac{1}{30} = \frac{1}{v} + \frac{1}{(-30)}$$
$$\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$$
$$\Rightarrow \; v = +15 \text{ cm}$$

The positive sign means the convex mirror forms a virtual image $$I_2$$ $$15 \text{ cm}$$ behind the mirror (on the right side of the pole).

Now place a plane mirror between the object and the convex mirror. Let the distance between the two mirrors be $$d \text{ cm}$$. With the pole of the convex mirror as origin, the coordinates along the principal axis are:

• Pole of convex mirror: $$x = 0$$
• Plane mirror surface: $$x = -d$$
• Object: $$x = -30 \text{ cm}$$

The object is $$30 - d \text{ cm}$$ in front of the plane mirror, so the plane mirror produces a virtual image $$I_1$$ the same distance behind it:

Coordinate of $$I_1$$:
$$x(I_1) = (-d) + (30 - d) = 30 - 2d$$

For the two images to coincide, $$I_1$$ and $$I_2$$ must lie at the same coordinate:

$$30 - 2d = +15$$
$$2d = 30 - 15 = 15$$
$$\therefore \; d = 7.5 \text{ cm}$$

Hence, the distance between the convex mirror and the plane mirror must be $$7.5 \text{ cm}$$.

Option B is correct.

$$\sin \theta_C = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \frac{\mu_1}{\mu_2}$$

$$\sin \theta_C = \frac{1.0}{1.25} = \frac{4}{5}$$

$$\cos \theta_C = \sqrt{1 - \sin^2 \theta_C} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$$

Using the geometric relation $$r = 90^\circ - \theta_C$$: $$\sin r = \sin(90^\circ - \theta_C) = \cos \theta_C = \frac{3}{5}$$

$$\mu_1 \sin \theta = \mu_2 \sin r$$

$$1.0 \times \sin \theta = 1.25 \times \frac{3}{5}$$

$$\theta = \sin^{-1}\left(\frac{3}{4}\right)$$

For a prism of refracting angle $$A$$ made of material having refractive index $$n$$, when the deviation is minimum the ray of light follows a symmetrical path.

Conditions at minimum deviation:
(i) Angle of incidence $$i_1$$ equals angle of emergence $$i_2$$.
(ii) Angle of refraction at first surface $$r_1$$ equals angle of refraction at second surface $$r_2$$, say $$r$$.
(iii) From the geometry of the prism, $$r_1 + r_2 = A$$, therefore at minimum deviation $$2r = A \Rightarrow r = \dfrac{A}{2}$$.

The deviation-incidence graph is a U-shaped curve with its lowest point at the minimum deviation $$D_m$$. Using Snell’s law at minimum deviation one writes

$$n = \dfrac{\sin\left(\dfrac{A + D_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$$ $$-(1)$$

Now each statement can be checked one by one.

Statement A: “The refracted ray inside prism becomes parallel to the base.”
In the isosceles triangle formed by the two refracting faces and the base, the internal ray makes the angle $$r = A/2$$ with each face. Hence it is parallel to the side opposite the apex, i.e. the base. So Statement A is TRUE.

Statement B: “Larger angle prisms provide smaller angle of minimum deviation.”
From equation $$(1)$$, keeping $$n$$ fixed, increase in $$A$$ increases the numerator $$\sin\bigl(\tfrac{A+D_m}{2}\bigr)$$ as well as the denominator $$\sin\bigl(\tfrac{A}{2}\bigr)$$, but the net result is that $$D_m$$ rises with $$A$$. A quick numerical check with $$n = 1.50$$ gives
for $$A = 30^\circ,\; D_m \approx 15.7^\circ$$;
for $$A = 60^\circ,\; D_m \approx 37.2^\circ$$.
Thus larger $$A$$ leads to larger, not smaller, $$D_m$$. So Statement B is FALSE.

Statement C: “Angle of incidence and angle of emergence become equal.”
This is the defining condition of minimum deviation (symmetrical path). Hence Statement C is TRUE.

Statement D: “There are always two sets of angle of incidence for which deviation will be the same except at minimum deviation setting.”
Because the deviation curve is symmetric about the minimum, every value of deviation greater than $$D_m$$ is obtained for two equal and opposite departures of $$i$$ from the symmetry point. Therefore Statement D is TRUE.

Statement E: “Angle of refraction becomes double of prism angle.”
We have already derived $$r = \dfrac{A}{2}$$, not $$2A$$. Hence Statement E is FALSE.

The TRUE statements are A, C, and D only.

Correct choice: Option A (A, C and D Only).

Take the pole of the spherical surface as the origin and measure all distances from this point along the principal axis (Cartesian sign convention).

• Incident medium (air): $$n_1 = 1$$.
• Refracting medium (glass): $$n_2 = 1.5$$.
• Radius of curvature: the centre of curvature lies inside the glass, i.e. on the side to which light travels, so $$R = +50\,\text{cm}$$.
• Image is formed inside the glass at a distance $$v = +200\,\text{cm}$$ (positive because it is on the side of the refracted light).

For refraction at a spherical surface, the Cartesian-form equation is

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \quad -(1)$$

Substitute the given data into $$(1)$$:

$$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50}$$

Simplify each term:
$$\frac{1.5}{200} = 0.0075$$,
$$\frac{1.5 - 1}{50} = \frac{0.5}{50} = 0.01$$.

Hence,

$$0.0075 - \frac{1}{u} = 0.01$$

Rearrange to solve for $$\frac{1}{u}$$:

$$-\frac{1}{u} = 0.01 - 0.0075 = 0.0025$$

Therefore,

$$\frac{1}{u} = -0.0025 \;\Rightarrow\; u = -\frac{1}{0.0025} = -400\,\text{cm}$$

The negative sign confirms that the object (light source) is on the side opposite to the direction of positive measurement, i.e. in air in front of the surface.

Magnitude of the object distance:
$$|u| = 400\,\text{cm} = 4\,\text{m}$$.

Hence, the light source is $$4\,\text{m}$$ in front of the glass surface.

We are given a convex mirror with radius of curvature $$R = 2$$ m, so the focal length is $$f = +\frac{R}{2} = +1$$ m (positive for convex mirror by sign convention).

An object (car) approaches from behind at uniform speed $$v_0 = 90$$ km/hr $$= 25$$ m/s. When the car is at a distance of 24 m, the object distance is $$u = -24$$ m (negative by sign convention, since the object is in front of the mirror).

Using the mirror formula: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1} - \frac{1}{-24} = 1 + \frac{1}{24} = \frac{25}{24}$$

So $$v = \frac{24}{25}$$ m (positive, confirming virtual image behind the mirror).

To find the acceleration of the image, we use the relation between image position and object position. From the mirror formula:

$$v = \frac{uf}{u - f} = \frac{u \cdot 1}{u - 1} = \frac{u}{u - 1}$$

Differentiating $$v$$ with respect to $$u$$:

$$\frac{dv}{du} = \frac{(u - 1) - u}{(u - 1)^2} = \frac{-1}{(u - 1)^2}$$

Differentiating again:

$$\frac{d^2v}{du^2} = \frac{2}{(u - 1)^3}$$

The velocity of the image is:

$$\frac{dv}{dt} = \frac{dv}{du} \cdot \frac{du}{dt}$$

The acceleration of the image is:

$$\frac{d^2v}{dt^2} = \frac{d^2v}{du^2} \left(\frac{du}{dt}\right)^2 + \frac{dv}{du} \cdot \frac{d^2u}{dt^2}$$

Since the object moves with uniform speed, $$\frac{d^2u}{dt^2} = 0$$. The object approaches the mirror, so $$u$$ increases from $$-24$$ toward $$0$$. Therefore $$\frac{du}{dt} = +25$$ m/s.

At $$u = -24$$:

$$\frac{d^2v}{du^2} = \frac{2}{(-24 - 1)^3} = \frac{2}{(-25)^3} = \frac{2}{-15625} = -\frac{2}{15625}$$

$$\frac{d^2v}{dt^2} = -\frac{2}{15625} \times (25)^2 = -\frac{2 \times 625}{15625} = -\frac{1250}{15625} = -\frac{2}{25}$$

The magnitude of the acceleration of the image is:

$$|a| = \frac{2}{25}$$ m/s$$^2$$

Therefore:

$$100a = 100 \times \frac{2}{25} = 8$$

The answer is $$8$$.

The bottom of the vessel is first covered by liquid 1 of refractive index $$\mu_1 = 1.2$$ and thickness $$t_1 = 60\; \text{cm}$$. Above it a second liquid of refractive index $$\mu_2 = 1.6$$ and thickness $$t_2 = H$$ is poured. The observer is in air, whose refractive index is $$\mu_a = 1$$.

Step 1 : Apparent depth through one plane layer
For an observer in a medium of refractive index $$\mu_o$$ looking normally through a layer of thickness $$t$$ and refractive index $$\mu$$, the apparent thickness is $$t_{\text{app}} = t \;\frac{\mu_o}{\mu}$$ $$-(1)$$ (The small-angle relation $$\mu_o\,\theta_o = \mu\,\theta$$ at a plane interface leads to this result.)

Step 2 : Apparent depth of the bottom
The light from the bottom travels successively through the two liquids, so the total apparent depth in air is the sum of the apparent thicknesses of each layer: $$d_{\text{app}} = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2}$$ Substituting $$t_1 = 60\;\text{cm}$$ and $$t_2 = H$$, $$d_{\text{app}} = \frac{60}{1.2} + \frac{H}{1.6} = 50 + \frac{H}{1.6}$$ $$-(2)$$

Step 3 : Relating real depth, apparent depth and shift
Real depth of the bottom below the top surface is $$d_{\text{real}} = t_1 + t_2 = 60 + H$$ $$-(3)$$
Given apparent shift $$S = 40\;\text{cm}$$, $$S = d_{\text{real}} - d_{\text{app}}$$ $$-(4)$$

Step 4 : Insert the expressions and solve for $$H$$
Using $$(2)$$ and $$(3)$$ in $$(4)$$: $$40 = (60 + H) - \left(50 + \frac{H}{1.6}\right)$$ $$40 = 10 + H - \frac{H}{1.6}$$ Bring the constant to the left: $$30 = H\left(1 - \frac{1}{1.6}\right)$$ Since $$1 - \frac{1}{1.6} = 1 - 0.625 = 0.375 = \frac{3}{8}$$, $$30 = H \times \frac{3}{8}$$ $$H = 30 \times \frac{8}{3} = 80\;\text{cm}$$.

Hence the height of the second liquid is 80 cm.

Solution :

For prism at minimum deviation :

$$\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

Given :

$$A = 60^\circ$$

$$\mu = \sqrt2$$

Therefore,

$$\sqrt2 = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\sin30^\circ}$$

$$\sqrt2 = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{1/2}$$

$$\sin\left(\frac{60^\circ+\delta_m}{2}\right) = \frac{1}{\sqrt2}$$

$$= \sin45^\circ$$

Therefore,

$$\frac{60^\circ+\delta_m}{2} = 45^\circ$$

$$60^\circ+\delta_m = 90^\circ$$

$$\delta_m = 30^\circ$$

At minimum deviation :

$$i = e$$

and

$$\delta_m = i+e-A$$

$$= 2i-A$$

Therefore,

$$30^\circ = 2i-60^\circ$$

$$2i = 90^\circ$$

$$i = 45^\circ$$

Final Answer :

$$45^\circ$$

For a spherical mirror the linear magnification is given by
$$m = -\dfrac{v}{u}$$
where $$u$$ is the object distance (from the pole) and $$v$$ is the image distance. Distances are measured from the pole, positive to the right (toward the reflecting side), negative to the left (toward the incident light).

The magnification in the question is $$m = -\dfrac{1}{3}$$. Using the formula:
$$-\dfrac{v}{u} = -\dfrac{1}{3}$$
gives
$$\dfrac{v}{u} = \dfrac{1}{3} \quad -(1)$$

This means the image distance is one-third of the object distance, with the same sign. Let the magnitude of the object distance be $$a$$ centimetres, so
$$u = -a \quad (\text{object is on the left})$$
Then from $$(1)$$:
$$v = -\dfrac{a}{3}$$

The actual separation between the object and its image is the difference of their magnitudes:
$$|\,u - v\,| = \bigl|\,(-a) - \!\left(-\dfrac{a}{3}\right)\bigr| = a - \dfrac{a}{3} = \dfrac{2a}{3}$$

This separation is given to be 30 cm, so
$$\dfrac{2a}{3} = 30 \;\Longrightarrow\; a = 45\text{ cm}$$
Hence
$$u = -45\text{ cm}, \qquad v = -15\text{ cm}$$

Apply the mirror equation
$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$
Substituting $$v = -15$$ cm and $$u = -45$$ cm:
$$\dfrac{1}{f} = \dfrac{1}{-15} + \dfrac{1}{-45} = -\dfrac{1}{15} - \dfrac{1}{45} = -\dfrac{3 + 1}{45} = -\dfrac{4}{45}$$
Therefore
$$f = -\dfrac{45}{4}\text{ cm}$$

The focal length is written in the form $$\dfrac{x}{4}\text{ cm}$$, so $$x = -45$$. The question asks for the magnitude of $$x$$, hence

$$|x| = 45$$

Answer: 45

The angles $$\theta_1$$ and $$\theta_2$$ are given with respect to the surface interface.

$$\text{Angle of incidence}, \quad i_1 = i_2 = 90^\circ - \theta_1$$

$$\theta_1 = \cos^{-1}\left(\frac{n_2}{2n_1}\right) \implies \cos\theta_1 = \frac{n_2}{2n_1} \implies \sin i_1 = \frac{n_2}{2n_1}$$

Applying Snell's law at the first interface:

$$n_1 \sin i_1 = n_2 \sin r_1 \implies n_1 \left(\frac{n_2}{2n_1}\right) = n_2 \sin r_1$$

$$\frac{n_2}{2} = n_2 \sin r_1 \implies \sin r_1 = \frac{1}{2} \implies r_1 = 30^\circ$$

Let $$t$$ be the thickness of the block. From the symmetric ray geometry converging at point 3:

$$\tan r_1 = \frac{d/2}{t}$$

$$\tan(30^\circ) = \frac{4\sqrt{3}/2}{t} \implies \frac{1}{\sqrt{3}} = \frac{2\sqrt{3}}{t}$$

$$t = 2\sqrt{3} \times \sqrt{3} = 6\text{ cm}$$

Concept:

• A ray passing through the focus of a convex lens emerges parallel.
• Parallel rays entering a convex lens converge at its focus.

Condition for parallel output:
For rays emerging from $$L_1$$​ to become parallel after $$L_2$$, the image formed by $$L_1$$​ must lie at the focus of L_2​.

Hence, separation between lenses:

D = $$f_1 + f_2$$ = 10 + 15 = 25

Final Answer:

25 cm

The critical angle is $$45°$$. Find the ratio of refractive indices $$n_1 : n_2$$.

Total internal reflection occurs when light travels from a denser to a rarer medium. The critical angle $$C$$ is given by:

$$\sin C = \frac{n_2}{n_1}$$

where $$n_1$$ is the refractive index of the denser (first) medium and $$n_2$$ is that of the rarer (second) medium.

$$\sin 45° = \frac{n_2}{n_1} \implies \frac{1}{\sqrt{2}} = \frac{n_2}{n_1}$$

$$\frac{n_1}{n_2} = \sqrt{2}$$

$$n_1 : n_2 = \sqrt{2} : 1$$

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

$$\text{First lens: } u_1 = -30\text{ cm},\ f_1 = +10\text{ cm} \implies \frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \implies v_1 = +15\text{ cm}$$

$$\text{Second lens: } u_2 = +(15 - 5) = +10\text{ cm},\ f_2 = -10\text{ cm} \implies \frac{1}{v_2} - \frac{1}{10} = \frac{1}{-10} \implies v_2 = \infty$$

$$\text{Third lens: } u_3 = \infty,\ f_3 = +30\text{ cm} \implies \frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30} \implies v_3 = +30\text{ cm}$$

$$R = +30\text{ cm} \implies f = \frac{R}{2} = +15\text{ cm}$$

$$m = -\frac{v}{u} = +\frac{1}{2} \implies v = -\frac{u}{2}$$

Mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

$$\frac{1}{15} = \frac{1}{\left(-\frac{u}{2}\right)} + \frac{1}{u} = -\frac{2}{u} + \frac{1}{u} = -\frac{1}{u}$$

$$u = -15\text{ cm}$$

So distance will be 15cm 

We need to find the refractive index of the lens material given the radii of curvature and focal length.

The Lensmaker's equation relates the focal length of a thin lens to the radii of curvature and the refractive index:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a convex lens, $$R_1 = +15$$ cm (first surface is convex, center of curvature on the right), $$R_2 = -30$$ cm (second surface is convex, center of curvature on the left, so negative by sign convention), and $$f = 20$$ cm.

Substituting these values into the Lensmaker's equation gives:

$$\frac{1}{20} = (\mu - 1)\left(\frac{1}{15} - \frac{1}{-30}\right) = (\mu - 1)\left(\frac{1}{15} + \frac{1}{30}\right)$$

The term in parentheses simplifies as follows:

$$\frac{1}{15} + \frac{1}{30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$$

Substituting this back yields:

$$\frac{1}{20} = (\mu - 1)\times\frac{1}{10}$$

Solving for the refractive index:

$$\mu - 1 = \frac{10}{20} = \frac{1}{2} = 0.5$$

$$\mu = 1.5$$

The correct answer is Option C: 1.5.

Statement I: A concave lens is a diverging lens. For any real object position, it forms a diminished image located between the optical center and the principal focus on the same side as the object. It does not form an image at the center of curvature on the other side $$\implies \text{Statement I is false}$$.

Statement II: For any real object, a concave lens always diverges the incident light rays, forming an image from back-projected virtual rays $$\implies \text{Statement II is true}$$.

We need to evaluate two statements about the dispersion of white light through a prism.

Analysis of Statement I: "When white light passes through a prism, the red light bends lesser than yellow and violet."

This statement is true. When white light undergoes dispersion through a prism, different colours bend (refract) by different amounts. The degree of bending depends on the refractive index, which in turn depends on the wavelength. Red light has the longest wavelength in the visible spectrum and therefore has the smallest refractive index in a dispersive medium. Since a smaller refractive index means less bending (by Snell's law: $$n_1 \sin \theta_1 = n_2 \sin \theta_2$$), red light bends the least. The order of increasing bending is: red, orange, yellow, green, blue, indigo, violet (ROYGBIV).

Analysis of Statement II: "The refractive indices are different for different wavelengths in dispersive medium."

This statement is also true. This is the fundamental property that defines a dispersive medium. In such a medium, the refractive index is a function of wavelength, described by Cauchy's equation:

$$n(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + \ldots$$

where $$A$$, $$B$$, $$C$$ are material-dependent constants. Since $$n$$ decreases with increasing $$\lambda$$, violet light (shortest wavelength) has the highest refractive index and red light (longest wavelength) has the lowest. This variation of refractive index with wavelength is the cause of dispersion.

Conclusion:

Both statements are true, and Statement II provides the underlying reason for the phenomenon described in Statement I.

The correct answer is Option (3): Both Statement I and Statement II are true.

$$|m| = 2$$ (Hence, mirror is concave)

$$\text{Distance between object and image} = 15\text{ cm}$$

For a virtual image in a mirror, the image is formed behind the mirror while the object is in front:

$$\text{Distance} = |v| + |u| = 15\text{ cm}$$

Magnification relation for mirrors: $$m = -\frac{v}{u} \implies |v| = |m||u| = 2|u|$$

$$2|u| + |u| = 15 \implies 3|u| = 15 \implies |u| = 5\text{ cm}$$

$$|v| = 2 \times 5 = 10\text{ cm}$$

$$u = -5\text{ cm},\quad v = +10\text{ cm}$$

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = -\frac{1}{10} \implies f = -10\text{ cm}$$

$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\mu = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \implies \cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$

$$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right) \implies \frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_m}{2}$$

$$\pi - A = A + \delta_m \implies \delta_m = \pi - 2A$$

We need to find the focal length of a biconvex lens when immersed in a liquid, given its focal length in air.

For a biconvex lens with equal radii of curvature $$R$$:

$$\frac{1}{f} = (\mu_{rel} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu_{rel} - 1)\frac{2}{R}$$

where $$\mu_{rel}$$ is the refractive index of the lens relative to the surrounding medium.

In air ($$\mu_{medium} = 1$$), $$\mu_{rel} = 1.5/1 = 1.5$$:

$$\frac{1}{20} = (1.5 - 1)\frac{2}{R} = 0.5 \times \frac{2}{R} = \frac{1}{R}$$

$$R = 20 \text{ cm}$$

In liquid ($$\mu_{liquid} = 1.6$$), $$\mu_{rel} = 1.5/1.6 = 15/16$$:

$$\frac{1}{f'} = \left(\frac{15}{16} - 1\right)\frac{2}{20} = \left(-\frac{1}{16}\right)\frac{1}{10} = -\frac{1}{160}$$

$$f' = -160 \text{ cm}$$

The negative sign indicates that the lens behaves as a diverging lens in this liquid. This happens because the refractive index of the liquid (1.6) is greater than that of the lens (1.5), so the lens bends light in the opposite direction compared to air.

The correct answer is Option (2): $$-160$$ cm.

For lenses in contact, the effective power is the sum of individual powers:

$$P_{total} = 5P$$ where P is the power of each lens.

$$25 = 5P \Rightarrow P = 5$$ D

Focal length: $$f = \frac{1}{P} = \frac{1}{5}$$ m = 20 cm

The correct answer is Option 1: 20 cm.

A prism has apex angle $$A$$ and refractive index $$\mu = \cot\frac{A}{2}$$. We need to find the angle of minimum deviation $$\delta_m$$.

We begin by recalling the prism formula for minimum deviation.

For a prism, the refractive index is related to the apex angle and minimum deviation by:

$$ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Next, we substitute $$\mu = \cot\frac{A}{2}$$ into this formula:

$$ \cot\frac{A}{2} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Since $$\cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$$, we can write:

$$ \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} $$

Multiplying both sides by $$\sin\frac{A}{2}$$ yields:

$$ \cos\frac{A}{2} = \sin\left(\frac{A + \delta_m}{2}\right) $$

Then, using the identity $$\cos\theta = \sin\left(90° - \theta\right)$$, we have:

$$ \sin\left(90° - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right) $$

Equating the arguments leads to:

$$ 90° - \frac{A}{2} = \frac{A + \delta_m}{2} $$

Multiplying both sides by 2 gives:

$$ 180° - A = A + \delta_m $$

Solving for $$\delta_m$$, we find:

$$ \delta_m = 180° - 2A $$

The correct answer is Option (4): $$\delta_m = 180° - 2A$$.

Using the power of a combined silvered thin lens system:

$$\frac{1}{F} = 2\left(\frac{n - 1}{R_1} + \frac{\mu_g - n}{R_2}\right) + \frac{1}{f_m}$$

Given parameters: flat air-liquid interface ($$R_1 = \infty$$), liquid-glass convex interface ($$R_2 = +9\text{ cm}$$), and flat mirror base ($$f_m = \infty$$):

$$\frac{1}{F} = 2\left(0 + \frac{1.60 - n}{9}\right) + 0 = \frac{2(1.60 - n)}{9}$$

Using the coincidence condition where the image reflects back to the object position ($$u = -h$$, $$v = -h$$):

$$\frac{1}{-h} + \frac{1}{-h} = -\frac{1}{F} \implies \frac{2}{h} = \frac{2(1.60 - n)}{9}$$

$$h = \frac{9}{1.60 - n}$$

For Option (A): $$n = 1.42 \implies h = \frac{9}{1.60 - 1.42} = \frac{9}{0.18} = 50\text{ cm}$$

For Option (B): $$n = 1.35 \implies h = \frac{9}{1.60 - 1.35} = \frac{9}{0.25} = 36\text{ cm}$$

For Option (C): $$n = 1.45 \implies h = \frac{9}{1.60 - 1.45} = \frac{9}{0.15} = 60\text{ cm}$$

For Option (D): $$n = 1.48 \implies h = \frac{9}{1.60 - 1.48} = \frac{9}{0.12} = 75\text{ cm}$$

For each equilateral-triangular prism the refracting (apex) angle is $$A = 60^\circ = \dfrac{\pi}{3}$$.

Step 1: Condition for minimum deviation in prism $$P_2$$
For minimum deviation in any prism, the path of the ray is symmetrical, so $$r = r' = \dfrac{A}{2} = 30^\circ$$ inside the prism, and the angles of incidence and emergence are equal: $$i = e$$.

Applying Snell’s law at the first face of $$P_2$$ (outside medium is vacuum, $$n = 1$$):
$$\sin i = n_2 \sin r$$

Given $$n_2 = \sqrt{3}$$, therefore
$$\sin i = \sqrt{3}\,\sin 30^\circ = \sqrt{3}\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}}{2}$$
so $$i = 60^\circ = \dfrac{\pi}{3}$$.

Hence the light ray must strike the first face of $$P_2$$ with an incidence angle of $$60^\circ$$.

Step 2: This same ray is the emergent ray from prism $$P_1$$
The two prisms are arranged with their corresponding faces parallel; therefore the angle the ray makes with the normal when it leaves $$P_1$$ is the same $$60^\circ$$. Let that angle of emergence from $$P_1$$ be $$e_1 = 60^\circ$$.

Step 3: Ray path inside prism $$P_1$$
Let the angles of refraction at the two faces of $$P_1$$ be $$r_1$$ and $$r_2$$. For any prism
$$r_1 + r_2 = A = 60^\circ \qquad -(1)$$

Snell’s law at the second face (emergence) of $$P_1$$:
$$n_1 \sin r_2 = \sin e_1$$
Given $$e_1 = 60^\circ$$ and $$n_1 = \sqrt{\dfrac{3}{2}}$$, we get
$$\sin r_2 = \dfrac{\sin 60^\circ}{n_1} = \dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{\dfrac{3}{2}}} = \dfrac{\sqrt{2}}{2} = \sin 45^\circ$$
so $$r_2 = 45^\circ$$.

Using $$(1)$$:
$$r_1 = 60^\circ - 45^\circ = 15^\circ.$$

Step 4: Incidence angle $$\theta$$ on the first face of $$P_1$$
Apply Snell’s law at the first face of $$P_1$$:
$$\sin \theta = n_1 \sin r_1 = \sqrt{\dfrac{3}{2}}\;\sin 15^\circ$$

Since $$15^\circ = \dfrac{\pi}{12}$$,
$$\sin \theta = \sqrt{\dfrac{3}{2}}\;\sin\!\left(\dfrac{\pi}{12}\right).$$

The question states this result as $$\theta = \sin^{-1}\!\left[\sqrt{\dfrac{3}{2}}\;\sin\!\left(\dfrac{\pi}{\beta}\right)\right].$$ Comparing, we must have $$\dfrac{\pi}{\beta} = \dfrac{\pi}{12}\quad\Longrightarrow\quad \beta = 12.$$

Hence the required value is 12.

The thickness of the glass slab is $$t = 4\sqrt{3}\,\text{cm}$$ and its refractive index with respect to air is $$\mu = \sqrt{2}$$.

Step 1: Find the critical angle
For a glass-air boundary, the critical angle $$c$$ satisfies
$$\sin c = \frac{\mu_{\text{air}}}{\mu_{\text{glass}}} = \frac{1}{\sqrt{2}} \quad -(1)$$
Hence $$c = 45^{\circ}$$ because $$\sin 45^{\circ} = \frac{\sqrt{2}}{2}$$.

The question states that the light ray is incident on the slab at an angle equal to this critical angle, so
$$i = c = 45^{\circ}$$.

Step 2: Use Snell’s law to obtain the refraction angle inside the slab
Snell’s law for the air-glass interface is
$$\mu_{\text{air}}\sin i = \mu_{\text{glass}}\sin r \quad -(2)$$
With $$\mu_{\text{air}} = 1$$ and $$\mu_{\text{glass}} = \sqrt{2}$$, put $$i = 45^{\circ}$$ into $$(2)$$:
$$\sin 45^{\circ} = \sqrt{2}\,\sin r$$
$$\frac{\sqrt{2}}{2} = \sqrt{2}\,\sin r$$
$$\sin r = \frac{1}{2}$$
Therefore $$r = 30^{\circ}$$.

Step 3: Formula for lateral displacement
For a parallel-sided slab, the lateral displacement $$d$$ is
$$d = t\,\frac{\sin (i - r)}{\cos r} \quad -(3)$$

Step 4: Substitute the known values
$$i - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$$
Given $$\sin 15^{\circ} = 0.25$$ and $$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$$, insert all values into $$(3)$$:
$$d = 4\sqrt{3}\;\text{cm}\times \frac{\sin 15^{\circ}}{\cos 30^{\circ}}$$
$$d = 4\sqrt{3}\;\text{cm}\times \frac{0.25}{\frac{\sqrt{3}}{2}}$$
$$d = 4\sqrt{3}\;\text{cm}\times \frac{0.25 \times 2}{\sqrt{3}}$$
$$d = 4\sqrt{3}\;\text{cm}\times \frac{0.5}{\sqrt{3}}$$
$$d = 4 \times 0.5 \;\text{cm} = 2\;\text{cm}$$.

Final Answer:
The lateral displacement of the ray after emerging from the glass slab is $$2\;\text{cm}$$.

We need to find the image distance from the object when light from a point source in air falls on a convex curved surface of glass.

Refraction at a single spherical surface is governed by the formula $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$. Here $$\mu_1$$ is the refractive index of the medium containing the object, $$\mu_2$$ is the refractive index of the medium containing the image, $$u$$ is the object distance, $$v$$ is the image distance, and $$R$$ is the radius of curvature.

We know that the refractive index of air is $$\mu_1 = 1$$ and that of glass is $$\mu_2 = 1.5$$.

Since the object lies on the same side as the incident light, the object distance is $$u = -100\ \text{cm}$$. For a convex surface the center of curvature lies on the transmission side, so $$R = +20\ \text{cm}$$.

Substituting these values into the formula gives $$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$$ which simplifies to $$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}\,.$$

On the right-hand side we have $$\frac{0.5}{20} = 0.025$$ so that $$\frac{1.5}{v} = 0.025 - \frac{1}{100} = 0.025 - 0.01 = 0.015$$ and therefore $$v = \frac{1.5}{0.015} = 100\ \text{cm}\,.$$

The positive value of $$v$$ indicates the image is formed on the transmission side (inside the glass), 100 cm from the surface.

Hence, the distance from the object to the image is $$|u| + v = 100 + 100 = 200\ \text{cm}$$ since they lie on opposite sides of the refracting surface.

The answer is 200 cm.

$$m = \frac{v}{u} \implies 3 = \frac{v}{u} \implies v = 3u$$

For a virtual image in a convex lens: $$\text{Distance} = |v| - |u| = |3u| - |u| = 2|u|$$

$$2|u| = 20 \implies |u| = 10\text{ cm}$$

Using sign convention: $$u = -10\text{ cm} \implies v = -30\text{ cm}$$

Lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

$$\frac{1}{f} = \frac{1}{-30} - \frac{1}{-10} = -\frac{1}{30} + \frac{1}{10} = \frac{2}{30} = \frac{1}{15} \implies f = 15\text{ cm}$$

We need to find the focal length of a convex lens that produces a 2 times magnified real image, with the distance between the object and image being 45 cm.

For a real, inverted image with magnification $$|m| = 2$$, we have $$m = -2$$. Since $$m = \frac{v}{u}$$, it follows that $$v = -2u$$.

Adopting the sign convention by letting the object distance be $$u = -d$$ (negative because the object is on the left), we get $$v = -2(-d) = 2d\,. $$

The distance between object and image is given by $$|v - u| = |2d - (-d)| = 3d = 45\text{ cm}\,, $$ which yields $$d = 15\text{ cm}\,. $$ Hence $$u = -15\text{ cm}$$ and $$v = 30\text{ cm}\,. $$

Substituting these into the lens formula $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30} = \frac{1}{10}$$ gives $$f = 10\text{ cm}\,. $$

The focal length of the lens is $$\boxed{10}$$ cm.

For minimum deviation: $$\mu = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}$$. Given $$\delta_m = A$$:

$$\sqrt{3} = \frac{\sin A}{\sin(A/2)} = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$$.

$$\cos(A/2) = \frac{\sqrt{3}}{2} \Rightarrow A/2 = 30° \Rightarrow A = 60°$$.

The answer is 60.

$$d' = \sum \frac{d_i}{\mu_i}$$

$$d' = \frac{6}{\frac{8}{5}} + \frac{6}{\frac{3}{2}} = \frac{30}{8} + \frac{12}{3} = \frac{15}{4} + 4 = \frac{31}{4}\text{ cm}$$

$$\frac{\alpha}{4} = \frac{31}{4} \implies \alpha = 31$$

According to Newton's lens formula: $$xy = f^2$$

From the given $$y-x$$ hyperbolic plot, 

$$\text{At point C: } x = 20\text{ cm}, \quad y = 20\text{ cm}$$

$$\implies 20 \times 20 = f^2$$

$$\implies f^2 = 400 \implies f = 20\text{ cm}$$

The ray suffers two refractions (at the first and the last surface) and one total internal reflection (at the back surface).
Let the angle of incidence be $$\theta_0$$ (in air) and the corresponding angle of refraction inside the sphere be $$\phi_0$$.
From Snell’s law (air to glass):

$$\sin \theta_0 = n \sin \phi_0 \qquad -(1)$$

For a spherical drop the total deviation produced by “refraction-reflection-refraction’’ is a standard result (used while studying rainbows):

$$\alpha \;=\; 180^\circ \;+\; 2\theta_0 \;-\; 4\phi_0 \qquad -(2)$$

Using $$(1)$$ and $$(2)$$ we answer each item.

Putting in $$(2):\; 180^\circ = 180^\circ + 2\theta_0 - 4\phi_0 \Longrightarrow \theta_0 = 2\phi_0.$$ Substituting in $$(1):\; \sin(2\phi_0) = 2\sin\phi_0.$$ Since $$\sin(2\phi_0)=2\sin\phi_0\cos\phi_0,$$ we get $$2\sin\phi_0\cos\phi_0 = 2\sin\phi_0.$$ Either $$\sin\phi_0 = 0$$ or $$\cos\phi_0 = 1.$$ Both give $$\phi_0 = 0^\circ \Longrightarrow \theta_0 = 0^\circ.$$ Hence all possible $$\theta_0$$ values = $$0^\circ$$ (List-II → 5).

Again $$\theta_0 = 2\phi_0.$$ Put in $$(1):\; \sin(2\phi_0) = \sqrt{3}\,\sin\phi_0.$$ $$2\sin\phi_0\cos\phi_0 = \sqrt{3}\,\sin\phi_0 \; \Longrightarrow\; 2\cos\phi_0 = \sqrt{3}.$$ So $$\cos\phi_0 = \dfrac{\sqrt{3}}{2} \Longrightarrow \phi_0 = 30^\circ.$$ Hence $$\theta_0 = 2\phi_0 = 60^\circ.$$ The trivial solution $$\phi_0 = 0^\circ,\; \theta_0 = 0^\circ$$ also satisfies both equations. Thus all possible $$\theta_0$$ values = $$60^\circ$$ and $$0^\circ$$ (List-II → 2).

The two $$\phi_0$$ values found above are $$30^\circ$$ and $$0^\circ$$ (List-II → 1).

From $$(1):\; \sin45^\circ = \sqrt{2}\,\sin\phi_0 \;\Longrightarrow\; \dfrac{1}{\sqrt{2}} = \sqrt{2}\,\sin\phi_0 \;\Longrightarrow\; \sin\phi_0 = \dfrac{1}{2} \;\Longrightarrow\; \phi_0 = 30^\circ.$$ Now use $$(2):\; \alpha = 180^\circ + 2(45^\circ) - 4(30^\circ) = 180^\circ + 90^\circ - 120^\circ = 150^\circ.$$ Thus all possible $$\alpha$$ values = $$150^\circ$$ (List-II → 4).

Collecting the matches:
P → 5,  Q → 2,  R → 1,  S → 4.

Hence the correct option is Option A: P → 5; Q → 2; R → 1; S → 4.

We have angle of incidence = 45° and angle of refraction = 30°. Using Snell's law:

$$\mu = \frac{\sin 45°}{\sin 30°} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}$$

Now, the frequency of a wave does not change when it passes from one medium to another, so $$\nu_2 = \nu_1$$.

The wavelength in the new medium is related to the wavelength in air by:

$$\lambda_2 = \frac{\lambda_1}{\mu} = \frac{\lambda_1}{\sqrt{2}}$$

Hence, $$\lambda_2 = \frac{1}{\sqrt{2}}\lambda_1$$ and $$\nu_2 = \nu_1$$. So, the correct answer is $$\lambda_2 = \dfrac{1}{\sqrt{2}}\lambda_1, \nu_2 = \nu_1$$.

Object at 80 cm mark, lens at 100 cm mark, image at 180 cm mark. Least count = 0.2 cm.

Object distance: $$u = 80 - 100 = -20$$ cm (taking sign convention).

Image distance: $$v = 180 - 100 = 80$$ cm.

Using the lens formula: $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{80} - \dfrac{1}{-20} = \dfrac{1}{80} + \dfrac{1}{20} = \dfrac{1 + 4}{80} = \dfrac{5}{80} = \dfrac{1}{16}$$

So $$f = 16$$ cm.

Differentiating $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$:

$$\dfrac{\Delta f}{f^2} = \dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$$

The error in each measurement: $$\Delta u = \Delta v = 0.2$$ cm (least count).

$$\dfrac{\Delta f}{f^2} = \dfrac{0.2}{80^2} + \dfrac{0.2}{20^2} = \dfrac{0.2}{6400} + \dfrac{0.2}{400} = 0.00003125 + 0.0005 = 0.00053125$$

$$\Delta f = f^2 \times 0.00053125 = 256 \times 0.00053125 = 0.136$$ cm.

$$\% \text{ error} = \dfrac{\Delta f}{f} \times 100 = \dfrac{0.136}{16} \times 100 = 0.85\%$$

The answer is Option 1: $$\boxed{0.85}\%$$.

A microscope focused on an object at the bottom of a bucket must be raised 30 cm when a liquid of refractive index 5/3 is poured in. We need the height of the liquid.

Understand apparent depth and real depth.

When light travels from a denser medium (liquid) to a rarer medium (air), the apparent position of an object is closer to the surface than its actual position. The relationship is:

$$ \text{Apparent depth} = \frac{\text{Real depth}}{n} $$

where $$n$$ is the refractive index of the liquid.

Calculate the apparent shift.

The apparent shift (the amount the object appears to rise) is:

$$ \text{Shift} = \text{Real depth} - \text{Apparent depth} = h - \frac{h}{n} = h\left(1 - \frac{1}{n}\right) $$

This is the distance the microscope must be raised to refocus.

Substitute and solve.

Given: shift = 30 cm, $$n = 5/3$$:

$$ 30 = h\left(1 - \frac{1}{5/3}\right) = h\left(1 - \frac{3}{5}\right) = h \times \frac{2}{5} $$

$$ h = \frac{30 \times 5}{2} = 75\;\text{cm} $$

The height of the liquid is 75 cm.

The correct answer is Option 1: 75 cm.

The least distance of distinct vision (near point) is obtained from the data of the reading (converging) glasses, because these spectacles are chosen so that an object kept at the ordinary reading distance appears at the patient’s near point.

Step 1: Write the basic formulas
Power: $$P = \frac{1}{f}$$  (with $$f$$ in metres)
Lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$-(1)$$
Here $$u$$ = object distance from the lens, $$v$$ = image distance from the lens (measured from the same side as the light is incident; a virtual image has negative $$v$$).

Step 2: Focal length of the reading glass
Given power of reading glass, $$P_r = +2.0\ \text{D}$$.
Therefore $$f_r = \frac{1}{P_r} = \frac{1}{2.0}\ \text{m} = 0.50\ \text{m}$$.

Step 3: Choose the usual reading distance
While reading, a book is normally held at about $$25\ \text{cm}$$ from the spectacles.
Hence for the lens, $$u = -25\ \text{cm} = -0.25\ \text{m}$$ (negative by the Cartesian sign convention).

Step 4: Locate the virtual image (near point) using Eq. $$(1)$$
Substituting $$f = +0.50\ \text{m}$$ and $$u = -0.25\ \text{m}$$:
$$\frac{1}{0.50} = \frac{1}{v} - \left(-\frac{1}{0.25}\right)$$
$$2 = \frac{1}{v} + 4$$
$$\frac{1}{v} = 2 - 4 = -2$$
$$v = -0.50\ \text{m}$$.

The negative sign confirms that the image is virtual and lies on the same side as the object (towards the eye). The magnitude gives the distance of the near point:

$$|v| = 0.50\ \text{m} = 50\ \text{cm}$$.

Step 5: State the result
The person’s least distance of distinct vision is $$50\ \text{cm}$$.

Therefore, the correct option is Option D (50 cm).

We need to determine what improves the resolving power of a compound microscope.

The resolving power formula:

The resolving power (RP) of a microscope is given by:

$$RP = \frac{2\mu \sin\theta}{\lambda}$$

where:

- $$\mu$$ is the refractive index of the medium between the object and the objective lens

- $$\theta$$ is the half-angle of the maximum cone of light entering the objective

- $$\lambda$$ is the wavelength of light used

The term $$\mu \sin\theta$$ is called the numerical aperture (NA) of the objective lens.

Analyzing each option:

Option A: Increase the wavelength of light — From the formula, RP is inversely proportional to $$\lambda$$. Increasing wavelength would decrease the resolving power. This is incorrect.

Option B: Increase the refractive index of the medium between the object and objective lens — From the formula, RP is directly proportional to $$\mu$$. Increasing $$\mu$$ (for example, by using oil immersion instead of air) would increase the resolving power. This is correct.

Option C: Decrease the focal length of the eyepiece — The focal length of the eyepiece affects the magnifying power of the microscope, not its resolving power. Resolving power depends on the objective lens and the medium, not the eyepiece. This is incorrect.

Option D: Decrease the diameter of the objective lens — A smaller objective lens aperture would reduce $$\sin\theta$$, thereby decreasing the resolving power. This is incorrect.

The correct answer is Option B: Increase the refractive index of the medium between the object and objective lens.

Find the apparent depth of a vessel of depth $$d$$ half filled with oil (refractive index $$n_1$$) and half filled with water (refractive index $$n_2$$).

For a layer of real depth $$t$$ and refractive index $$n$$, the apparent depth is $$\frac{t}{n}$$. Each layer here has real depth $$\frac{d}{2}$$, so the apparent depth of the oil layer is $$\frac{d/2}{n_1}$$ and that of the water layer is $$\frac{d/2}{n_2}$$. Adding these, the total apparent depth becomes $$\frac{d}{2n_1} + \frac{d}{2n_2} = \frac{d}{2}\left(\frac{1}{n_1} + \frac{1}{n_2}\right) = \frac{d}{2}\cdot\frac{n_1 + n_2}{n_1 n_2} = \frac{d(n_1 + n_2)}{2n_1 n_2}$$.

The correct answer is Option B: $$\frac{d(n_1 + n_2)}{2n_1 n_2}$$.

$$d_{\text{real1}} = \mu \cdot d_{\text{apparent1}}, \quad d_{\text{real2}} = \mu \cdot d_{\text{apparent2}}$$

$$\text{Total real side length: } L = d_{\text{real1}} + d_{\text{real2}} = \mu(d_{\text{apparent1}} + d_{\text{apparent2}})$$

$$24 = \mu(12 + 4) \implies 24 = 16\mu \implies \mu = \frac{24}{16} = \frac{3}{2}$$

An object is 12 cm in front of a plane mirror. The mirror moves 4 cm towards the stationary object. Find the shift in image position.

To begin,

For a plane mirror, the image is as far behind the mirror as the object is in front. Initially, the image is 12 cm behind the mirror, i.e., 24 cm from the object.

Next,

When the mirror moves 4 cm towards the object, the new object distance = $$12 - 4 = 8$$ cm.

The new image is 8 cm behind the new mirror position.

From this,

Original image position (from a fixed reference, say the object): 24 cm from object (or 12 cm behind mirror's original position).

New image position: The new mirror is at 8 cm from object. Image is 8 cm behind this new position = 16 cm from object.

Shift in image = $$24 - 16 = 8$$ cm towards the mirror (or towards the object).

Alternatively: When mirror shifts by $$d$$, image shifts by $$2d = 2 \times 4 = 8$$ cm in the same direction as the mirror (towards the object/mirror).

The correct answer is Option B: 8 cm towards mirror.

Statement I: Brewster's angle air→glass: $$\tan\theta_B = \mu_g$$. Glass→air: $$\tan\theta'_B = 1/\mu_g = \cot\theta_B = \tan(\pi/2-\theta_B)$$. So $$\theta'_B = \pi/2-\theta_B$$. TRUE.

Statement II: Brewster's angle glass→air should be $$\tan^{-1}(1/\mu_g)$$, not $$\tan^{-1}(\mu_g)$$. FALSE.

Statement I is true but Statement II is false.

In a reflecting telescope, a secondary mirror is used to redirect the light collected by the primary concave mirror.

Understand the structure of a reflecting telescope. A reflecting telescope uses a large concave mirror (primary mirror) to collect and focus light. However, the focal point of the primary mirror lies in front of the mirror, which means the observer's head would block incoming light if the eyepiece were placed there.

Role of the secondary mirror. The secondary mirror (a small flat or convex mirror) is placed near the focal point of the primary mirror to redirect the light to a more convenient location. In a Cassegrain telescope, it reflects light back through a hole in the primary mirror. In a Newtonian telescope, it deflects light to the side of the tube.

Evaluate the options. Option A: "Reduce the problem of mechanical support" — The secondary mirror helps avoid the need to place heavy equipment at the prime focus, reducing mechanical support issues. This is correct.

Option B: "Make chromatic aberration zero" — Reflecting telescopes already have zero chromatic aberration (mirrors don't cause dispersion). The secondary mirror doesn't contribute to this.

Option C: "Move the eyepiece outside the telescopic tube" — This is also a benefit of the secondary mirror.

Option D: "Remove spherical aberration" — The secondary mirror doesn't remove spherical aberration.

The primary purpose is to reduce the mechanical support problem by redirecting light so equipment doesn't need to be at the prime focus.

The correct answer is Option A.

The critical angle $$\theta_c$$ for a denser-rarer interface is related to the refractive indices by:

$$\sin \theta_c = \frac{n_r}{n_d} = \frac{v_d}{v_r}$$

where $$v_d$$ and $$v_r$$ are the speeds of light in the denser and rarer media respectively.

We are given that $$\theta_c = 45°$$, $$v_r = 3 \times 10^8$$ m/s

$$\sin 45° = \frac{v_d}{v_r}$$

$$\frac{1}{\sqrt{2}} = \frac{v_d}{3 \times 10^8}$$

$$v_d = \frac{3 \times 10^8}{\sqrt{2}} = \frac{3 \times 10^8 \times \sqrt{2}}{2} = \frac{3 \times 1.414 \times 10^8}{2}$$

$$v_d = \frac{4.243 \times 10^8}{2} = 2.12 \times 10^8 \text{ m/s}$$

The speed of light in the denser medium is $$2.12 \times 10^8$$ m/s.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies v = \frac{uf}{u - f}$$

$$v_A = \frac{u_A f}{u_A - f}$$

$$v_A = \frac{(-15)(-20)}{-15 - (-20)}$$

$$v_A = \frac{300}{-15 + 20}$$

$$v_A = \frac{300}{5} = +60 \text{ cm}$$

$$v_B = \frac{u_B f}{u_B - f}$$

$$v_B = \frac{(-25)(-20)}{-25 - (-20)}$$

$$v_B = \frac{500}{-5} = -100 \text{ cm}$$

$$d = |v_A - v_B| = |60 - (-100)| = 160 \text{ cm}$$

When a beam of white light is allowed to pass through a convex lens, parallel to principal axis, the different colors of light converge at different points on the principle axis after refraction. This is called Chromatic aberration.

$$\text{By the law of reflection, the angle of incidence equals the angle of reflection: } i = r = 30^\circ$$

$$\delta = 180^\circ - (i + r)$$

$$\delta = 180^\circ - (30^\circ + 30^\circ) = 180^\circ - 60^\circ = 120^\circ$$

A plane mirror forms a virtual and erect image of the same size as the object. The image is also laterally inverted (left and right are interchanged).

Therefore:

• A. Real → Incorrect
• B. Erect → Correct
• C. Smaller in size than object → Incorrect
• D. Laterally inverted → Correct

When a ray of a single, plane-polarised colour is incident on a glass surface and no part of it is reflected, the following two facts must hold simultaneously.
  • The incident light has to be polarised in the plane of incidence (otherwise the perpendicular component would still be partially reflected).
  • The angle of incidence must be the Brewster (polarising) angle, $$i_B$$, defined by $$\tan i_B = \mu$$, where $$\mu$$ is the refractive index of the second medium with respect to air.

For the given prism $$\mu_{\,\text{blue}} = \sqrt3$$, hence

$$i_B = \tan^{-1}\!\left( \sqrt3 \right) = 60^\circ.$$(1)

Because the incident ray is already plane-polarised and the incidence is at the polarising angle, exactly zero reflection is possible only when the electric field vector lies in the plane of incidence. Therefore, statement A is correct.

No other statement can be asserted solely from the Brewster condition, so options B, C and D cannot be concluded as correct on the basis of the given information.

Hence the only correct statement is:
Option A - The blue light is polarised in the plane of incidence.

Let the lower face of the slab be taken as the reference (height $$y = 0$$) and the upper face be at $$y = h$$. The refractive index varies linearly with the height, so we write

$$n(y)=n_1+\left(\dfrac{n_2-n_1}{h}\right)\,y\qquad 0\le y\le h$$

Because the incident wave is normal to the slab, every ray travels the same geometrical length $$d$$ inside the glass. However, owing to the variation of $$n(y)$$, different rays accumulate different phases (or optical paths) inside the slab. For a ray that enters the slab at a height $$y$$, the optical path (OP) inside the glass is

$$\text{OP}(y)=n(y)\,d$$

The emerging beam will become a plane wave only when the phase difference produced inside the glass is exactly compensated by the geometrical path difference produced in air because of the deflection. Suppose the emergent beam makes a small angle $$\theta$$ (measured from the original direction, i.e. the normal to the slab). For two rays separated by a vertical distance $$\Delta y$$ at the exit surface, the extra geometrical distance travelled in air is $$\Delta y\,\tan\theta\;(\approx\Delta y\,\theta)$$. Hence, to obtain a single plane wave after the slab we must have

$$n(y)\,d+\;\Delta y\,\tan\theta=\text{constant}$$

Differentiating with respect to $$y$$ gives the required condition for compensation:

$$\dfrac{d}{dy}\Big[n(y)\,d\Big]+\,\tan\theta=0 \;\;\Longrightarrow\;\; \theta=\;d\;\dfrac{dn}{dy}$$ $$-(1)$$

Using the linear variation of $$n$$, $$\dfrac{dn}{dy}= \dfrac{n_2-n_1}{h}$$. Substituting this in $$(1)$$ gives

$$\theta=\dfrac{d\,(n_2-n_1)}{h}$$

The slab as a whole therefore behaves like a very thin prism of small apex angle $$A=\dfrac{d\,(n_2-n_1)}{h}$$. For a thin prism, the angular deviation $$\delta$$ produced when light emerges into air is given by the well-known formula

$$\delta=\mu_{\text{mean}}\;A,\qquad \mu_{\text{mean}}=\dfrac{n_1+n_2}{2}$$

Hence

$$\delta =\left(\dfrac{n_1+n_2}{2}\right)\left[\dfrac{d\,(n_2-n_1)}{h}\right] =\dfrac{(n_2^2-n_1^2)\,d}{2h}$$

Because the angles involved are small, $$\delta\approx\tan\delta$$, so the emergent beam is deflected upward by

$$\boxed{\; \tan^{-1}\!\left[\dfrac{(n_2^2-n_1^2)\,d}{2h}\right]\;}$$

This matches the expression given in Option A. Option B gives only $$\dfrac{(n_2-n_1)d}{h}$$ (missing the factor $$\dfrac{n_1+n_2}{2}$$), Option C is incorrect because a finite gradient always deflects the beam, and Option D is incorrect since the deflection also depends on the individual values of $$n_1$$ and $$n_2$$ through the factor $$\dfrac{n_1+n_2}{2}$$.

Therefore, the correct statement is:
Option A which is: It will deflect up by an angle $$\tan^{-1}\!\left[\dfrac{(n_2^2-n_1^2)\,d}{2h}\right]$$

For reflection from $$M_2$$:

$$\frac{1}{v} + \frac{1}{(-10)} = \frac{1}{(-12)}$$

$$\frac{1}{v} = \frac{1}{10} - \frac{1}{12}$$

$$v = +60\text{ cm}$$ (for $$I_1$$)

For refraction from $$L$$:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} - \frac{1}{(-80)} = \frac{1}{10}$$

$$v = +\frac{80}{7}$$ (For $$I_2$$)

This image should be at focus of $$\text{M}_1$$ $$\implies$$ $$\therefore \frac{20}{2} + \frac{80}{7} = \frac{n}{7}$$ $$\implies$$ $$n = 150$$

Also, If $$I_2$$ is formed at pole of $$M_1$$ then $$\frac{n}{7} = \frac{80}{7}$$ $$\implies$$ $$n = 80$$

And further if $$\text{I}_2$$ is formed at centre of curvature of $$\text{M}_1$$ then $$\frac{n}{7} = \frac{80}{7} + 20$$ $$\implies$$ $$\therefore n = 220$$

Solution :

For a converging mirror,

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

Given :

$$u = 40\text{ cm}$$

$$v = 120\text{ cm}$$

Therefore,

$$\frac{1}{f} = \frac{1}{120} + \frac{1}{40}$$

$$= \frac{1+3}{120}$$

$$= \frac{1}{30}$$

$$f = 30\text{ cm}$$

Modified scale has :

$$20\text{ divisions} = 1\text{ cm}$$

Least count :

$$LC = \frac{1}{20}\text{ cm}$$

$$= 0.05\text{ cm}$$

Hence,

$$\Delta u = \Delta v = 0.05\text{ cm}$$

Using error propagation formula :

$$\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$$

Substituting values :

$$\Delta f = (30)^2\left(\frac{0.05}{(120)^2} + \frac{0.05}{(40)^2}\right)$$

$$= 900 \times 0.05 \left(\frac{1}{14400} + \frac{1}{1600}\right)$$

$$= 45\left(\frac{1+9}{14400}\right)$$

$$= 45 \times \frac{10}{14400}$$

$$= \frac{450}{14400}$$

$$= \frac{1}{32}\text{ cm}$$

Given,

$$\Delta f = \frac{1}{K}\text{ cm}$$

Therefore,

$$K = 32$$

Final Answer :

$$32$$

Using apparent position relation: $$y_{\text{apparent}} = \mu y$$

Total apparent distance observed by fish: $$s = x + \mu y$$

Differentiating with respect to time for velocities:

$$\frac{ds}{dt} = \frac{dx}{dt} + \mu \frac{dy}{dt}$$

$$12 = 8 + \frac{4}{3} v_{\text{bird}}$$

$$4 = \frac{4}{3} v_{\text{bird}} \implies v_{\text{bird}} = 3\text{ m s}^{-1}$$

A biconvex lens of focal length 10 cm is cut perpendicular to the principal axis into two identical halves. Each half becomes a plano-convex lens.

For the original biconvex lens (assuming symmetric with equal radii of curvature $$R$$):

$$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (n-1)\left(\frac{2}{R}\right)$$

When cut perpendicular to the principal axis, each half is a plano-convex lens with one curved surface (radius $$R$$) and one flat surface (radius $$\infty$$):

$$\frac{1}{f'} = (n-1)\left(\frac{1}{R}\right) = \frac{1}{2} \times \frac{1}{f} = \frac{1}{2 \times 10} = \frac{1}{20}$$

$$f' = 20$$ cm $$= 0.2$$ m

Power: $$P = \frac{1}{f'} = \frac{1}{0.2} = 5$$ D

The power of each lens is $$\mathbf{5}$$ D.

$$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{1}{f_a} = (\mu_l - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{1}{f_w} = \left(\frac{\mu_l}{\mu_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{f_w}{f_a} = \frac{\mu_l - 1}{\frac{\mu_l}{\mu_w} - 1}$$

$$\frac{f_w}{f_a} = \frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1}$$

$$\frac{f_w}{f_a} = 4$$

$$f_w = 4 \times 35 = 140\text{ cm}$$

$$\Delta f = 140 - 35$$

$$\Delta f = 105\text{ cm}$$

We have two coaxial convex lenses: $$L_1$$ with focal length $$f_1 = 24$$ cm and $$L_2$$ with focal length $$f_2 = 9$$ cm, separated by 10 cm. An object is placed 6 cm from $$L_1$$, and we seek the distance between the object and the final image.

We apply the lens formula $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$ to the first lens, $$L_1$$. Here the object distance is $$u_1 = -6$$ cm (using the sign convention that left of the lens is negative) and the focal length is $$f_1 = 24$$ cm. Substituting these values gives $$\dfrac{1}{v_1} - \dfrac{1}{-6} = \dfrac{1}{24}$$, which simplifies to $$\dfrac{1}{v_1} + \dfrac{1}{6} = \dfrac{1}{24}$$. Therefore, $$\dfrac{1}{v_1} = \dfrac{1}{24} - \dfrac{1}{6} = \dfrac{1-4}{24} = \dfrac{-3}{24} = \dfrac{-1}{8}$$, giving $$v_1 = -8$$ cm. Hence the image formed by $$L_1$$ is 8 cm to the left of $$L_1$$ (a virtual image on the same side as the object).

Since the second lens $$L_2$$ is 10 cm to the right of $$L_1$$, this virtual image lies $$8 + 10 = 18$$ cm to the left of $$L_2$$ and thus acts as an object for $$L_2$$ with $$u_2 = -18$$ cm.

Applying the lens formula $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$ to $$L_2$$, we substitute $$u_2 = -18$$ cm and $$f_2 = 9$$ cm to obtain $$\dfrac{1}{v_2} - \dfrac{1}{-18} = \dfrac{1}{9}$$. This becomes $$\dfrac{1}{v_2} + \dfrac{1}{18} = \dfrac{1}{9}$$, so $$\dfrac{1}{v_2} = \dfrac{1}{9} - \dfrac{1}{18} = \dfrac{2-1}{18} = \dfrac{1}{18}$$, and hence $$v_2 = 18$$ cm. Consequently, the final image is 18 cm to the right of $$L_2$$.

Finally, since the object lies 6 cm to the left of $$L_1$$ and the final image lies 18 cm to the right of $$L_2$$, which itself is 10 cm to the right of $$L_1$$, the total separation between object and final image is $$6 + (10 + 18) = 6 + 28 = 34$$ cm. Thus the required distance is $$\boxed{34}$$ cm.

The angle of incidence $$i$$ with the normal is calculated from the angle made with the water surface:  $$i = 90^\circ - 30^\circ = 60^\circ$$

Applying Snell's law at the air-water interface to find the angle of refraction $$r$$:

$$\sin i = n_w \sin r$$

$$\sin 60^\circ = \frac{4}{3} \sin r \implies \frac{\sqrt{3}}{2} = \frac{4}{3} \sin r \implies \sin r = \frac{3\sqrt{3}}{8} \approx 0.65$$

$$\tan r = \frac{\sin r}{\sqrt{1 - \sin^2 r}} = \frac{\frac{3\sqrt{3}}{8}}{\sqrt{1 - \left(\frac{3\sqrt{3}}{8}\right)^2}} = \frac{3\sqrt{3}}{\sqrt{64 - 27}} = \frac{3\sqrt{3}}{\sqrt{37}} \approx 0.854$$

The total shadow length within the water is composed of the shadow component from the air section ($$x_1$$) and the water section ($$x_2$$):

$$x_{\text{total}} = x_1 + x_2$$ $$\implies$$ $$2.15 = h_a \tan 60^\circ + h_w \tan r$$

$$2.15 = h_a \sqrt{3} + 1.5 \times 0.854$$ $$\implies$$ $$2.15 = h_a \sqrt{3} + 1.281$$ $$\implies$$ $$h_a \sqrt{3} = 2.15 - 1.281 = 0.869$$

$$h_a = \frac{0.869}{\sqrt{3}} = \frac{0.869}{1.732} \approx 0.50\ \text{m}$$

$$h_a = 0.50 \times 100 = 50\ \text{cm}$$

Given: Glass plate thickness $$t = \sqrt{3}$$ cm, refractive index $$n = \sqrt{2}$$, angle of incidence = critical angle for glass-air interface.

The critical angle $$i_c$$ satisfies:

$$\sin i_c = \frac{1}{n} = \frac{1}{\sqrt{2}}$$

$$i_c = 45°$$

The angle of incidence $$i = 45°$$. Using Snell's law to find the angle of refraction inside the glass:

$$\sin i = n \sin r$$

$$\frac{1}{\sqrt{2}} = \sqrt{2} \sin r$$

$$\sin r = \frac{1}{2} \implies r = 30°$$

The lateral displacement formula for a parallel-sided glass slab is:

$$d = \frac{t \sin(i - r)}{\cos r}$$

Substituting the values:

$$d = \frac{\sqrt{3} \times \sin(45° - 30°)}{\cos 30°} = \frac{\sqrt{3} \times \sin 15°}{\cos 30°}$$

Using $$\sin 15° = 0.26$$ and $$\cos 30° = \frac{\sqrt{3}}{2}$$:

$$d = \frac{\sqrt{3} \times 0.26}{\frac{\sqrt{3}}{2}} = \frac{0.26 \times 2}{1} = 0.52 \text{ cm}$$

Converting to the required form:

$$d = 52 \times 10^{-2} \text{ cm}$$

Therefore, the answer is $$\mathbf{52}$$.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies v = \frac{uf}{u - f}$$

Focal length of concave mirror $$f = -20 \text{ cm}$$

Length of rod $$L = 10 \text{ cm}$$

Midpoint of rod is at $$u_{\text{mid}} = -40 \text{ cm}$$

Since the midpoint is at $$40 \text{ cm}$$ from the pole, the rod extends $$5 \text{ cm}$$ on either side.

Position of the closer end ($$A$$): $$u_1 = -40 + 5 = -35 \text{ cm}$$

Position of the farther end ($$B$$): $$u_2 = -40 - 5 = -45 \text{ cm}$$

For end $$A$$ ($$u_1 = -35 \text{ cm}$$):

$$v_1 = \frac{(-35)(-20)}{-35 - (-20)}$$

$$v_1 = \frac{700}{-15}$$

$$v_1 = -\frac{140}{3} \text{ cm}$$

For end $$B$$ ($$u_2 = -45 \text{ cm}$$):

$$v_2 = \frac{(-45)(-20)}{-45 - (-20)}$$

$$v_2 = \frac{900}{-25}$$

$$v_2 = -36 \text{ cm}$$

The length of the image is the distance between these two points:

$$L' = |v_1 - v_2|$$

$$L' = \left| -\frac{140}{3} - (-36) \right|$$

$$L' = \left| -\frac{140}{3} + \frac{108}{3} \right|$$

$$L' = \left| -\frac{32}{3} \right| = \frac{32}{3} \text{ cm}$$

Lens maker formula used:

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

For the plano-concave lens,

$$\frac{1}{f_1}=(1.75-1)\left(-\frac{1}{30}\right)$$

$$\frac{1}{f_1}=0.75\left(-\frac{1}{30}\right)$$

$$f_1=-40\ \text{cm}$$

For the plano-convex lens,

$$\frac{1}{f_2}=(1.75-1)\left(\frac{1}{30}\right)$$

$$\frac{1}{f_2}=0.75\left(\frac{1}{30}\right)$$

$$f_2=40\ \text{cm}$$

The object is at infinity, so after refraction from the concave lens ($$L_1$$), the image is formed at its focal point.

Since $$f_1=-40$$ cm, the image formed by $$L_1$$ is virtual and $$40$$ cm to the left of $$L_1$$.

Distance between lenses:

$$L_1L_2 = 40\ \text{cm}$$

Therefore, object distance for $$L_2$$:

$$u_2 = 40+40=80\ \text{cm}$$

Since,

$$u_2=2f_2$$

Now we use the standard lens property:

When an object is placed at $$2f$$ of a convex lens, the image is formed at $$2f$$ on the other side.

So,

$$v_2=2f_2=80\ \text{cm}$$

You can also verify using lens formula:

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$

Putting,

$$f=40,\qquad u=-80$$

$$\frac{1}{40}=\frac{1}{v}+\frac{1}{80}$$

$$\frac{1}{v}=\frac{1}{40}-\frac{1}{80} = \frac{1}{80}$$

$$v=80\ \text{cm}$$

So image is formed at $$80\ \text{cm}$$ on the right side of the convex lens.

Hence total distance from concave lens:

x = 40 + 80

$$\boxed{x=120\ \text{cm}}$$

The power of a lens is given by the lensmaker's equation:

$$P = \left(\frac{n_l}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a convex lens with both surfaces having radius of curvature 20 cm: $$R_1 = +20$$ cm, $$R_2 = -20$$ cm.

Power in air ($$n_m = 1$$):

$$P_{air} = (1.8 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = 0.8 \times \frac{2}{20} = 0.8 \times 0.1 = 0.08 \text{ cm}^{-1}$$

Power in liquid ($$n_m = 1.5$$):

$$P_{liq} = \left(\frac{1.8}{1.5} - 1\right)\left(\frac{1}{20} + \frac{1}{20}\right) = (1.2 - 1) \times 0.1 = 0.2 \times 0.1 = 0.02 \text{ cm}^{-1}$$

Ratio:

$$\frac{P_{air}}{P_{liq}} = \frac{0.08}{0.02} = 4$$

Therefore, $$x = 4$$.

For an equilateral prism ($$A = 60°$$), the refractive index and angle of minimum deviation are related by:

$$n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\sqrt{2} = \frac{\sin\left(\frac{60° + D_m}{2}\right)}{\sin 30°} = \frac{\sin\left(\frac{60° + D_m}{2}\right)}{0.5}$$

$$\sin\left(\frac{60° + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin 45°$$

$$\frac{60° + D_m}{2} = 45°$$

$$60° + D_m = 90°$$

$$D_m = 30°$$

The angle of minimum deviation is $$\mathbf{30}$$°.

$$\text{For the first lens: } u_1 = -\infty, \ f_1 = 20\text{ cm} \implies \frac{1}{v_1} - \frac{1}{-\infty} = \frac{1}{20} \implies v_1 = 20\text{ cm}$$

$$\text{The first image forms } 20\text{ cm to the right of the first lens, acting as the object for the second lens.}$$

$$\text{For the second lens: } u_2 = v_1 - d = 20 - 60 = -40\text{ cm}, \ f_2 = 20\text{ cm}$$

$$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \implies \frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20} \implies \frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \implies v_2 = 40\text{ cm}$$

$$\text{Distance of the final image from the first lens: } D = d + v_2 = 60 + 40 = 100\text{ cm}$$