Thinking Process

The ray refractive by first surface falls normally on second surface, in order to emerges from the other face normally.

Answer

(a) Since, deviation $\delta=(\mu-1) A=(1.5-1) \times 5^{\circ}=2.5^{\circ}$

By geometry, angle of refraction by first surface is $5^{\circ}$.

But $\delta=\theta-r$, so, we have, $2.5^{\circ}=\theta-5^{\circ}$ on solving $\theta=7.5^{\circ}$.

Thinking Process

When light ray goes from one medium to other medium, the frequency of light remains unchanged.

Answer

(d) Since $v \propto \lambda$, the light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerge first.

Thinking Process

This problem has link with the formation of image when object is at different positions.

Answer

(c) When an object approaches a convergent lens from the left of the lens with a uniform speed of $5 m / s$, the image away from the lens with a non-uniform acceleration.

Answer

(b) A passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

Thinking Process

This problem is based on the critical angle of total internal reflection.

Answer

(c) According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength.According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle.

So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

Thinking Process

By lens maker’s formula for plano-convex lens, focal length is given by $f=\frac{R}{\mu-1}$. This is always positive for $\mu>1$ or optically denser medium of material of lens placed in air.

Answer

(c) Here, $R=20 cm, \mu=1.5$, on substituting the values in $f=\frac{R}{\mu-1}=\frac{20}{1.5-1}=40 cm$ of converging nature as $f>0$. Therefore, lens act as a convex lens irrespective of the side on which the object lies.

Answer

(b) The phenomenon involved in the reflection of radiowaves by ionosphere is similar to total internal reflection of light in air during a mirage i.e., angle of incidence is greater than critical angle.

Answer

(b) The $P Q$ ray of light passes through focus $F$ and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray-2 in the figure.

Thinking Process

When a light ray goes from (optically) rarer medium to (optically) denser medium, then it bends towards the normal i.e., i>r and vice-versa.

Answer

(b) Here, light ray goes from (optically) rarer medium air to optically denser terpentine, then it bends towards the normal i.e., $i>r$ whereas when it goes from to optically denser medium terpentine to rarer medium water. then it bends away the normal i.e., $i<r$.

Thinking Process

The image formed by convex mirror does not depend on the relative position of object w.r.t. mirror.

Answer

(d) The speed of the image of the car would appear to increase as the distance between the cars decreases.

Answer

(a) The negative refractive index metamaterials are those in which incident ray from air (Medium 1) to them refract or bend differently to that of positive refractive index medium.

Thinking Process

This problem is based on the phenomenon of reflection when it goes from denser to rare medium

Answer

$(a, b, c)$

When immersed object is seen from close to the edge of the trough the object looks distorted because the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.

The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air and some of the points of the object far away from the edge may not be visible because of total internal reflection.

Thinking Process

This problem is associated with the phenomenon of total internal reflection.

Answer

(d) For $\mu=1.6$, the critical angle, $\mu=1 / \sin C$, we have $C=38.7^{\circ}$, when viewed from $A D$, as long as angle of incidence on $A D$ of the ray emanating from pin is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through $A D$.

Answer

$(a, d)$

Alexandar’s dark band lies between the primary and secondary rainbows, forms due to light scattered into this region interfere destructively.

Since, primary rainbows subtends an angle nearly $41^{\circ}$ to $42^{\circ}$ at observer’s eye, whereas, secondary rainbows subtends an angle nearly $51^{\circ}$ to $54^{\circ}$ at observer’s eye w.r.t. incident light ray.

So, the scattered rays with respect to the incident light of the sun lies between approximately $42^{\circ}$ and $50^{\circ}$.

Answer

$(a, b)$

A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Morever, the formation of a virtual erect and enlarged image, takes place.

Thinking Process

The magnifying power $m$ is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\boldsymbol{\alpha}$ which the object subtends at the lens or the eye. Hence, in normal adjustment

$$ m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_{e}} \frac{f_{o}}{h}=\frac{f_{o}}{f_{e}} $$

In this case, the length of the telescope tube is $f_{o}+f_{e}$.

Answer

$(a, b, c)$

The length of the telescope tube is $f_{o}+f_{e}=20+(0.02)=20.02 m$

Also,

$m=20 / 0.02=1000$

Also, the image formed is inverted.

Answer

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

In other words, $\mu_{b}>\mu_{r}$

By lens maker’s formula

$$ \frac{1}{f}=(n_{21}-1) \frac{1}{R_1}-\frac{1}{R_2} $$

Therefore, $f_{b}<f_{r}$.

Thus, the focal length for blue light will be smaller than that for red.

Answer

The least distance of distinct vision of an average person (i.e., $D$ ) is $25 cm$, in order to view an object with magnification 10 ,

Here, $v=D=25 cm$ and $u=f$

But the magnification $m=v / u=D / f$

$$ \begin{aligned} & m=\frac{D}{f} \\ & f=\frac{D}{m}=\frac{25}{10}=2.5=0.025 m \\ & P=\frac{1}{0.025}=40 D \end{aligned} $$

This is the required power of lens

Thinking Process

One lens have unique focal length irrespective of its face or geometry taken for use.

Answer

No, the reversibility of the lens maker’s equation.

Thinking Process

The image formed by first medium act as an object for second medium .

Answer

Let the apparent depth be $O_1$ for the object seen from $m_2$, then

$$ O_1=\frac{\mu_2}{\mu_1} \frac{h}{3} $$

Ray Optics and Optical Instruments

Since, apparent depth $=$ real depth $/$ refractive index $\mu$.

Since, the image formed by Medium $1, O_2$ act as an object for Medium 2.

If seen from $\mu_3$, the apparent depth is $O_2$.

Similarly, the image formed by Medium 2, $O_2$ act as an object for Medium 3

$$ \begin{aligned} O_2 & =\frac{\mu_3}{\mu_2} \frac{h}{3}+O_1 \\ & =\frac{\mu_3}{\mu_2} \frac{h}{3}+\frac{\mu_2 h}{\mu_1 3}=\frac{h}{3} \frac{\mu_3}{\mu_2}+\frac{\mu_2}{\mu_1} \end{aligned} $$

Seen from outside, the apparent height is

$$ \begin{aligned} O_3 & =\frac{1}{\mu_3} \frac{h}{3}+O_2=\frac{1}{\mu_3} \frac{h}{3}+\frac{h}{3} \frac{\mu_3}{\mu_2}+\frac{\mu_3}{\mu_1} \\ & =\frac{h}{3} \frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3} \end{aligned} $$

This is the required expression of apparent depth.

Answer

The relationship between refractive index, prism angle $A$ and angle of minimum deviation is given by

$$ \mu=\frac{\sin \frac{(A+D_{m})}{2}}{\sin \frac{A}{2}} $$

Here,

$\therefore$ Given,

$$ D_{m}=A $$

Substituting the value, we have

$\therefore \quad \mu=\frac{\sin A}{\sin \frac{A}{2}}$

On solving, we have

$$ \begin{aligned} & =\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2} \\ \mu & =\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2} \end{aligned} $$

For the given value of refractive index,

we have

$$ \therefore \cos \frac{A}{2} =\frac{\sqrt{3}}{2} $$ $$ \text{or} \quad \quad s\frac{A}{2} =30^{\circ}$$ $$\therefore A =60^{\circ}$$

This is the required value of prism angle.

Thinking Process

The length of image is the separation between the images formed by mirror of the extremities of object.

Answer

Since, the object distance is $u$. Let us consider the two ends of the object be at distance $u_1=u-L / 2$ and $u_2=u+L / 2$, respectively so that $|u_1-u_2|=L$. Let the image of the two ends be formed at $v_1$ and $v_2$, respectively so that the image length would be

$$ L^{\prime}=|v_1-v_2| $$

Applying mirror formula, we have

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

On solving, the positions of two images are given by

$$ v_1=\frac{f(u-L / 2)}{u-f-L / 2}, v_2=\frac{f(u+L / 2)}{u-f+L / 2} $$

For length, substituting the value in (i), we have

$$ L^{\prime}=|v_1-v_2|=\frac{f^{2} L}{(u-f)^{2} \times L^{2} / 4} $$

Since, the object is short and kept away from focus, we have

$$ L^{2} / 4«(u-f)^2 \\ \text {Hence, finally} L^{\prime}=\frac{f^2}{(u-f)^2} $$

This is the required expression of length of image.

Thinking Process

This problem involves the wide application of geometry and Snell’s law.

Answer

Refering to the figure, $A M$ is the direction of incidence ray before liquid is filled. After liquid is filled in , $B M$ is the direction of the incident ray. Refracted ray in both cases is same as that along $A M$.

Let the disc is separated by $O$ at a distance $d$ as shown in figure. Also, considering angle

Here, in figure

$$ N=90^{\circ}, O M=a, C B=N B=a-R, A N=a+R $$

and

$$ \begin{aligned} \sin t & =\frac{a-R}{\sqrt{d^{2}+(a-R)^{2}}} \\ \sin \alpha & =\cos (90-\alpha)=\frac{a+R}{\sqrt{d^{2}+(a+R)^{2}}} \end{aligned} $$

But on applying Snell’s law,

$$ \frac{1}{\mu}=\frac{\sin t}{\sin r}=\frac{\sin t}{\sin \alpha} $$

On substituting the values, we have the separation

$$ d=\frac{\mu(a^{2}-b^{2})}{\sqrt{(a+r)^{2}-\mu(a-r)^{2}}} $$

This is the required expression.

Thinking Process

There is no effect on the focal length of the lens if it is cut as given in the question.

Answer

If there was no cut, then the object would have been at a height of $0.5 cm$ from the principal axis $OO^{\prime}$.

Applying lens formula, we have

$$ \begin{aligned} & & \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ & \therefore & \frac{1}{v} & =\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50} \\ & & \text { Mangnification is } m & =\frac{v}{u}=-\frac{50}{50}=-1 \end{aligned} $$

Thus, the image would have been formed at $50 cm$ from the pole and $0.5 cm$ below the principal axis. Hence, with respect to the $X$-axis passing through the edge of the cut lens, the coordinates of the image are $(50 cm,-1 cm)$.

Thinking Process

This is also one of the methods for finding the focal length of the lens in laboratory and known as displacement method.

Answer

Principal of reversibility is states that the position of object and image are interchangeable. So, by the versibility of $u$ and $v$, as seen from the formula for lens.

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

It is clear that there are two positions for which there shall be an image.

On the screen, let the first position be when the lens is at $O$. Finding $u$ and $v$ and substituting in lens formula.

Given,

$\Rightarrow$

Placing it in the lens formula

$$ \frac{1}{D-v}+\frac{1}{v}=\frac{1}{f} $$

On solving, we have

$$ \begin{aligned} \Rightarrow \frac{v+D-v}{(D-v) v} =\frac{1}{f} \newline \newline\\ \Rightarrow v^{2}-D v+D f =0 \newline \newline\\ \Rightarrow v =\frac{D}{2} \pm \frac{\sqrt{D^{2}-4 D f}}{2} \end{aligned} $$

Hence, finding $u$

$$ u=-(D-v)=-\frac{D}{2} \pm \frac{\sqrt{D^{2}-4 D f}}{2} $$

When, the object distance is $\quad \frac{D}{2}+\frac{\sqrt{D^{2}-4 D f}}{2}$

the image forms at

$$ \frac{D}{2}-\frac{\sqrt{D^{2}-4 D f}}{2} $$

Similarly, when the object distance is

The image forms at

$$ \begin{aligned} & \frac{D}{2}-\frac{\sqrt{D^{2}-4 D f}}{2} \\ & \frac{D}{2}+\frac{\sqrt{D^{2}-4 D f}}{2} \end{aligned} $$

The distance between the poles for these two object distance is

$$ \frac{D}{2}+\frac{\sqrt{D^{2}-4 D f}}{2}-\frac{D}{2}-\frac{\sqrt{D^{2}-4 D f}}{2}=\sqrt{D^{2}-4 D f} $$

$$ \begin{aligned} & \text { Let } \\ & \text { If } u=\frac{D}{2}+\frac{d}{2} \text {, then the image is at } v=\frac{\sqrt{D^{2}-4 D f}}{2}-\frac{d}{2} . \\ & \therefore \quad \text { The magnification } m_1=\frac{D-d}{D+d} \\ & \text { If } u=\frac{D-d}{2} \text {, then } v=\frac{D+d}{2} \\ & \therefore \text { The magnification } m_2=\frac{D+d}{D-d} \\ & \text { Thus, } \quad \frac{m_2}{m_1}=\frac{D+d}{D-d} . \end{aligned} $$

This is the required expression of magnification.

Thinking Process

The problem is based on the principle of total internal reflection and area of visibility.

Answer

Let $d$ be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at $O$ to the surface at $d / 2$ at the critical angle.

Let $i$ be the angle of incidence.

Using relationship between refractive index and critical angle,

then,

$$ \sin t=\frac{1}{\mu} $$

Using geometry and trigonometry.

Now,

$$ \begin{aligned} \frac{d / 2}{h} & =\tan i \\ \frac{d}{2} & =h \tan i=h[\sqrt{\mu^{2}-1}]^{-1} \\ d & =\frac{2 h}{\sqrt{\mu^{2}-1}} \end{aligned} $$

$$ \Rightarrow $$

This is the required expression of $d$.

Thinking Process

If two thin lenses of focal length $f_1$ and $f_2$ are in contact, the effective focal length of the combination is given by,

in terms of power $P=P_1+P_2$

$$ \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} $$

Answer

(i) Let the power at the far point be $P_{f}$ for the normal relaxed eye of an average person. The required power

$$ P_{f}=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60 D $$

By the corrective lens the object distance at the far point is $\infty$.

The power required is

$$ P_f^{\prime}=\frac{1}{f^{\prime}}=\frac{1}{\infty}+\frac{1}{0.02}=50 D $$

So for eye + lens system,

we have the sum of the eye and that of the glasses $P_{g}$

$$ \begin{matrix} \therefore & P_f^{\prime}=P_{f}+P_{g} \\ \therefore & P_{g}=-10 D \end{matrix} $$

(ii) His power of accomodatlon is $4 D$ for the normal eye. Let the power of the normal eye for near vision be $P_{n}$.

$$ \text { Then, } \quad 4=P_{n}-P_{f} \text { or } P_{n}=64 D $$

Let his near point be $x_{n}$, then

$$ \begin{aligned} \frac{1}{x_{n}}+\frac{1}{0.02} & =64 \text { or } \frac{1}{x_{n}}+50=64 \\ \frac{1}{x_{n}} & =14, \\ \therefore \quad x_{n} & =\frac{1}{14} ; 0.07 m \end{aligned} $$

(iii) With glasses $P_n^{\prime}=P_f^{\prime}+4=54$

$$ \begin{aligned} 54 & =\frac{1}{x_n^{\prime}}+\frac{1}{0.02}=\frac{1}{x_n^{\prime}}+50 \\ \frac{1}{x_n^{\prime}} & =4 \\ \therefore \quad x_n^{\prime} & =\frac{1}{4}=0.25 m \end{aligned} $$

Answer

Any ray entering at an angle $i$ shall be guided along $A C$ if the angle ray makes with the face $A C(\varphi)$ is greater than the critical angle as per the principle of total internal reflection $\varphi+r=90^{\circ}$, therefore $\sin \varphi=\cos r$

$\begin{aligned} \Rightarrow \qquad \sin \varphi & \geq \frac{1}{\mu} \\ \Rightarrow \qquad \cos r & \geq \frac{1}{\mu} \\ \text{or} \qquad 1-\cos ^2 r & \leq 1-\frac{1}{\mu^2} \\ \text{i.e.,} \qquad \sin ^2 r & \leq \frac{1}{\mu^2} \end{aligned}$

$\begin{aligned} & \text { i.e., } \qquad \sin ^2 r \leq 1-\frac{1}{\mu^2} \\ \qquad & \text { since, } \qquad \sin i=\mu \sin r \\ & \frac{1}{\mu_2} \sin ^2 i \leq 1-\frac{1}{\mu^2} \\ & \text { when } \qquad i=\frac{\pi}{2} \\ & \end{aligned}$

Then,we have smallest angle $\varphi$.

If that is greater than the critical angle,then all other the critical angle.

$\begin{array}{ll} \text { Thus, }\qquad & 1 \leq \mu^2-1 \\ \text { or } & \mu^2 \geq 2 \\ \Rightarrow & \mu \geq \sqrt{2} \end{array}$

This is the required result.

Answer

Let us consider a portion of a ray between $x$ and $x+d x$ inside the liquid. Let the angle of incidence at $x$ be $\theta$ and let it enter the thin column at height $y$. Because of the bending it shall emerge at $x+d x$ with an angle $\theta+d \theta$ and at a height $y+d y$. From Snell’s law,

$\mu(y) \sin \theta=\mu(y+d y) \sin (\theta+d \theta)$

$\mu(y) \sin \theta ; \mu(y)+\frac{d \mu}{d y} d y \quad(\sin \theta \cos d \theta+\cos \theta \sin d \theta) \newline$

$\mu(y) \sin \theta+\mu(y) \cos \theta d \theta+\frac{d \mu}{d y} d y \sin \theta \newline$

$\mu(y) \cos \theta d \theta ; \frac{-d \mu}{d y} d y \sin \theta$ $d \theta ; \frac{-d \mu}{\mu d y} d y \tan \theta$

$$ \tan \theta=\frac{d x}{d y} $$

(from the figure)

On solving, we have

$$ \therefore \quad d \theta=\frac{-1 d \mu}{\mu d y} d x $$

Solving this variable separable form of differential equation.

$$ \therefore \quad \theta=\frac{-1 d \mu}{\mu d y} \int_0^{d} d x=\frac{-1 d \mu}{\mu d y} d $$

Answer

Let us consider two planes at $r$ and $r+d r$. Let the light be incident at an angle $\theta$ at the plane at $r$ and leave $r+d r$ at an angle $\theta+d \theta$. Then from Snell’s law,

$\Rightarrow \quad n(r) \sin \theta ; n(r)+\frac{d n}{d r} d r \quad(\sin \theta \cos d \theta+\cos \theta \sin d \theta)$

$$ n(r)+\frac{d n}{d r} d r \quad(\sin \theta+\cos \theta d \theta) $$

Ignoring the product of differentials

or we have,

$$ \begin{aligned} n(r) \sin \theta ; n(r) \sin \theta & +\frac{d n}{d r} d r \sin \theta+n(r) \cos \theta d \theta \\ -\frac{d n}{d r} \tan \theta & =n(r) \frac{d \theta}{d r} \\ \frac{2 G M}{r^{2} c^{2}} \tan \theta & =1+\frac{2 G M}{r^{2}} \frac{d \theta}{d r} \approx \frac{d \theta}{d r} \\ \int_0^{\theta_0} d \theta & =\frac{2 G M}{c^{2}} \int_{-\infty}^{\infty} \frac{\tan \theta d r}{r^{2}} \end{aligned} $$

Now substitution for integrals, we have

$$ \text { Now, } \quad \begin{aligned} r^{2} & =x^{2}+R^{2} \text { and } \tan \theta=\frac{R}{x} \\ 2 r d r & =2 x d x \\ \text { Put } \quad \int_0^{\theta_0} d \theta & =\frac{2 G M}{c^{2}} \int_{-\infty}^{\infty} \frac{R}{x} \frac{x d x}{(x^{2}+R^{2})^{\frac{3}{2}}} \\ x & =R \tan \varphi \\ d x & =R \sec ^{2} \varphi d \varphi \\ \theta_0 & =\frac{2 G M R}{c^{2}} \int_{-\pi / 2}^{\pi / 2} \frac{R \sec ^{2} \varphi d \varphi}{R^{3} \sec ^{3} \varphi} \\ & =\frac{2 G M}{R c^{2}} \int_{-\pi / 2}^{\pi / 2} \cos \varphi d \varphi=\frac{4 G M}{R c^{2}} \end{aligned} $$

This is the required proof.

Answer

Since, the material is of refractive index $-1, \theta_{r}$, is negative and $\theta_r^{\prime}$ positive.

Now,

$$ |\theta_{t}|=|\theta_{r}|=|\theta_r^{\prime}| $$

The total deviation of the outcoming ray from the incoming ray is $4 \theta$. Rays shall not reach the recieving plate if

$$ .\frac{\pi}{2} \leq 4 \theta_{t} \leq \frac{3 \pi}{2} \quad \text { [angles measured clockwise from the } y \text {-axis }] $$

On solving,

$$ \frac{\pi}{8} \leq \theta_{t} \leq \frac{3 \pi}{8} $$

Now,

or

$$ \begin{aligned} \sin \theta_{t} & =\frac{x}{R} \\ \frac{\pi}{8} & \leq \sin ^{-1} \frac{x}{R} \leq \frac{3 \pi}{8} \\ \frac{\pi}{8} & \leq \frac{x}{R} \leq \frac{3 \pi}{8} \end{aligned} $$

Thus, for light emitted from the source shall not reach the receiving plate. If $\frac{R \pi}{8} \leq x \leq \frac{R 3 \pi}{8}$.

Answer

(i) The time elapsed to travel from $S$ to $P_1$ is

$$ t_1=\frac{S P_1}{c}=\frac{\sqrt{u^{2}+b^{2}}}{c} $$

or

$$ \frac{u}{c} 1+\frac{1}{2} \frac{b^{2}}{u^{2}} \text { assuming } b«u_0 \text {. } $$

The time required to travel from $P_1$ to $O$ is

$$ t_2=\frac{P_1 O}{c}=\frac{\sqrt{v^{2}+b^{2}}}{c} ; \frac{v}{c} 1+\frac{1}{2} \frac{b^{2}}{v^{2}} $$

The time required to travel through the lens is

$$ t_1=\frac{(n-1) w(b)}{c} $$

where $n$ is the refractive index.

Thus, the total time is

$$ t=\frac{1}{c} u+v+\frac{1}{2} b^{2} \frac{1}{u}+\frac{1}{v}+(n-1) w(b) $$

Put

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

Then,

$$ t=\frac{1}{c} u+v+\frac{1}{2} \frac{b^{2}}{D}+(n-1) w_0+\frac{b^{2}}{\alpha} $$

Fermet’s principle gives the time taken should be minimum.

For that first derivative should be zero.

$$ \begin{aligned} \frac{d t}{d b} & =0=\frac{b}{C D}-\frac{2(n-1) b}{c \alpha} \\ \alpha & =2(n-1) D \end{aligned} $$

Thus, a convergent lens is formed if $\alpha=2(n-1) D$. This is independant of and hence, all paraxial rays from $S$ will converge at $O$ i.e., for rays

and

$$ (b«v .) $$

Since, $\frac{1}{D}=\frac{1}{u}+\frac{1}{V}$, the focal length is $D$.

(ii) In this case, differentiating expression of time taken $t$ w.r.t. $b$

$$ t =\frac{1}{c} u+v+ \frac{1}{2} \frac{b^2}{D}+(n-1) k_1 \ln \frac{k_2}{b} \newline $$ $$ \Rightarrow \frac{d t}{d b} =0=\frac{b}{D}-(n-1) \frac{k_1}{b} \newline \therefore b^{2} =(n-1) k_1 D \newline \therefore b =\sqrt{(n-1) k_1 D} \newline $$

Thus, all rays passing at a height $b$ shall contribute to the image. The ray paths make an angle.

$$ \beta ; \frac{b}{v}=\frac{\sqrt{(n-1) k_1 D}}{v^{2}}=\sqrt{\frac{(n-1) k_1 u v}{v^{2}(u+v)}}=\sqrt{\frac{(n-1) k_1 u}{(u+v) v}} $$

This is the required expression.