Geometrical Optics

For a single spherical surface of radius $R$ separating two media of refractive indices $n_1$ and $n_2$, the relation between object distance $u$, image distance $v$, and radius of curvature $R$ is given by:

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

A thin lens consists of two such spherical surfaces placed very close to each other.

Let $n_L$ be the refractive index of the surrounding liquid medium.

$n_g$ be the refractive index of the glass material of the lens $n_g = 1.5$.

$R_1$ and $R_2$ be the radii of curvature of the first and second surfaces, respectively.

Using Lens Maker's formula, the focal length of the lens is,

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

Substituting $R_1 = +R$ and $R_2 = -R$:

$\frac{1}{f} = \left(\frac{n_g}{n_L} - 1\right)\left(\frac{1}{R} - \left(-\frac{1}{R}\right)\right)$

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

The optical power $P$ of a lens in diopters (D) is defined as the reciprocal of its focal length when measured in meters.

$P = \frac{1}{f_{\text{meters}}} = \left(\frac{n_g}{n_L} - 1\right)\frac{2}{R_{\text{cm}}/100} = \left(\frac{n_g}{n_L} - 1\right)\frac{200}{R_{\text{cm}}}$

The refractive index of glass is $n_g = 1.5$

Radius of curvature, $R_{\text{cm}} = 20\ \text{cm}$

$P(n_L) = \left(\frac{1.5}{n_L} - 1\right)\frac{200}{20}$

$P(n_L) = 10\left(\frac{1.5}{n_L} - 1\right)$

$P(n_L)=\frac{15}{n_L}-10$

Case 1: when the setup is immersed in air $n_L=1.0$

$P(1.0)=\frac{15}{1.0}-10=15-10=+5\ \mathrm{D}$

Thus, the graph must pass through the coordinate point $(1.0,5)$.

Case 2: when the setup is immersed in a matching medium $n_L=1.5$

$P(1.5)=\frac{15}{1.5}-10=10-10=0\ \mathrm{D}$

So, the graph crosses the horizontal axis at $(1.5,0)$.

Case 3: when the setup is immersed in a higher index liquid $n_L=2.0)$

$P(2.0)=\frac{15}{2.0}-10=7.5-10=-2.5\ \mathrm{D}$

When $n_L>n_g$, the lens changes its nature, turning from a converging lens to a diverging lens, so the power becomes negative.

The mathematical relation is of power of lens and the refractive index of medium is,

$P=\frac{15}{n_L}-10 \rightarrow P+10=\frac{15}{n_L} \rightarrow (P+10)n_L=15$

This equation is of the form $(y-y_0)x=c$, which represents a rectangular hyperbola shifted vertically downward by 10 units.

As $n_L\geq 1.0$, $P\leq +5\mathrm{D}$

As $n_L\rightarrow \infty$, $P\rightarrow -10$.

Therefore, the correct option is (A).

$ \begin{aligned} & \frac{S-t}{2}+\frac{t}{2 n_0}=2 f \\ & S+t\left(\frac{1}{n_0}-1\right)=4 f \\ & S=4 f+\left(1-\frac{1}{n_0}\right) t \end{aligned} $

Also

$ \begin{aligned} & \frac{S-t}{2}+\frac{t}{2 n_0}=f \\ & S=2 f+\left(1-\frac{1}{n_0}\right) t \end{aligned} $

Since STU is a plane mirror, we can take mirror image of the whole situation about it and final image can be assumed to be at a distance h below the base.

Since object and image are at same distance from equivalent lens, hence $\mathrm{h}=2 \mathrm{~F}_{\mathrm{qq}}$

$\begin{aligned} & \frac{1}{\mathrm{~F}_{\mathrm{eq}}}=\left(\frac{1.6-1}{1}\right)\left(\frac{2}{9}\right)+\frac{(\mathrm{n}-1)}{1}\left(\frac{-2}{9}\right) \\ & \frac{1}{\frac{\mathrm{h}}{2}}=\frac{1.2}{9}+\frac{2(1-\mathrm{n})}{9} \\ & \frac{2}{\mathrm{~h}}=\frac{3.2-2 \mathrm{n}}{9} \\ & \mathrm{~h}=\frac{9}{1.6-\mathrm{n}} \mathrm{cm} \end{aligned}$

(A) for $\mathrm{n}=1.42, \mathrm{~h}=50 \mathrm{~cm}$

(B) for $\mathrm{n}=1.35, \mathrm{~h}=36 \mathrm{~cm}$

(C) for $\mathrm{n}=1.45, \mathrm{~h}=60 \mathrm{~cm}$

(D) for $\mathrm{n}=1.48, \mathrm{~h}=75 \mathrm{~cm}$

$\begin{aligned} &\begin{aligned} & \alpha=\left(\theta_0-\phi_0\right)+\left(180-2 \phi_0\right)+\left(\theta_0-2 \phi_0\right) \\ & \alpha=180+2 \theta_0-4 \phi_0 \end{aligned}\\ &\begin{aligned} & \text { (P) } \alpha=180+2 \theta_0-4 \phi_0 \\ & 180=180+2 \theta_0-4 \phi_0 \Rightarrow \theta_0=2 \phi_0 \quad \text{... (i}\\ & \sin \theta_0=2 \sin \phi_0 \quad \text{... (ii} \end{aligned} \end{aligned}$

From (i) & (ii)

$\begin{aligned} & \sin \theta_0=2 \sin \left(\theta_0 / 2\right) \Rightarrow \cos \left(\frac{\theta_0}{2}\right)=1 \\ & \frac{\theta_0}{2}=0 \\ & \Rightarrow \theta_0=0 \end{aligned}$

$\begin{aligned} & \text { (Q) } \theta_0=2 \phi_0 \quad \text{... (i)}\\ & \sin \theta_0=\sqrt{3} \sin \phi_0 \quad \text{... (ii)} \end{aligned}$

From (i) & (ii)

$\begin{aligned} & \sin \theta_0=\sqrt{3} \sin \left(\frac{\theta_0}{2}\right) \\ & \Rightarrow \cos \left(\frac{\theta_0}{2}\right)=\frac{\sqrt{3}}{2} \\ & \frac{\theta_0}{2}=30,150 \\ & \theta_0=60,300 \text { (Rejected) } \\ & \theta_0=60 \& 0 \end{aligned}$

$\begin{aligned} & \text { (R) } \theta_0=2 \phi_0 \\ & \sin \theta_0=\sqrt{3} \sin \phi_0 \\ & \sin 2 \theta_0=\sqrt{3} \sin \phi_0 \\ & \cos \phi_0=\frac{\sqrt{3}}{2} \\ & \phi_0=30,150 \text { (Rejected) } \end{aligned}$

$\begin{aligned} & \phi_0=30 \& 0 \quad \text{... (i)}\\ & (S) \sin 45=\sqrt{2} \cos \phi_0 \\ & \cos \phi_0=1 / 2 \\ & \phi_0=60 \\ & \alpha=180+2 \theta_0-4 \phi_0 \end{aligned}$

$\begin{aligned} & \alpha=180+90-120 \quad \text{... (iv)}\\ & =180-30 ; \alpha=150^{\circ} \end{aligned}$

At point A on one end of the slab, the wavefront takes a time $ t $ to reach the other end. The time $ t $ can be expressed as the ratio of the distance traveled, $ d $, to the speed of the wave in the medium, which is $ \frac{c}{n_1} $ (where $ c $ is the speed of light in vacuum and $ n_1 $ is the refractive index of the medium at A). Therefore, the time taken $ t $ is given by $ t = \frac{d}{\frac{c}{n_1}} = \frac{n_1 d}{c} $.

Simultaneously, consider the wavefront at point B, located in a different medium with a refractive index $ n_2 $. In the same time $ t $, the wavefront at B travels a distance $ s_1 $. The distance $ s_1 $ can be calculated as the product of the speed of light in the medium at B, $ \frac{c}{n_2} $, and the time $ t $, yielding $ s_1 = \frac{c}{n_2} \times \frac{n_1 d}{c} = \frac{d n_1}{n_2} $.

Once the wavefront at B reaches the end of the slab, an additional time $ t_1 $ elapses. The time $ t_1 $ can be determined by considering the remaining distance the wavefront must travel in the slab, which is $ d - s_1 $, and dividing it by the speed of light in the medium at B. This gives us $ t_1 = \left(\frac{d - s_1}{c}\right) n_2 = \left(d - \frac{d n_1}{n_2}\right) \frac{n_2}{c} = \frac{d(n_2 - n_1)}{c} $.

During this time $ t_1 $, the wavefront at A will travel a distance $ s_2 $. The distance $ s_2 $ is simply the speed of light in vacuum multiplied by the time $ t_1 $, leading to $ s_2 = c t_1 = d(n_2 - n_1) $.

Finally, the angle $ \theta $ can be determined using the tangent function, where $ \tan \theta $ is the ratio of the difference in distances traveled by the wavefronts in the two media to the height $ h $ of the slab.

Therefore, $ \tan \theta = \frac{(n_2 - n_1) d}{h} $.

When light strikes a surface at Brewster's angle, it becomes polarized perpendicular to the plane of incidence.

Therefore, there is no reflection of blue light observed.

This implies that the light must be polarized within the plane of incidence.



$\begin{aligned} \tan i & =\sqrt{3} \\\\ i & =60^{\circ} \\\\ \delta & =i+e-A \\\\ 60 & =60+e-A \\\\ \Rightarrow A & =e\end{aligned}$

$\begin{aligned} \frac{\sin i}{\sin r_1} & =\sqrt{3} \\\\ \Rightarrow \frac{\sin 60}{\sin r_1} & =\sqrt{3} \Rightarrow r_1=30^{\circ}\end{aligned}$

$\begin{array}{rlrl}r_2 =A-r_1 \\\\ r \sin \left(A-r_1\right) =\sin e \\\\ \Rightarrow \sqrt{3} \sin (A-30) =\sin A \\\\ r_3\left[\sin A \cos 30^{\circ}-\cos A\right. \sin 30]=\sin A \\\\ \Rightarrow \frac{\sqrt{3}}{2}(\sqrt{3}-\cot A) =1 \\\\ \Rightarrow \cot A =\sqrt{3}-\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\\\ \Rightarrow \cot A =\frac{1}{\sqrt{3}} \\\\ \Rightarrow A =60^{\circ}\end{array}$

So, $\quad e=60^{\circ}$ (for blue)

For red,

$ \begin{aligned} \mu & =\frac{\sin \left(\frac{\delta_m+A}{2}\right)}{\sin \frac{A}{2}} \\\\ & =\frac{\sin \left(\frac{30+60}{2}\right)}{\sin \frac{60}{2}}=\frac{\sin 45}{\sin 30}=\frac{1 / \sqrt{2}}{1 / 2} \\\\ \mu_{\text {red }} & =\sqrt{2} \end{aligned} $

(A), (C), (D) are correct.

As we can see, for $\theta=30^\circ$, the ray will incident normally and hence will retrace its path.

$\Rightarrow$ (A) is correct

Considering the symmetry of the situation, we can have :

$\Rightarrow$ (B) is correct

As is clear from the above diagram, ray comes out.

$\Rightarrow$ (C) is not correct

Also, as is clear from the above diagram, total number of reflections = 5.

(D) is not correct.

(I) $u = - 20$ cm

$f = + 10$ cm

${1 \over v} + {1 \over {20}} = {1 \over {10}}$

${1 \over v} = {1 \over {10}} - {1 \over {20}}$

${1 \over v} = {1 \over {20}}$

$v = 20$ cm

$u = + 15$ cm

$f = + 15$ cm

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

$ \Rightarrow {1 \over v} - {1 \over {15}} = {1 \over {15}}$

${1 \over v} = {2 \over {15}}$

$v = 7.5$ cm (from lens 2)

I $\to$ P

(II) $u = - 20$ cm, $f = + 10$ cm

${1 \over v} + {1 \over {20}} = {1 \over {10}}$

${1 \over v} = {1 \over {10}} - {1 \over {20}} \Rightarrow v = 20$ cm

$u = + 15$ cm

$f = - 10$ cm

${1 \over v} - {1 \over u} = {1 \over f} \Rightarrow {1 \over v} - {1 \over {15}} = {{ - 1} \over {10}}$

${1 \over v} = {{ - 1} \over {10}} + {1 \over {15}} = {{ - 3 + 2} \over {30}}$

${1 \over v} = - {1 \over {30}}$

$v = - 30$ cm

II $\to$ R

(III) $u = - 20$ cm

$f = + 10$ cm

${1 \over v} + {1 \over {20}} = {1 \over {10}}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$

$v = 20$ cm

$ \Rightarrow u = 15$ cm

$f = - 20$ cm

${1 \over v} - {1 \over {15}} = {{ - 1} \over {20}}$

${1 \over v} = - {1 \over {20}} + {1 \over {15}} = {{ - 3 + 4} \over {60}} = {1 \over {60}}$

$v = 60$ cm

III $\to$ Q

(IV) $u = - 20$ cm

$f = - 20$ cm

$ \Rightarrow {1 \over v} - {1 \over u} = {1 \over f}$

$ \Rightarrow {1 \over v} + {1 \over {20}} = {{ - 1} \over {20}}$

$v = - 10$ cm

$u = - 15$ cm

$f = 10$ cm

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

${1 \over v} + {1 \over {15}} = {1 \over {10}}$

$ \Rightarrow {1 \over v} = {1 \over {10}} - {1 \over {15}}$

$ = {{3 - 2} \over {30}} = {1 \over {30}}$

$v = 30$ cm

IV $\to$ T

${n_1} = {n_2} = n = {3 \over 2}$ for minimum deviation.

${r_m} = {A \over 2} = {{60} \over 2} = 30^\circ $ and $i = e = \theta $

$n = {{\sin \left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}} $

$\Rightarrow {3 \over 2} = {{\sin \theta } \over {\sin \left( {{{60} \over 2}} \right)}}$

$ \Rightarrow \sin \theta = {3 \over 4} \Rightarrow \cos \theta = {{\sqrt 7 } \over 4}$

If ${n_1} = n = {3 \over 2}$ and ${n_2} = n + \Delta n$

$e = \theta + \Delta \theta $

$(1)\sin \theta = n\sin 30^\circ $

$(n + \Delta n)\sin 30^\circ = (1)\sin (\theta + \Delta \theta )$

Solving ${1 \over 2}(\Delta n) = \sin (\theta + \Delta \theta ) - \sin \theta $

${{\Delta n} \over 2} = {{\sin (\theta + \Delta \theta ) - sin\theta } \over {\Delta \theta }} \times \Delta \theta $

${{d(\sin \theta )} \over {d\theta }} = \cos \theta $

${{\Delta n} \over 2} = (\cos \theta )(\Delta \theta ) \Rightarrow {{\Delta n} \over 2} = {{\sqrt 7 } \over 4}(\Delta \theta )$

$ \Rightarrow {{\Delta n} \over {\Delta \theta }} = {{\sqrt 7 } \over 2} = {{2.64} \over 2}$

${{\Delta n} \over {\Delta \theta }} = 1.34 > 1$

$\Rightarrow$ $\Delta$n > $\Delta$$\theta$, so option (a) is incorrect.

$\Delta n = (1.34)\Delta \theta $

$ \Rightarrow \Delta \theta \propto \Delta n \Rightarrow $ option (b) is correct.

${{\Delta n} \over {\Delta \theta }} = {{\sqrt 7 } \over 2} \Rightarrow {{2.8 \times {{10}^{ - 3}}} \over {\Delta \theta }} = {{\sqrt 7 } \over 2}$

$\Delta \theta = {{5.6 \times {{10}^{ - 3}}} \over {\sqrt 7 }} = \sqrt 7 \times 0.8 \times {10^{ - 3}}$

$\Delta \theta = (2.64 \times 0.8) \times {10^{ - 3}} = 2.11 \times {10^{ - 3}}$ rad

So, option (c) is correct.

$\sin \theta > {1 \over {{n_1}}}$ (Given)

i.e., $\sin {\theta _1} > {1 \over {{n_1}}}$

${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$

$\sin {\theta _2} = {{{n_1}\sin {\theta _1}} \over {{n_2}}}$

if ${n_1} = {n_2}$ then ${\theta _2} = {\theta _1}$

${n_2}\sin {\theta _2} = (1)\sin {\theta _3}$

$\sin {\theta _3} = {n_2}\sin {\theta _2}$

$\sin {\theta _3} = {n_1}\sin {\theta _1}$

$\sin {\theta _1} = {{\sin {\theta _3}} \over {{n_1}}} > {1 \over {{n_1}}}$

$\sin {\theta _3} > 1$

${\theta _3} > 90^\circ $

This means ray cannot enter air.

For ${n_1} > {n_2};\sin {\theta _1} = {{{n_2}} \over {{n_1}}}\sin {\theta _2} > {1 \over {{n_1}}}$

$\sin {\theta _2} > {1 \over {{n_2}}}$

For surface 2 - air interface

${n_2}\sin {\theta _2} = \sin {\theta _3}$

$\sin {\theta _2} = {{\sin {\theta _3}} \over {{n_2}}} > {1 \over {{n_2}}}$

${\theta _2} > 90^\circ $

It means ray is reflected back in medium-2


For surface 1 and surface 2 interface

${n_2}\sin {\theta _2} = {n_1}\sin {\theta _1}$

$\sin {\theta _{2C}} = {{{n_1}} \over {{n_2}}}$

${\theta _{2C}}$ : critical angle

For ray to enter medium - 1

${\theta _2} < {\theta _{2C}}$

$\sin {\theta _2} < \sin 2{\theta _C}$

${{{n_1}} \over {{n_2}}}\sin {\theta _1} < {{{n_1}} \over {{n_2}}}$

$\sin {\theta _1} < 1$

${\theta _1} < 90^\circ $, which is true.

Hence, ray enters medium - 1

For ${n_2} > {n_1}$

${{{n_2}} \over {{n_1}}}\sin {\theta _2} > {{{n_2}} \over {{n_1}}}$

$\sin {\theta _2} > {1 \over {{n_2}}}$

For surface 2 - air interface

${n_2}\sin {\theta _2} = \sin {\theta _3}$

$\sin {\theta _2} = {{\sin {\theta _3}} \over {{n_2}}} > {1 \over {{n_2}}}$

${\theta _2} > 90$

It means ray reflected back in medium - 2


${n_2}\sin {\theta _2} = {n_1}\sin {\theta _1}$

$\sin {\theta _1} = {{{n_2}} \over {{n_1}}}\sin {\theta _2}$

$\sin {\theta _{2C}} = {{{n_1}} \over {{n_2}}};{\theta _{2C}}$ $\to$ critical angle

For ray to enter medium - 1

${\theta _2} < {\theta _{2C}}$

$\sin {\theta _2} < \sin {\theta _{2C}}$

${{{n_1}} \over {{n_2}}}\sin {\theta _1} < {{{n_1}} \over {{n_2}}}$

$\sin {\theta _1} < 1$

${\theta _1} = 90^\circ $, which is true.

Hence, ray enters medium - 1

Let, n₂ = 1


${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2} \Rightarrow {n_2} = 1$

${n_1}\sin {\theta _1} = \sin {\theta _2}$

$\sin {\theta _1} = {{\sin {\theta _2}} \over {{n_1}}} > {1 \over {{n_1}}}$

$\sin {\theta _2} > 1 \Rightarrow {\theta _2} = 90^\circ $

$\therefore$ Ray is reflected back in medium.

Focal length of convex lens f₁,

Therefore ${1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right] = (1.5 - 1)\left[ {{1 \over {20}} - \left( {{1 \over { - 20}}} \right)} \right]$

$\Rightarrow$ f₁ = + 20 cm

Focal length of concave lens f₂,

$\therefore$ ${1 \over {{f_2}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

${1 \over {{f_2}}} = (1.5 - 1)\left[ { - {1 \over {20}} - {1 \over {20}}} \right] = {1 \over { - 20}}$

$\Rightarrow$ f₂ = $-$ 20 cm

For lens 1


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

$\Rightarrow$ v = $-$ 20 cm

${m_1} = {v \over u} = {{ - 20} \over { - 10}} = 2$

For lens 2

u = $-$ 30n cm, f = $-$ 20 cm,

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

v = $-$ 12 cm

${m_2} = {v \over u} = {{ - 12} \over { - 30}} = {2 \over 5}$

Net magnification,

$m = {m_1}{m_2} = 2 \times {2 \over 5} = {4 \over 5} = 0.8$

${(R - h)^2} + {r^2} = {R^2}$

${R^2} + {h^2} - 2Rh + {r^2} = {R^2}$

${h^2} + {r^2} = 2Rh \Rightarrow {{{h^2} + {r^2}} \over {2h}} = R$ .....(i)

When r > > > h, then

$R = {{{r^2}} \over {2h}} \Rightarrow h = {{{\omega ^2}{r^2}} \over {2g}}$

$R = {{{r^2}2g} \over {2{\omega ^2}{r^2}}} \Rightarrow R = {g \over {{\omega ^2}}}$ ..... (ii)

${{{\mu _1}} \over v} - {{{\mu _2}} \over u} = {{{\mu _1} - {\mu _2}} \over R} $

$\Rightarrow {1 \over v} + {4 \over {3u}} = {{1 - {4 \over 3}} \over R}$

${1 \over v} = - \left( {{1 \over {3R}} + {4 \over {3(H - h)}}} \right)$

${1 \over v} = - \left( {{1 \over {3R}} + {4 \over {3H}}} \right)$

Putting the value of Eq. (ii) h < < H

${1 \over v} = - \left[ {{4 \over {3H}}\left( {1 + {{3H} \over 4}{{{\omega ^2}} \over {39}}} \right)} \right]$

$v = - \left[ {{{3H} \over 4}{{\left( {1 + {{{\omega ^2}H} \over 4}} \right)}^{ - 1}}} \right]$

Case - I

H = 30 cm

n = ${3 \over 2}$

H₁ = ${H \over n}$

$ \Rightarrow $ ${{30 \times 2} \over 3}$ = 20 cm

Case - II

R = 300 cm

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

${1 \over { - {H_2}}} - {3 \over { - 2 \times 30}} = {{1 - {3 \over 2}} \over { - 300}}$

${H_2} = {{600} \over {29}} = 20.684$ cm

Case - III

${{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_2} - {n_1}} \over R}$;

${1 \over { - {H_3}}} - {3 \over { - 2 \times 30}} = {{1 - {3 \over 2}} \over {300}}$

${H_3} = {{600} \over {31}} = 19.354$ cm

Given, thin convex lens made of two materials.

Radius of curvature of left spherical surface $=$ radius of curvature of right spherical surface

When $n_1=n_2=n$, focal length $=f$

When $n_1=n$ and $n_2=n+\Delta n$, focal length $=f+\Delta f$

Here, $\Delta n \ll(n-1)$ and $1< n < 2$

Now, when $n_1=n_2=n$, from lens maker's formula

$ \begin{aligned} \frac{1}{f} & =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(n-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \\\\ \Rightarrow \frac{1}{f} & =(n-1)\left(\frac{2}{R}\right) ........(1) \end{aligned} $

When $n_1=n$ and $n_2=n+\Delta n$

$ \frac{1}{f+\Delta f}=(n-1) \frac{1}{R_1}-(n+\Delta n-1) \frac{1}{+R_2} $

$ \begin{aligned} & \quad=(n-1) \frac{1}{R}-(n+\Delta n-1) \frac{1}{(-R)} \\\\ & \Rightarrow \frac{1}{f+\Delta f}=\frac{1}{R}(n-1+n+\Delta n-1)=\frac{1}{R}(2 n+\Delta n-2) \\\\ & \Rightarrow \frac{1}{f+\Delta f}=\frac{1}{R}(2 n+\Delta n-2) ........(2) \end{aligned} $

Subtracting Eq. (2) from Eq. (1), we get

$ \begin{aligned} & \frac{1}{f}-\frac{1}{f+\Delta f}=(n-1)\left(\frac{2}{R}\right)-\frac{1}{R}(2 n+\Delta n-2) \\\\ & =\frac{1}{R}(2 n-2-2 n-\Delta n+2)=-\frac{\Delta n}{R} \\\\ & \Rightarrow \frac{1}{f}-\frac{1}{f+\Delta f}=-\frac{\Delta n}{R} \Rightarrow \frac{(f+\Delta f)-f}{f(f+\Delta f)}=-\frac{\Delta n}{R} \\\\ & \Rightarrow \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R} \sim \frac{\Delta f}{f}=-\frac{\Delta n}{R} \\\\ & \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{R} \times f \end{aligned} $

Using Eq. (1), we get $f=\frac{R}{2(n-1)}$

$ \begin{aligned} & \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{R} \times \frac{R}{2(n-1)} \Rightarrow \frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)} \\\\ & \Rightarrow\left|\frac{\Delta f}{f}\right|=\frac{1}{2}\left|\frac{\Delta n}{n-1}\right| \\\\ & \Rightarrow\left|\frac{\Delta f}{f}\right|>\left|\frac{\Delta n}{n}\right| \end{aligned} $

Hence, option (C) is incorrect.

Now, for $n=1.5, \Delta n=10^{-3}$ and $f=20 \mathrm{~cm}$. Using relation $\frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R}$ and Eq. (1), we get

$ \begin{aligned} & \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{2(n-1) f} \Rightarrow \frac{\Delta f}{f+\Delta f}=-\frac{\Delta n}{2 n-2} \\\\ & \Rightarrow \Delta f(2 n-2)=-\Delta n(f+\Delta f) \\\\ & \Rightarrow \Delta f(2 n-2)=-\Delta n f-\Delta n \Delta f \\\\ & \Rightarrow \Delta f(2 n-2+\Delta n)=-\Delta n f \\\\ & \Rightarrow \Delta f=\frac{-\Delta n \times f}{2 n-2+\Delta n} \\\\ & \Rightarrow|\Delta f|=\frac{\Delta n+f}{2 n-2+\Delta n} \end{aligned} $

Substituting all values, we get

$ |\Delta f|=\frac{10^{-3} \times 20}{2 \times 1.5-2+10^{-3}}=0.02 \mathrm{~cm} $

Hence, option (B) is correct.

Since $\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$, so if $\frac{\Delta n}{n}<0$ then $\frac{\Delta f}{f}>0$.

Hence, option $(\mathrm{A})$ is correct.

Since the relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ is independent of radius of curvature so the relation remains unchanged only the sign of focal length changes.

Hence, option (D) is correct.

Consider the point F, the focus of the concave mirror. All rays emanating from F will be reflected back parallel to the principal axis PF. Thus, virtual image F' of the point F is formed at $\infty$.

Consider the point L at an object distance u = $-$ | f | / 2. For convenience, we use symbol $\widehat f$ for | f | i.e., $\widehat f$ = | f |. Use mirror formula, 1/v + 1/u = 1/f, with f = $-$ $\widehat f$ to get the image distance v = $\widehat f$. Thus, the virtual image L' of the point L is formed on the left side of the mirror at a distance $\widehat f$.

The image of a point on LF will lie on the line L'F'. Now, consider the point M. Its object distance is u = $-$ $\widehat f$/2. Thus, the virtual image M' of the point M is also formed at a distance $\widehat f$ from the mirror. The distance of M' above the principal axis is given by h_M' = $-$ (v/u) (LM) = $-$ ($-$2) ($\widehat f$/2) = $\widehat f$ (magnification by a spherical mirror).

Now consider a point N on the line segment FM. Let N₁ be the projection of N on the principal axis PF such that FN₁ = x. The object distance is u = $-$ ($\widehat f$ $-$ x). The mirror formula gives the image distance v = $\widehat f$ ($\widehat f$ $-$ x) / x. The distance of the image N' above the principal axis is given by h_N' = $-$ (v/u) (N₁N) = ($\widehat f$/x) (x) = $\widehat f$ (it is independent of x). Since, N is an arbitrary point on FM, the image of all points on FM will be formed at a distance $\widehat f$ above the optic axis.

We discuss the options as follows:

$\bullet$ For option (A) : For the angle of incidence i₁ = A, we have minimum deviation and the ray inside the prism is parallel to the base of the prism.

Hence, option (A) is correct.

$\bullet$ For option (C) : The angle of deviation is given as

$\delta$ = i₁ + i₂ $-$ A

At minimum deviation, i₁ = i₂ and r₁ = r₂. Therefore,

i₁ = A ....... (1)

Also, we know that

r₁ + r₂ = A

As r₁ = r₂, we get

2r₁ = A $\Rightarrow$ r₁ = ${A \over 2}$

From Eq. (1), we get

${r_1} = {{{i_1}} \over 2}$

Hence, option (C) is correct.

According to Snell's law, we have

${{\sin {i_2}} \over {\sin {r_2}}} = \mu $

For emergent ray to be tangential to the surface, we have

i₂ = 90$^\circ$

$\Rightarrow$ sin90$^\circ$ = $\mu$sinr₂ $\Rightarrow$ $\mu$sinr₂ = 1

sinr₂ = ${1 \over \mu }$ $\Rightarrow$ r₂ = sin^$-$1$\left( {{1 \over \mu }} \right)$ ....... (2)

From the relation r₁ + r₂ = A, we get

r₁ = A $-$ r₂

$\Rightarrow$ sinr₁ = sin(A $-$ r₂)

Using sin(a $-$ b) = sinacosb $-$ sinbcosa, we have

$\Rightarrow$ sinr₁ = sinAcosr₂ $-$ sinr₂cosA ......... (3)

Using Eq. (2), we have

sinr₂ = ${{1 \over \mu }}$

Squaring it, we get

sin²r₂ = ${{1 \over {{\mu ^2}}}}$

Using the relation sin²x + cos²x = 1, we get

sin²x = 1 $-$ cos²x

$\Rightarrow$ 1 $-$ cos² r₂ = ${{1 \over {{\mu ^2}}}}$

$\Rightarrow$ cos² r₂ = 1 $-$ ${{1 \over {{\mu ^2}}}}$

That is,

cosr₂ = $\sqrt {1 - {1 \over {{\mu ^2}}}} $

Substituting the values of sinr₂ and cosr₂ in Eq. (3), we get

sinr₁ = sin A$\sqrt {1 - {1 \over {{\mu ^2}}}} $ $-$ ${{1 \over \mu }}$cos A

According to Snell's law, we have

${{\sin {i_1}} \over {\sin {r_1}}} = \mu $

That is,

$\sin {i_1} = \mu \sin {r_1} = \mu \left( {\sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - {1 \over \mu }\cos A} \right)$

$\sin {i_1} = \mu \left( {\sin A{{\sqrt {{\mu ^2} - 1} } \over \mu } - {{\cos A} \over \mu }} \right)$

$\sin {i_1} = \sin A\sqrt {{\mu ^2} - 1} - \cos A$ ....... (4)

Now, at minimum deviation, we have A = i₁ and r₁ = ${A \over 2}$. Thus, from Snell's law, we get

$\mu = {{\sin {i_1}} \over {\sin {r_1}}} = {{\sin A} \over {\sin (A/2)}}$

Using $\sin x = 2\sin {x \over 2}\cos {x \over 2}$, we get

$\mu = {{2\sin A/2\cos A/2} \over {\sin A/2}} \Rightarrow \mu = 2\cos {A \over 2}$ ....... (5)

Now, Eq. (4) becomes

$\sin {i_1} = \sin A\sqrt {{{\left( {2\cos {A \over 2}} \right)}^2} - 1 - \cos A} $

$ = \sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A$

${i_1} = {\sin ^{ - 1}}\left[ {\sin A\sqrt {4{{\cos }^2}{A \over 2} - 1} - \cos A} \right]$

Hence, option (D) is correct.

$\bullet$ For option (B) : From Eq. (5), we have

$\mu = 2\cos {A \over 2}$

$\cos {A \over 2} = {\mu \over 2}$

$ \Rightarrow {A \over 2} = {\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$

$ \Rightarrow A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$

Hence, option (B) is incorrect.

For refraction through plano - Convex lens,

$ m=-2=\frac{v}{u} $

$ \begin{array}{ll} \Rightarrow v=-2 u \\\\ \text { As, } u=-30 \mathrm{~cm} \\\\ \Rightarrow v=60 \mathrm{~cm} \end{array} $

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$ \begin{aligned} \frac{1}{60}-\frac{1}{-30} & =\frac{1}{f} \\\\ \Rightarrow f =20 \mathrm{~cm} \end{aligned} $



By lens maker's formula,

$ \begin{aligned} \frac{1}{f} & =(n-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\\\ \Rightarrow \frac{1}{20} & =(n-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\infty}\right)=\frac{n-1}{\mathrm{R}} \\\\ \Rightarrow n & =1+\left(\frac{\mathrm{R}}{20}\right) ........(i) \end{aligned} $

for poor reflection from convex surface,

$ \begin{aligned} & u=-30 \mathrm{~cm}, \\\\ & v=10 \mathrm{~cm}, f=\mathrm{R} / 2 \end{aligned} $



Using mirror formula,

$ \begin{aligned} \frac{1}{v}+\frac{1}{u} & =\frac{1}{f}=\frac{2}{\mathrm{R}} ; \\\\ \Rightarrow \frac{1}{10}+\frac{1}{-30} & =\frac{2}{\mathrm{R}} \end{aligned} $

$ \begin{array}{llrl} \Rightarrow \frac{2}{\mathrm{R}} =\frac{2}{30} \\\\ \therefore \mathrm{R} =30 \mathrm{~cm} \end{array} $

Substitute $R=30 \mathrm{~cm}$ in equation (i) to get $n=$ 2.5 for $f=20 \mathrm{~cm}$ (inverted image) and $n=1.5$ for $f=60 \mathrm{~cm}$ (erect image).

From Snell's law,

$ \begin{aligned} n_1 \sin \theta_i=n\left(z_1\right) \sin \theta_{z_1}=n\left(z_2\right) \sin \theta_{z_2} & =\ldots \ldots \\\\ \ldots & =n(d) \sin \theta_d=n_2 \sin \theta_f \end{aligned} $

The lateral displacement of a ray by a parallel slab depends on the angle of incidence $\theta_i$, refractive index $n_1$, slab thickness $d$ and the refractive index of the slab $n$ (it is independent of $n_2$ because $l$ is measured before the start of medium 2, there is no effect of $n_2$ on $l$ ). Consider the parallel slab shown in the figure.


By Snell's law, $n_1 \sin i_1=n \sin r_1$ and $n \sin i_2=$ $n_2 \sin r_2$. By geometry, $i_2=r_1$ and the lateral displacement is given by

$ l=d \frac{\sin r_1}{\cos r_1}=\frac{d n_1 \sin i_1}{\sqrt{n^2-n_1^2 \sin ^2 i_1}} . $

Hence, $l$ is independent of $n_2$ but it is dependent of $n(z)$.

For convex lens, u = $-$50 cm, f = 30 cm

$\therefore$ ${1 \over v} = {1 \over f} + {1 \over u} = {1 \over {30}} - {1 \over {50}} = {{5 - 3} \over {150}} = {2 \over {150}}$ or v = 75 cm

Image formed by convex lens acts as a virtual object for mirror.

If we consider that the mirror is not tilted, then for mirror,

${u_1} = 75 - 50 = 25$ cm, ${f_1} = {R \over 2} = {{100\,cm} \over 2} = 50$ cm

$\therefore$ ${1 \over {{v_1}}} = {1 \over {{f_1}}} - {1 \over {{u_1}}} = {1 \over {50}} - {1 \over {25}} = {{ - 1} \over {50}}$ or ${v_1} = 50$ cm

So, the co-ordinates of final image formed (I₂) if the mirror is not tilted are (0 cm, 0 cm)

If mirror is not tilted then this ray starts from the object, refracts through the lens without deviation, incident normally on the mirror and finally comes to the image I₂ after reflection from the mirror. If the mirror is tilted by an angle $\theta$ then this ray is incident on the mirror at an angle of reflection $\theta$ and finally forms a new image at I'₂ (see figure). Thus, the axis on which image lies makes an angle 2$\theta$ = 60$^\circ$ with the principal axis of the lens. Also, the image distance will remain same upto the first order of approximation (image distance will remain exactly same in case of plane mirror). Thus, (x, y) coordinates of the image are

$x = 50 - 50\cos 60^\circ = 25$ cm,

$y = 50\sin 60^\circ = 25\sqrt 3 $ cm.

Applying Snell's law at M,

$n = {{\sin \alpha } \over {\sin {r_1}}} \Rightarrow \sqrt 2 = {{\sin 45^\circ } \over {\sin {r_1}}}$

$ \Rightarrow \sin {r_1} = {{\sin 45^\circ } \over {\sqrt 2 }} = {{1/\sqrt 2 } \over {\sqrt 2 }} = {1 \over 2}$

${r_1} = 30^\circ $

$\sin {\theta _c} = {1 \over n} = {1 \over {\sqrt 2 }} \Rightarrow {\theta _c} = 45^\circ $

Let us take ${r_2} = {\theta _c} = 45^\circ $ for just satisfying the condition of TIR.

In $\Delta PNM$,

$\theta + 90 + {r_1} + 90 - {r_2} = 180^\circ $

or $\theta = {r_2} - {r_1} = 45^\circ - 30^\circ = 15^\circ $

Note If $\alpha$ = 45$^\circ $ (the given value). Then, r₁ > 30$^\circ $ (the obtained value)

${r_2} > {\theta _c}$ (as r₂ $-$ r₁ = $\theta$ or r₂ = $\theta$ + r₁)

or TIR will take place. So, for taking TIR under all conditions $\alpha$ should be greater than 45$^\circ$ or this is the minimum value of $\alpha$.

Let the whole structure be placed in the medium of refractive index n₀. From geometry, if the angle of incidence at Q is $\theta$ then the angle of refraction at P is 90$^\circ$ $-$ $\theta$.

The ray will undergo total internal reflection at Q if the angle of incidence at Q is greater than or equal to the critical angle i.e.,

$\theta$ $\ge$ $\theta$_c = sin^$-$1(n₂ / n₁). ........ (1)

Apply Snell's law for refraction at P to get

n₀ sin i = n₁ sin(90$^\circ$ $-$ $\theta$) = n₁ cos$\theta$

= ${n_1}\sqrt {1 - {{\sin }^2}\theta } $. ...... (2)

From equation (2), the angle of incidence is maximum (i = i_m) when $\theta$ is minimum i.e., when $\theta$ = $\theta$_c (from equation (1)). Thus, the numerical aperture is given by

$NA = \sin {i_m} = {{{n_1}\sqrt {1 - {{\sin }^2}{\theta _c}} } \over {{n_0}}} = {{\sqrt {n_1^2 - n_2^2} } \over {{n_0}}}$ ........ (3)

Substitute ${n_1} = \sqrt {45} /4$ and ${n_2} = 3/2$ in equation (3) to get the numerical aperture for the structure S₁ is

$N{A_1} = {{\sqrt {45/16 - 9/4} } \over {{n_0}}} = {3 \over {4{n_0}}}$ ...... (4)

Similarly, substitute n₁ = 8/4 and n₂ = 7/5 in equation (3) to get the numerical aperture for the structure S₂ as

$N{A_2} = {{\sqrt {64/25 - 49/25} } \over {{n_0}}} = {{\sqrt {15} } \over {5{n_0}}}$ ...... (5)

In case (A), substitute n₀ = 4/3 in equation (4) to get NA₁ = 9/16 and substitute n₀ = 16/3$\sqrt15$ in equation (5) to get NA₂ = 9/16.

In case (B), substitute n₀ = 6/$\sqrt15$ in equation (4) to get NA₁ = $\sqrt15$/8 and substitute n₀ = 4/3 in equation (5) to get NA₂ = 3$\sqrt15$/20.

In case (C), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/$\sqrt15$ equation (5) to get NA₂ = 3/4.

In case (D), substitute n₀ = 1 in equation (4) to get NA₁ = 3/4 and substitute n₀ = 4/3 equation (5) to get NA₂ = 3$\sqrt15$/20.

$ \sin i_m=n_1 \sin \left(90-\theta_c\right) $

$ \Rightarrow \sin i_m=n_1 \cos \theta_c $

$ \begin{aligned} \Rightarrow N A & =n_1 \sqrt{1-\sin ^2 \theta_c} \\\\ & =n_1 \sqrt{1-\frac{n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2} \end{aligned} $

Substituting the values we get,

$ N A_1=\frac{3}{4} $

$ \text { and } N A_2=\frac{\sqrt{15}}{5}=\sqrt{\frac{3}{4}} $

and $ N A_2 < N A_1 $

Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical aperture, which is $N A_2$.

Apply the formula for the refraction at the spherical surface of S₁, ${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$, to get

${{1.0} \over v} - {{1.5} \over {( - 50)}} = {{1 - 1.5} \over {( - 10)}}$.

Solve to get the image distance v = 50 cm (point Q in the figure).

The image Q acts as an object for refraction at the spherical surface of S₂. The object distance is $u = - (d - 50)$. The image distance is v = $\infty$ (because rays are parallel in S₂). Apply the formula for refraction at the spherical surface of S₂, to get

${{1.5} \over \infty } - {1 \over { - (d - 50)}} = {{1.5 - 1} \over {(10)}}$.

Solve to get d = 70 cm.

As the film has R₁ = R₂ = R (say), its focal length

${1 \over f} = ({n_1} - 1)\left( {{1 \over R} - {1 \over R}} \right)$


${1 \over f} = 0$ or $f = \infty $. It will not cause refraction.

Refraction will take place at air and glass cylinder. Given u = $\infty$

$ - {{{n_1}} \over u} + {{{n_2}} \over v} = {{{n_2} - {n_1}} \over R}$

$ - {{{n_1}} \over \infty } + {1.5 \over {{f_1}}} = {{1.5 - 1} \over R}$ or ${f_1} = 3R$

For glass-air refraction, we have

$ - {{{n_2}} \over u} + {{{n_1}} \over v} = {{{n_1} - {n_2}} \over R}$

$ - {{{n_2}} \over \infty } + {1 \over {{b_2}}} = {{1 - 1.5} \over R}$

or ${f_2} = - 2R$ or $\left| {{f_2}} \right| = 2R$.

In $\Delta$SAB

$\tan \theta = {{AB} \over {AS}} = {{11.54/2} \over {10}}$

tan$\theta$ = 0.577

$\theta$ = 30$^\circ$

Using ${{\sin {\theta _l}} \over {\sin 90}} = {{{n_l}} \over {{n_B}}}$, we get

$\sin {\theta _l} = {{{n_l}} \over {{n_B}}}$ or ${n_l} = {n_B} \times \sin 30 = 2.72 \times {1 \over 2} = 1.36$

The lens maker's formula gives focal length of the bi-convex lens as

${1 \over {{f_1}}} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right]$

$ = (1.5 - 1)\left[ {{1 \over r} - {1 \over { - r}}} \right] = {1 \over r}$ ..... (1)

The focal length of a plano-convex (also convex-plane) lens is given by

${1 \over {{f_2}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over { - r}}} \right] = {1 \over {2r}}$ ...... (2)

and that of a plano-concave (also concave-plano) lens is given by

${1 \over {{f_3}}} = (1.5 - 1)\left[ {{1 \over \infty } - {1 \over r}} \right] = - {1 \over {2r}}$ ..... (3)

Case P:

Therefore, ${1 \over {{f_{eff}}}} = {1 \over r} + {1 \over r}$

${f_{eff}} = {r \over 2}$

(P) $\to$ (2)

Case Q :

${1 \over {{f_{eff}}}} = {1 \over {2r}} + {1 \over {2r}}$

${f_{eff}} = r$

(Q) $\to$ (4)

Case R :

${1 \over {{f_{eff}}}} = - {1 \over {2r}} - {1 \over {2r}}$

${f_{eff}} = - r$

(R) $\to$ (3)

Case S : ${1 \over {{f_{eff}}}} = {1 \over r} - {1 \over {2r}} = {1 \over {2r}}$

${f_{eff}} = 2r$

(S) $\to$ (1)

For e $\to$ f

$\mu$₂ > $\mu$₁, as ray bends towards the normal.

$\mu$₂ > $\mu$₃, as ray bends away from the normal.

P $\to$ 2

For e $\to$ g

$\mu$₁ = $\mu$₂ as there is no deviation.

Q $\to$ 3

For e $\to$ h

$\mu$₂ < $\mu$₁, as ray bends away from the normal.

$\mu$₂ > $\mu$₃, as ray bends away from the normal.

Also, $\mu$₁ < $\sqrt2$$\mu$₂ (No total internal reflection)

R $\to$ 4

For e $\to$ i

Total internal reflection takes place

$\therefore$ sin45$^\circ$ > sinC

But $\sin C = {{{\mu _2}} \over {{\mu _1}}}$

$\therefore$ ${1 \over {\sqrt 2 }} > {{{\mu _2}} \over {{\mu _1}}} \Rightarrow {\mu _1} > \sqrt 2 {\mu _2}$

$S \to 1$

The speed of light in a medium is given by the equation :

$ \frac{C}{V}=\frac{\lambda_0}{\lambda}=\mu $

Where :

• V is the velocity of light in the medium
• λ is the wavelength of light in the medium
• λ₀ is the wavelength in free space
• μ is the refractive index of the medium

Given :

$ \lambda=\frac{2}{3} \lambda_0 $

Using the given data :

$\begin{aligned} u & = -24 \text{ m},\\\\ h_i & = \frac{1}{3}h_0\end{aligned}$

To find the magnification (m) :

$ m = \frac{h_i}{h_0} = \frac{1}{3} = \frac{v}{u} $

Hence,

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

Given :

$ v = 8 \text{ m},\ u = -24 \text{ m} $

Applying the lens formula :

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

Since the plane surface has an infinite radius :

$ \frac{1}{\infty} = 0 $

Therefore :

$\frac{1}{8} - \frac{1}{-24} = (1.5 - 1)\left(\frac{1}{R} - 0\right)$

$\Rightarrow \frac{1}{6} = \frac{0.5}{R}$

Solving for R :

$ \Rightarrow R = 3 \text{ m} $

Let ${\widehat v_i} = {1 \over 2}\left( {\widehat i + \sqrt 3 \widehat j} \right)$ and ${\widehat v_r} = {1 \over 2}\left( {\widehat i - \sqrt 3 \widehat j} \right)$ be the unit vectors along the incident and the reflected rays. From Snell's law, angle of incidence (say $\theta$) is equal to the angle of reflection. Thus, angle between $ - {\widehat v_i}$ and ${\widehat v_r}$ is 2$\theta$.

Using, dot products of vectors, we get

$\cos (2\theta ) = - {\widehat v_i}\,.\,{\widehat v_r} = 1/2$.

Hence, $\theta = 30^\circ $.

For meta-material, the refractive index is negative. Let n₁ is refractive index of air and n₂ is refractive index of meta-material.

$\therefore$ From Snell's law, ${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$

Since, n₂ is negative, therefore $\theta$₂ is also negative. Hence, appropriate diagram (c) is correct.

Refractive index for a medium

$n = \left( {{c \over v}} \right)$

For meta-material, $n = \,|n|$

$\therefore$ $v = {c \over {|n|}}$

According to lens maker's formula

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

For the first lens

n = 1.5, R₁ = + 14 cm, R₂ = $\infty$

$\therefore$ ${1 \over {{f_1}}} = (1.5 - 1)\left( {{1 \over {14}} - {1 \over \infty }} \right) = {{0.5} \over {14}}$

For the second lens,

n = 1.2, R₁ = $\infty$, R₂ = $-$14 cm

$\therefore$ ${1 \over {{f_2}}} = (1.2 - 1)\left( {{1 \over \infty } - {1 \over { - 14}}} \right) = {{0.2} \over {14}}$

The focal length of the bi-convex lens is

${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} = {{0.5} \over {14}} + {{0.2} \over {14}} = {{0.7} \over {14}} = {1 \over {20}}$

According to thin lens formula

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

Here, u = $-$40 cm

$\therefore$ ${1 \over v} - {1 \over { - 40}} = {1 \over {20}}$ or ${1 \over v} = {1 \over {20}} - {1 \over {40}}$ or v = 40 cm

Consider the refraction at the face AB. Snell's law,

$\sin i/\sin r = \sin 60^\circ /\sin r = \sqrt 3 $,

gives $r = 30^\circ $.

The geometry in BCQP gives

$\angle BPQ = 30^\circ + 90^\circ = 120^\circ $

$\angle CQP = 360^\circ - (135^\circ + 60^\circ + 120^\circ ) = 45^\circ $.

Thus, the angle of incidence at Q is ${i_1} = 45^\circ $. The critical angle for prism to air refraction is given by $\sin {i_c} = 1/\sqrt 3 $. Since $\sin {i_1} = 1/\sqrt 2 > 1\sqrt 3 $, we get ${i_1} > {i_c}$ i.e., the angle of incidence is greater than the critical angle. Thus, the ray undergoes total internal reflection at Q. The laws of reflection gives ${r_1} = {i_1} = 45^\circ $. In triangle QRD, $\angle QRD = 60^\circ $ and hence the angle of incidence at R is ${i_2} = 30^\circ $.

Applying Snell's law at face AD, we get

$\sqrt 3 \times \sin 30^\circ = 1 \times \sin e$

or, $\sqrt 3 \times {1 \over 2} = \sin e$

$\sin e = {{\sqrt 3 } \over 2}$

or, $e = {\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) = 60^\circ $

From figure,

The angle between the incident ray and the emergent ray is 90$^\circ$.

Hence, option (c) is correct and option (d) is incorrect.

Note : Angle between incident and emergent rays is the same as the angle between the two faces = 90$^\circ$.