Geometrical Optics

Total internal reflection (TIR) occurs at the interface between the prism ( $n_p=\sqrt{3}$ ) and the film ( $n_f=1.5$ ).

$ \sin C=\frac{n_{\text {rarer }}}{n_{\text {denser }}}=\frac{1.5}{\sqrt{3}}=\frac{\sqrt{3}}{2} \Rightarrow C=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^{\circ} $

So, the critical angle is $C=60^{\circ}$.

For TIR to occur at the second surface, the angle of incidence $r_2$ must satisfy:

$ r_2>C \Rightarrow r_2>60^{\circ} $

For a prism with angle $\mathrm{A}=75^{\circ}$ :

$ \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 \Rightarrow 75^{\circ}=\mathrm{r}_1+\mathrm{r}_2 $

Substituting the condition $\mathrm{r}_2>60^{\circ}$ :

$ r_1=75^{\circ}-r_2<75^{\circ}-60^{\circ} $

$\Rightarrow $ $ \mathrm{r}_1<15^{\circ} $

Apply Snell's Law at the first surface (assuming air to prism) :

$ \mathrm{n}_{\text {air }} \sin \mathrm{i}=\mathrm{n}_{\mathrm{p}} \sin \mathrm{r}_1 $

$\Rightarrow $ $ \sin \mathrm{i}=\sqrt{3} \sin \mathrm{r}_1 $

Since $\mathrm{r}_1<15^{\circ}$, then $\sin \mathrm{i}<\sqrt{3} \sin 15^{\circ}$ :

$ \sin i<1.732 \times 0.25 $

$\Rightarrow $ $ \sin i<0.433 $

Given that $\sin 25^{\circ}=0.43$, we can conclude:

$ \mathrm{i}<25^{\circ} $

For the angle $r_2$ to remain above $60^{\circ}$, i.e., the value of $r_1$ will be between 0 and $15^{\circ}$. The maximum value of $r_2$ can be $75^{\circ}$ for which $\mathrm{r}_1=0$.

$ \sin \mathrm{i}=\sqrt{3} \sin 0^{\circ}=0 $

So, the value of i will be between $0^{\circ}$ and $25^{\circ}$.

$ \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 $\mathrm{n}=1.48, \mathrm{~h}=75 \mathrm{~cm}$

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.

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

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

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

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

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 $n(z)$.

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.

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

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$^\circ$.

Using mirror formula for the concave mirror, we have

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

Option (A) : When the value is (42, 56), we have

${1 \over { - 56}} + {1 \over { - 42}} = {{ - 98} \over {56 \times 42}} \Rightarrow f = {{ - 56 \times 42} \over {98}} = - 24$ cm

Option (B) : When the value is (48, 48), we have

${1 \over { - 48}} + {1 \over { - 48}} = {{ - 1} \over {24}} \Rightarrow f = - 24$ cm

Option (C) : When the value is (66, 33), we have

${1 \over { - 33}} + {1 \over { - 66}} = {{ - 1} \over {22}} \Rightarrow f = - 22$ cm

Option (D) : When the value is (78, 39), we have

${1 \over { - 39}} + {1 \over { - 78}} = {{ - 1} \over {26}} \Rightarrow f = - 26$ cm

Since options (A) and (B) give $f = - 24$ cm, which is the same as that given in the problem, we conclude that the options (C) and (D) as the correct options.