Refraction Through Prism Questions in English

(C) At minimum deviation,$\delta_m = 2i - A = A$,which implies $i = A$. Since $i_1 = i_2 = i$ and $r_1 = r_2 = r$,we have $2r = A$,so $r = A/2$.
Thus,$r_1 = i_1/2$,making option [$A$] correct.
Using Snell's law,$\sin i = \mu \sin r \implies \sin A = \mu \sin(A/2) \implies 2 \sin(A/2) \cos(A/2) = \mu \sin(A/2) \implies \mu = 2 \cos(A/2)$.
Rearranging gives $\cos(A/2) = \mu/2$,so $A = 2 \cos^{-1}(\mu/2)$. Option [$B$] is incorrect.
For the emergent ray to be tangential,$i_2 = 90^\circ$,so $r_2 = \theta_c = \sin^{-1}(1/\mu)$. Then $r_1 = A - r_2 = A - \sin^{-1}(1/\mu)$.
Using $\sin i_1 = \mu \sin r_1 = \mu \sin(A - \sin^{-1}(1/\mu)) = \mu [\sin A \cos(\sin^{-1}(1/\mu)) - \cos A \sin(\sin^{-1}(1/\mu))] = \mu [\sin A \sqrt{1 - 1/\mu^2} - \cos A (1/\mu)] = \sin A \sqrt{\mu^2 - 1} - \cos A$.
Substituting $\mu = 2 \cos(A/2)$,$\mu^2 - 1 = 4 \cos^2(A/2) - 1$. Thus,$i_1 = \sin^{-1}[\sin A \sqrt{4 \cos^2(A/2) - 1} - \cos A]$. Option [$C$] is correct.
At minimum deviation,the ray inside the prism is parallel to the base,which occurs when $i_1 = i_2 = A$. Option [$D$] is correct.

(A) $1$. At face $AB$: Using Snell's law,$1 \cdot \sin(60^{\circ}) = \sqrt{3} \cdot \sin(r_1)$.
$\sin(r_1) = \frac{\sqrt{3}/2}{\sqrt{3}} = 1/2$,so $r_1 = 30^{\circ}$.
$2$. In the triangle formed inside the prism,the angle of incidence at face $CD$ is $r_2 = 180^{\circ} - 60^{\circ} - (90^{\circ} - 30^{\circ}) - 45^{\circ} = 45^{\circ}$.
$3$. Critical angle $C = \sin^{-1}(1/\sqrt{3}) \approx 35.26^{\circ}$. Since $r_2 = 45^{\circ} > C$,the ray undergoes total internal reflection at face $CD$.
$4$. After reflection,the ray strikes face $AD$ at an angle of incidence $i' = 30^{\circ}$.
$5$. Using Snell's law at face $AD$: $\sqrt{3} \cdot \sin(30^{\circ}) = 1 \cdot \sin(e)$.
$\sin(e) = \sqrt{3}/2$,so $e = 60^{\circ}$.
$6$. The total deviation $\delta = 90^{\circ}$. Thus,$(A)$ and $(B)$ are correct,and the angle between incident and emergent ray is $90^{\circ}$,so $(C)$ is correct.

(C) For total internal reflection $(TIR)$ to occur at the second surface for all $\theta \leq 60^{\circ}$,the condition must be satisfied at the minimum angle of incidence within the prism.
Let $r_1$ be the angle of refraction at the first surface and $r_2$ be the angle of incidence at the second surface.
From the geometry of the prism,$r_1 + r_2 = A = 75^{\circ}$.
For $TIR$ at the second surface,the angle of incidence $r_2$ must be greater than or equal to the critical angle $C$,where $\sin C = \frac{n}{n_0}$.
As $\theta$ decreases,$r_1$ decreases,and consequently $r_2 = 75^{\circ} - r_1$ increases.
The condition for $TIR$ is most easily satisfied at the maximum value of $r_2$,which occurs at the minimum value of $r_1$. However,the problem states $TIR$ occurs for $\theta \leq 60^{\circ}$. The boundary condition is at $\theta = 60^{\circ}$.
At $\theta = 60^{\circ}$,by Snell's law at the first surface: $1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r_1 \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \Rightarrow r_1 = 30^{\circ}$.
Then,$r_2 = 75^{\circ} - 30^{\circ} = 45^{\circ}$.
For $TIR$ at this surface,$r_2 \geq C$,so $45^{\circ} \geq C$,which means $\sin 45^{\circ} \geq \sin C = \frac{n}{\sqrt{3}}$.
$\frac{1}{\sqrt{2}} \geq \frac{n}{\sqrt{3}} \Rightarrow n \leq \sqrt{\frac{3}{2}}$.
The limiting value is $n = \sqrt{1.5}$,so $n^2 = 1.5$.

(C) For minimum deviation in a symmetric prism $(n_1=n_2=n)$,the angle of incidence $i$ is given by $\sin i = n \sin(\theta/2)$.
Given $\theta=60^{\circ}$,$\sin i = 1.5 \times \sin 30^{\circ} = 1.5 \times 0.5 = 0.75 = 3/4$.
At the second surface,the angle of incidence is $r_2 = 30^{\circ}$. The refraction at the second surface is $n_2 \sin r_2 = 1 \sin e$.
For $n_1=n$ and $n_2=n+\Delta n$,the light enters the first surface at angle $i$ and refracts at $r_1=30^{\circ}$. It then travels to the second surface. Since the prism is symmetric,the ray hits the second surface at $r_2=30^{\circ}$.
Thus,$(n+\Delta n) \sin 30^{\circ} = \sin e$.
Since $e = i + \Delta e$,we have $\sin(i + \Delta e) = (n+\Delta n) \sin 30^{\circ} = n \sin 30^{\circ} + \Delta n \sin 30^{\circ} = \sin i + \Delta n \sin 30^{\circ}$.
Using $\sin(i + \Delta e) \approx \sin i + \Delta e \cos i$,we get $\Delta e \cos i = \Delta n \sin 30^{\circ}$.
Since $\sin i = 3/4$,$\cos i = \sqrt{1 - (3/4)^2} = \sqrt{7}/4$.
$\Delta e = \Delta n \frac{\sin 30^{\circ}}{\cos i} = \Delta n \frac{0.5}{\sqrt{7}/4} = \Delta n \frac{2}{\sqrt{7}} \approx 0.756 \Delta n$.
Since $0.756 < 1$,$\Delta e < \Delta n$. Thus,$(A)$ is incorrect.
Since $\Delta e = (2/\sqrt{7}) \Delta n$,$\Delta e$ is proportional to $\Delta n$. Thus,$(B)$ is correct.
For $\Delta n = 2.8 \times 10^{-3}$,$\Delta e = (2/\sqrt{7}) \times 2.8 \times 10^{-3} \approx 0.756 \times 2.8 \times 10^{-3} \approx 2.11 \times 10^{-3} \text{ rad} = 2.11 \text{ mrad}$.
This value lies between $2.0$ and $3.0$ mrad. Thus,$(C)$ is correct.

(B) Applying Snell's Law on the $1^{\text{st}}$ surface: $\sin(60^{\circ}) = n \sin(r_1) \Rightarrow \sin(r_1) = \frac{\sqrt{3}}{2n}$.
For an equilateral prism,the prism angle $A = 60^{\circ}$,so $r_1 + r_2 = 60^{\circ}$,which implies $r_2 = 60^{\circ} - r_1$.
Applying Snell's Law on the $2^{\text{nd}}$ surface: $n \sin(r_2) = \sin(\theta)$.
Substituting $r_2$: $\sin(\theta) = n \sin(60^{\circ} - r_1) = n [\sin(60^{\circ}) \cos(r_1) - \cos(60^{\circ}) \sin(r_1)]$.
Since $\sin(r_1) = \frac{\sqrt{3}}{2n}$,then $\cos(r_1) = \sqrt{1 - \frac{3}{4n^2}} = \frac{\sqrt{4n^2 - 3}}{2n}$.
Substituting these into the equation for $\sin(\theta)$:
$\sin(\theta) = n [\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{4n^2 - 3}}{2n} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2n}] = \frac{\sqrt{3}}{4} (\sqrt{4n^2 - 3} - 1)$.
Differentiating with respect to $n$: $\cos(\theta) \frac{d\theta}{dn} = \frac{\sqrt{3}}{4} \cdot \frac{1}{2\sqrt{4n^2 - 3}} \cdot 8n = \frac{\sqrt{3} n}{\sqrt{4n^2 - 3}}$.
At $n = \sqrt{3}$,$\theta = 60^{\circ}$,so $\cos(60^{\circ}) \frac{d\theta}{dn} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{4(3) - 3}} = \frac{3}{\sqrt{9}} = 1$.
$\frac{1}{2} \cdot \frac{d\theta}{dn} = 1 \Rightarrow \frac{d\theta}{dn} = 2$. Thus,$m = 2$.

(A) For minimum deviation in prism $P_2$,the angle of incidence at the second surface of $P_2$ must be equal to the angle of emergence,and the ray inside $P_2$ must be parallel to the base. For an equilateral prism,this implies the angle of refraction $r_2$ at the first surface of $P_2$ is $30^{\circ}$.
Applying Snell's law at the interface between $P_1$ and $P_2$ (refractive index of $P_1$ is $\mu_1 = \sqrt{3/2}$ and $P_2$ is $\mu_2 = \sqrt{3}$):
$\mu_1 \sin r_2' = \mu_2 \sin r_2$
$\sqrt{\frac{3}{2}} \sin r_2' = \sqrt{3} \sin 30^{\circ}$
$\sqrt{\frac{3}{2}} \sin r_2' = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}$
$\sin r_2' = \frac{\sqrt{3}}{2} \times \sqrt{\frac{2}{3}} = \frac{1}{\sqrt{2}}$
$r_2' = 45^{\circ}$
Since the prism $P_1$ is equilateral,the sum of refraction angles $r_1 + r_2' = A = 60^{\circ}$.
$r_1 = 60^{\circ} - 45^{\circ} = 15^{\circ}$.
Applying Snell's law at the first surface of $P_1$ (refractive index of vacuum is $1$):
$1 \sin \theta = \sqrt{\frac{3}{2}} \sin 15^{\circ}$
$\theta = \sin^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{12}\right)\right]$
Comparing this with the given expression,we find $\beta = 12$.

(B) The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin \left(\frac{A + \delta_{min}}{2}\right)}{\sin \frac{A}{2}}$.
Given that the angle of minimum deviation $\delta_{min}$ is equal to the angle of the prism $A$,we substitute $\delta_{min} = A$ into the formula.
$\mu = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$.
Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get $\mu = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$.
Given $\mu = \sqrt{3}$,we have $\sqrt{3} = 2 \cos \frac{A}{2}$,which implies $\cos \frac{A}{2} = \frac{\sqrt{3}}{2}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $\frac{A}{2} = 30^{\circ}$,so $A = 60^{\circ}$.

(B) For dispersion without deviation,the net deviation produced by the combination of two thin prisms must be zero.
The net deviation $\delta_{\text{net}}$ is given by $\delta_{\text{net}} = \delta_1 - \delta_2 = 0$.
For a thin prism,the deviation is $\delta = (\mu - 1)A$.
Thus,$(\mu_1 - 1)A_1 - (\mu_2 - 1)A_2 = 0$.
Given $\mu_1 = 1.54$,$A_1 = 4^{\circ}$,and $\mu_2 = 1.72$.
Substituting the values: $(1.54 - 1) \times 4^{\circ} - (1.72 - 1) \times A_2 = 0$.
$0.54 \times 4 = 0.72 \times A_2$.
$2.16 = 0.72 \times A_2$.
$A_2 = \frac{2.16}{0.72} = 3^{\circ}$.

(C) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.
Given $A = 60^{\circ}$ and $\mu = \sqrt{2}$,we substitute these values: $\sqrt{2} = \frac{\sin((60^{\circ} + \delta_m)/2)}{\sin(30^{\circ})}$.
Since $\sin(30^{\circ}) = 0.5$,we have $\sqrt{2} \times 0.5 = \sin((60^{\circ} + \delta_m)/2)$,which simplifies to $\sin(45^{\circ}) = \sin((60^{\circ} + \delta_m)/2)$.
Thus,$45^{\circ} = (60^{\circ} + \delta_m)/2$,which gives $90^{\circ} = 60^{\circ} + \delta_m$,so $\delta_m = 30^{\circ}$.
At the condition of minimum deviation,the angle of incidence $i$ is related to the angle of the prism $A$ and the angle of minimum deviation $\delta_m$ by the formula: $i = (A + \delta_m)/2$.
Substituting the values: $i = (60^{\circ} + 30^{\circ})/2 = 90^{\circ}/2 = 45^{\circ}$.

(A) The deviation produced by a prism is given by $\delta = i + e - A$.
At the condition of minimum deviation $(\delta_{\text{min}})$,the angle of incidence equals the angle of emergence $(i = e)$.
Under this condition,the refracted ray inside the prism is parallel to the base of the prism.
Statement $(A)$ is correct because the ray inside is parallel to the base.
Statement $(C)$ is correct because $i = e$ at minimum deviation.
Statement $(D)$ is correct because for any deviation $\delta > \delta_{\text{min}}$,there exist two angles of incidence $i_1$ and $i_2$ such that $i_1 = e$ and $i_2 = i$,resulting in the same deviation.
Statement $(B)$ is incorrect because a larger prism angle $A$ generally leads to a larger angle of minimum deviation.
Statement $(E)$ is incorrect because at minimum deviation,$r_1 = r_2 = A/2$,so the angle of refraction is half the prism angle,not double.
Therefore,the correct statements are $(A), (C),$ and $(D)$.

(C) The formula for the refractive index of a prism is $\mu = \frac{\sin \left(\frac{\delta_{\min} + A}{2}\right)}{\sin(A/2)}$.
Given that the angle of minimum deviation $\delta_{\min} = A$,we substitute this into the formula:
$\mu = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.
Using the trigonometric identity $\sin A = 2 \sin(A/2) \cos(A/2)$,we get:
$\mu = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.
Given $\mu = 1.5 = \frac{3}{2}$,we have:
$\frac{3}{2} = 2 \cos(A/2) \Rightarrow \cos(A/2) = \frac{3}{4} = 0.75$.
Since $\cos 41^{\circ} = 0.75$,it follows that $A/2 = 41^{\circ}$.
Therefore,$A = 82^{\circ}$.

(D) From the graph,at minimum deviation,$\delta_m = 60^{\circ}$ and the corresponding angle of incidence $i = 60^{\circ}$.
For minimum deviation,$i = e$,so $\delta_m = 2i - A$.
$60^{\circ} = 2(60^{\circ}) - A \Rightarrow A = 60^{\circ}$. Thus,option $(A)$ is correct.
The refractive index $\mu$ is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)} = \frac{\sin((60^{\circ} + 60^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin(60^{\circ})}{\sin(30^{\circ})} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$. Thus,option $(B)$ is correct.
For a deviation $\delta = 65^{\circ}$,the graph shows two possible angles of incidence,$i_1$ and $70^{\circ}$. Since $\delta = i + e - A$,for $i = 70^{\circ}$,$65^{\circ} = 70^{\circ} + e - 60^{\circ} \Rightarrow e = 55^{\circ}$. By the principle of reversibility,if $i = 70^{\circ}$ gives $e = 55^{\circ}$,then $i = 55^{\circ}$ gives $e = 70^{\circ}$. Thus,$i_1 = 55^{\circ}$. Option $(C)$ is correct.
Therefore,all the statements are correct.

(C) Given: Angle of incidence $i = 60^{\circ}$,Prism angle $A = 30^{\circ}$,and Angle of deviation $\delta = 30^{\circ}$.
Using the prism formula for deviation: $\delta = i + e - A$.
Substituting the values: $30^{\circ} = 60^{\circ} + e - 30^{\circ}$.
Solving for the angle of emergence: $e = 30^{\circ} - 60^{\circ} + 30^{\circ} = 0^{\circ}$.
Since the angle of emergence $e = 0^{\circ}$,the ray emerges normally from the second surface,which implies $r_2 = 0^{\circ}$.
Using the relation $r_1 + r_2 = A$,we get $r_1 + 0^{\circ} = 30^{\circ}$,so $r_1 = 30^{\circ}$.
Applying Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r_1}$.
$\mu = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(D) For a thin prism,the angle of deviation is given by $\delta = (\mu - 1)A$,where $\mu$ is the relative refractive index of the prism material with respect to the surrounding medium.
$1$. When the prism is in air:
$\mu_1 = \frac{\mu_{\text{glass}}}{\mu_{\text{air}}} = \frac{3/2}{1} = \frac{3}{2}$
$\delta_1 = (\frac{3}{2} - 1)A = \frac{1}{2}A$
$2$. When the prism is in water:
$\mu_2 = \frac{\mu_{\text{glass}}}{\mu_{\text{water}}} = \frac{3/2}{4/3} = \frac{9}{8}$
$\delta_2 = (\frac{9}{8} - 1)A = \frac{1}{8}A$
$3$. Calculating the ratio:
$\frac{\delta_1}{\delta_2} = \frac{\frac{1}{2}A}{\frac{1}{8}A} = \frac{1}{2} \times 8 = 4$
Therefore,the ratio $\delta_1 : \delta_2$ is $4 : 1$.

(B) The angle of deviation $\delta$ for a thin prism is given by the formula $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.
When the prism is in air,the refractive index is ${ }_{a} \mu_{g} = \frac{3}{2}$. Thus,$\delta_1 = (\frac{3}{2} - 1)A = \frac{1}{2}A$.
When the prism is immersed in water,the refractive index of the prism relative to water is ${ }_{w} \mu_{g} = \frac{{ }_{a} \mu_{g}}{{ }_{a} \mu_{w}} = \frac{3/2}{4/3} = \frac{9}{8}$.
Thus,$\delta_2 = (\frac{9}{8} - 1)A = \frac{1}{8}A$.
The ratio $\delta_2 : \delta_1 = \frac{\frac{1}{8}A}{\frac{1}{2}A} = \frac{1}{8} \times 2 = \frac{1}{4}$.
Therefore,the ratio is $1: 4$.

(A) Given: Prism angle $A = 30^{\circ}$,angle of incidence $i_1 = 60^{\circ}$,and angle of deviation $\delta = 30^{\circ}$.
For a prism,the deviation formula is $\delta = (i_1 + i_2) - A$.
Substituting the values: $30^{\circ} = (60^{\circ} + i_2) - 30^{\circ}$.
$30^{\circ} = 30^{\circ} + i_2$,which gives $i_2 = 0^{\circ}$.
Since the emergent angle $i_2 = 0^{\circ}$,the emergent ray is normal to the second face,implying the angle of refraction at the second face $r_2 = 0^{\circ}$.
Using the relation $r_1 + r_2 = A$,we get $r_1 + 0^{\circ} = 30^{\circ}$,so $r_1 = 30^{\circ}$.
Applying Snell's law at the first face: $\mu_1 \sin i_1 = \mu_2 \sin r_1$.
Assuming the surrounding medium is air $(\mu_1 = 1)$: $1 \cdot \sin 60^{\circ} = \mu_2 \sin 30^{\circ}$.
$\frac{\sqrt{3}}{2} = \mu_2 \cdot \frac{1}{2}$.
Therefore,$\mu_2 = \sqrt{3} \approx 1.732$.

(B) For a prism,the condition for minimum deviation is given by the formula $i = \frac{A + \delta_m}{2}$.
In an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of incidence $i = 50^{\circ}$.
Substituting these values into the formula:
$50^{\circ} = \frac{60^{\circ} + \delta_m}{2}$
$100^{\circ} = 60^{\circ} + \delta_m$
$\delta_m = 100^{\circ} - 60^{\circ} = 40^{\circ}$.
Thus,the angle of minimum deviation is $40^{\circ}$.

(A) The general prism formula is given by $\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$.
For a symmetric (equilateral) prism,the prism angle $A = 60^{\circ}$.
Substituting $A = 60^{\circ}$ into the formula:
$\mu = \frac{\sin \left(\frac{60^{\circ}+\delta_m}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}$
$\mu = \frac{\sin \left(30^{\circ}+\frac{\delta_m}{2}\right)}{\sin 30^{\circ}}$
Since $\sin 30^{\circ} = \frac{1}{2}$,we have:
$\mu = \frac{\sin \left(30^{\circ}+\frac{\delta_m}{2}\right)}{1/2}$
$\mu = 2 \sin \left(30^{\circ}+\frac{\delta_m}{2}\right)$.

(D) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.
In air,the refractive index of glass relative to air is $_{a}\mu_{g} = \frac{3}{2}$. Thus,$\delta_1 = (_{a}\mu_{g} - 1)A = (\frac{3}{2} - 1)A = \frac{1}{2}A$.
When immersed in water,the refractive index of glass relative to water is $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}} = \frac{3/2}{4/3} = \frac{9}{8}$.
The new angle of minimum deviation is $\delta_2 = (_{w}\mu_{g} - 1)A = (\frac{9}{8} - 1)A = \frac{1}{8}A$.
Comparing $\delta_2$ with $\delta_1$: $\frac{\delta_2}{\delta_1} = \frac{\frac{1}{8}A}{\frac{1}{2}A} = \frac{2}{8} = \frac{1}{4}$.
Therefore,$\delta_2 = \frac{\delta_1}{4}$.

(C) The dispersive power $\omega$ is given by the formula $\omega = \frac{\delta_v - \delta_R}{\delta_y}$,where $\delta_y = \frac{\delta_v + \delta_R}{2}$ is the mean deviation.
For prism $A$:
$\delta_y = \frac{12^{\circ} + 10^{\circ}}{2} = 11^{\circ}$.
$\omega_A = \frac{12^{\circ} - 10^{\circ}}{11^{\circ}} = \frac{2}{11}$.
For prism $B$:
$\delta_y = \frac{10^{\circ} + 8^{\circ}}{2} = 9^{\circ}$.
$\omega_B = \frac{10^{\circ} - 8^{\circ}}{9^{\circ}} = \frac{2}{9}$.
The ratio of their dispersive powers is $\frac{\omega_A}{\omega_B} = \frac{2/11}{2/9} = \frac{9}{11}$.

(D) For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally.
In the prism,the angle of refraction at the first surface is '$r_1$'. From the geometry of the prism,the angle of incidence at the second surface is '$r_2$'. Since the ray strikes normally,'$r_2 = 0$'.
Using the relation '$A = r_1 + r_2$',we get '$A = r_1 + 0$',so '$r_1 = A$'.
Applying Snell's law at the first surface: '$\mu = \frac{\sin i}{\sin r_1}$'.
Given the angle of incidence '$i = 2A$',we have '$\mu = \frac{\sin 2A}{\sin A}$'.
Using the trigonometric identity '$\sin 2A = 2 \sin A \cos A$',we get '$\mu = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A$'.
Thus,the refractive index of the material of the prism is '$2 \cos A$'.

(A) The dispersive power $\omega$ of a prism is given by the formula $\omega = \frac{\delta_v - \delta_r}{\delta_y}$,where $\delta_v$ is the deviation for violet light,$\delta_r$ is the deviation for red light,and $\delta_y$ is the mean deviation,calculated as $\delta_y = \frac{\delta_v + \delta_r}{2}$.
For the first prism:
$\delta_v = 11^{\circ}$,$\delta_r = 9^{\circ}$.
$\delta_{y1} = \frac{11 + 9}{2} = 10^{\circ}$.
$\omega_1 = \frac{11 - 9}{10} = \frac{2}{10} = \frac{1}{5}$.
For the second prism:
$\delta_v = 13^{\circ}$,$\delta_r = 11^{\circ}$.
$\delta_{y2} = \frac{13 + 11}{2} = 12^{\circ}$.
$\omega_2 = \frac{13 - 11}{12} = \frac{2}{12} = \frac{1}{6}$.
The ratio of the dispersive power of the second prism to the first prism is $\frac{\omega_2}{\omega_1} = \frac{1/6}{1/5} = \frac{5}{6}$.
Thus,the ratio is $5: 6$.

(C) Given: The ray emerges normally from the second surface,so the angle of emergence $e = 0$.
Since $e = 0$,the angle of refraction at the second surface $r_2 = 0$.
For a prism,the angle of the prism $A = r_1 + r_2$. Substituting $r_2 = 0$,we get $A = r_1$.
According to Snell's law at the first surface,$\mu = \frac{\sin i}{\sin r_1}$.
Since the prism angle $A$ is small,$i$ and $r_1$ are also small,so $\sin i \approx i$ and $\sin r_1 \approx r_1$.
Thus,$\mu = \frac{i}{r_1}$.
Substituting $r_1 = A$,we get $\mu = \frac{i}{A}$,which implies $i = \mu A$.

(B) For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.
Given that the angle of incidence $(i)$ is equal to the angle of emergence $(e)$,and $e = \frac{3}{4} A$.
Substituting the value of $A$:
$e = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
Since $i = e$,we have $i = 45^{\circ}$.
The formula for the angle of deviation $(\delta)$ is $\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ} = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Therefore,the angle of deviation is $30^{\circ}$.

(B) Given,refractive index $\mu = \sqrt{3}$.
The condition for minimum deviation is $\delta = A$,where $A$ is the angle of the prism.
The formula for the refractive index of a prism is $\mu = \frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}$.
Substituting $\delta = A$ into the formula:
$\mu = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)} = \frac{\sin A}{\sin (A/2)}$.
Using the trigonometric identity $\sin A = 2 \sin (A/2) \cos (A/2)$:
$\mu = \frac{2 \sin (A/2) \cos (A/2)}{\sin (A/2)} = 2 \cos (A/2)$.
Given $\mu = \sqrt{3}$,we have $\sqrt{3} = 2 \cos (A/2) \Rightarrow \cos (A/2) = \frac{\sqrt{3}}{2}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we get $A/2 = 30^{\circ}$.
Therefore,$A = 60^{\circ}$.

(D) Given: Refracting angle $A = 30^{\circ}$,Refractive index $\mu = 1.5$,and angle of incidence $i_1 = 0^{\circ}$ (since the ray is perpendicular to the face).
Since $i_1 = 0^{\circ}$,the angle of refraction $r_1$ is also $0^{\circ}$.
Using the relation $r_1 + r_2 = A$,we get $0^{\circ} + r_2 = 30^{\circ}$,so $r_2 = 30^{\circ}$.
Applying Snell's Law at the second face: $\mu \sin(r_2) = 1 \cdot \sin(i_2)$.
$1.5 \cdot \sin(30^{\circ}) = \sin(i_2)$.
$1.5 \cdot 0.5 = \sin(i_2) \Rightarrow \sin(i_2) = 0.75$.
Given $\sin(48.6^{\circ}) = 0.75$,therefore $i_2 = 48.6^{\circ}$.
The angle of deviation $\delta$ is given by $\delta = (i_1 + i_2) - A$.
$\delta = (0^{\circ} + 48.6^{\circ}) - 30^{\circ} = 18.6^{\circ}$.

(A) For a prism,the angle of the prism is given by $A = r_1 + r_2$.
Since the ray emerges normally from the second face,the angle of emergence $e = 0$,which implies $r_2 = 0$.
Therefore,$A = r_1$.
Using Snell's law at the first surface: $n = \frac{\sin i}{\sin r_1}$.
Substituting $r_1 = A$,we get $n = \frac{\sin i}{\sin A}$.
For a thin prism,the angles $i$ and $A$ are very small,so $\sin i \approx i$ and $\sin A \approx A$.
Thus,$n = \frac{i}{A}$,which gives $i = An$.

(D) The emergent ray just grazes the second face,so the angle of emergence $e = 90^{\circ}$.
Using Snell's law at the second face: $\mu = \frac{\sin e}{\sin r_2} = \frac{\sin 90^{\circ}}{\sin r_2} = \frac{1}{\sin r_2}$.
Given $\mu = \sqrt{2}$,we have $\sin r_2 = \frac{1}{\sqrt{2}}$,which gives $r_2 = 45^{\circ}$.
For an equilateral prism,the prism angle $A = 60^{\circ}$.
Using the relation $A = r_1 + r_2$,we find $r_1 = A - r_2 = 60^{\circ} - 45^{\circ} = 15^{\circ}$.
Applying Snell's law at the first face: $\frac{\sin i}{\sin r_1} = \mu$.
Therefore,$\sin i = \mu \sin r_1 = \sqrt{2} \sin 15^{\circ}$.
Thus,$i = \sin ^{-1}(\sqrt{2} \sin 15^{\circ})$.

(D) For a thin prism,the angle of deviation is given by $\delta = A(n - 1)$.
For dispersion without deviation,the net deviation must be zero,which means the deviation produced by the first prism must be equal to the deviation produced by the second prism in the opposite direction.
Therefore,$\delta_{1} = \delta_{2}$.
Substituting the formula,we get $A_{1}(n_{1} - 1) = A_{2}(n_{2} - 1)$.
Given $A_{1} = 4^{\circ}$,$n_{1} = 1.54$,and $n_{2} = 1.72$.
$4(1.54 - 1) = A_{2}(1.72 - 1)$.
$4 \times 0.54 = A_{2} \times 0.72$.
$A_{2} = \frac{4 \times 0.54}{0.72} = \frac{2.16}{0.72} = 3^{\circ}$.

(C) For a thin prism with a small angle '$A$',the relationship between the angle of incidence '$i$',the angle of refraction '$r_1$',and the angle of emergence '$r_2$' is given by $A = r_1 + r_2$.
Since the ray emerges normally from the opposite surface,the angle of emergence is $0$,which implies $r_2 = 0$.
Therefore,$A = r_1 + 0$,so $r_1 = A$.
According to Snell's Law,$\mu = \frac{\sin i}{\sin r_1}$. For small angles,$\sin i \approx i$ and $\sin r_1 \approx r_1$.
Thus,$\mu = \frac{i}{r_1}$.
Substituting $r_1 = A$,we get $\mu = \frac{i}{A}$,which simplifies to $i = \mu A$.

(D) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and $i = e = \frac{3}{4}A$.
Substituting the value of $A$:
$i = e = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The formula for the angle of deviation $\delta$ is given by $\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ}$.
$\delta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Therefore,the angle of deviation is $30^{\circ}$.

(A) For a prism with a small angle '$A$',the angle of deviation '$\delta$' is given by $\delta = (\mu - 1)A$.
We know the relation for a prism: $i + e = A + \delta$.
Given that the ray emerges normally from the other surface,the angle of emergence '$e$' is $0$.
Substituting $e = 0$ and $\delta = (\mu - 1)A$ into the relation:
$i + 0 = A + (\mu - 1)A$
$i = A + \mu A - A$
$i = \mu A$.
Therefore,the angle of incidence is $\mu A$.

(D) For the light to retrace its path,it must strike the silvered surface normally.
Since the second surface is silvered,the angle of refraction at the second surface $r_2$ must be $0^{\circ}$.
Using the prism formula $A = r_1 + r_2$,where $A = 30^{\circ}$ is the refracting angle and $r_2 = 0^{\circ}$,we get $r_1 = 30^{\circ}$.
Applying Snell's Law at the first surface: $n = \frac{\sin i}{\sin r_1}$.
Given $n = \sqrt{2}$ and $r_1 = 30^{\circ}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.
$\sin i = \sqrt{2} \times \sin 30^{\circ} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

(A) For a thin prism,the relationship between the prism angle '$A$' and the angles of refraction '$r_1$' and '$r_2$' is given by: $A = r_1 + r_2$.
Since the ray emerges normally from the other face,the angle of emergence '$e = 0$',which implies that the second angle of refraction '$r_2 = 0$'.
Therefore,$A = r_1$.
According to Snell's Law at the first surface,$n = \frac{\sin i}{\sin r_1}$.
For a thin prism,the angles '$i$' and '$r_1$' are very small,so we can approximate $\sin i \approx i$ and $\sin r_1 \approx r_1$.
Thus,$n = \frac{i}{r_1}$.
Substituting $r_1 = A$,we get $n = \frac{i}{A}$,which implies $i = An$.

(A) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the emergent ray grazes the adjacent face,the angle of emergence $e = 90^{\circ}$.
From the condition of refraction at the second surface,$r_2 = C$,where $C$ is the critical angle.
Using Snell's law at the second surface: $\mu \sin r_2 = 1 \sin e \implies \sqrt{2} \sin r_2 = \sin 90^{\circ} = 1$.
Thus,$\sin r_2 = 1/\sqrt{2}$,which means $r_2 = 45^{\circ}$.
Since $A = r_1 + r_2$,we have $60^{\circ} = r_1 + 45^{\circ}$,so $r_1 = 15^{\circ}$.
Applying Snell's law at the first surface: $1 \sin i = \mu \sin r_1$.
$\sin i = \sqrt{2} \sin 15^{\circ}$.
Therefore,$i = \sin^{-1}(\sqrt{2} \sin 15^{\circ})$.

(B) For a thin prism,the angle of deviation $\delta$ is given by $\delta = (\mu - 1)A$. Given $\delta = 1.15^{\circ}$,we have $(\mu - 1)A = 1.15^{\circ}$ (Equation $1$).
When the ray is incident normally on the first face,it strikes the second face at an angle of incidence equal to the prism angle $A$. The ray is reflected internally at the second face,so the angle of reflection is $A$. This reflected ray then strikes the first face at an angle $2A$ with the normal. Upon emerging from the first face,the angle of refraction $r'$ is given by $\sin(r') = \mu \sin(2A)$. For small angles,$r' \approx \mu(2A)$.
The angle between the emerging ray and the incident ray is given as $6.3^{\circ}$. The emerging ray makes an angle $r'$ with the normal,and the incident ray is normal to the surface,so the angle between them is $r' = 6.3^{\circ}$.
Thus,$2\mu A = 6.3^{\circ}$ (Equation $2$).
Dividing Equation $2$ by Equation $1$: $\frac{2\mu A}{(\mu - 1)A} = \frac{6.3}{1.15} \implies \frac{2\mu}{\mu - 1} = 5.478$.
$2\mu = 5.478\mu - 5.478 \implies 3.478\mu = 5.478 \implies \mu \approx 1.575$.

(C) For a prism,the angle of deviation is given by $\delta = i + e - A$. When the refracted ray inside the prism is parallel to the base,the prism is in the position of minimum deviation. In the position of minimum deviation,the angle of incidence is equal to the angle of emergence,i.e.,$i = e$. Therefore,the correct option is $C$.

(B) For a thin prism,the angle of deviation $\delta$ is given by the formula: $\delta = (n - 1)A$,where $n$ is the refractive index of the material of the prism and $A$ is the refracting angle of the prism.
Given:
Refracting angle $A = 3^{\circ}$
Angle of deviation $\delta = 1^{\circ}$
Substituting these values into the formula:
$1^{\circ} = (n - 1) \cdot 3^{\circ}$
$n - 1 = \frac{1}{3}$
$n = 1 + \frac{1}{3} = \frac{4}{3}$
$n = 1.33$
Therefore,the refractive index of water is $1.33$.

(C) Given: Angle of prism $= A$,Angle of incidence $i = 2A$.
For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally.
Let $r$ be the angle of refraction at the first surface. In the triangle formed by the ray inside the prism,the angle at the silvered surface is $90^{\circ}$. Thus,the angle of refraction $r$ is related to the prism angle $A$ by $r + A = 90^{\circ}$ is incorrect here; rather,from the geometry,$r = A$.
Using Snell's law at the first surface:
$\mu = \frac{\sin i}{\sin r} = \frac{\sin 2A}{\sin A}$
Using the trigonometric identity $\sin 2A = 2 \sin A \cos A$:
$\mu = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A$

(C) For a thin prism,the angle of minimum deviation $(D_{m})$ is given by the formula:
$D_{m} = A(n - 1)$
where $A$ is the angle of the prism and $n$ is the refractive index.
Given:
$A = 4^{\circ}$
$n = 1.6$
Substituting the values into the formula:
$D_{m} = 4^{\circ}(1.6 - 1)$
$D_{m} = 4^{\circ}(0.6)$
$D_{m} = 2.4^{\circ}$
Therefore,the angle of minimum deviation is $2.4^{\circ}$.

(D) The formula for the refractive index of a prism is given by $\mu = \frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \frac{A}{2}}$,where $A$ is the angle of the prism and $D_m$ is the angle of minimum deviation.
Given that $D_m = A$,we substitute this into the formula:
$\mu = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}$
$\mu = \frac{\sin A}{\sin \frac{A}{2}}$
Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get:
$\mu = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$\mu = 2 \cos \frac{A}{2}$
$\frac{\mu}{2} = \cos \frac{A}{2}$
$\frac{A}{2} = \cos ^{-1}\left(\frac{\mu}{2}\right)$
$A = 2 \cos ^{-1}\left(\frac{\mu}{2}\right)$

(B) For a small angled prism,the angle of deviation $\delta$ is given by the formula: $\delta = A(n - 1)$,where $A$ is the angle of the prism and $n$ is the refractive index of the material of the prism.
Given:
Refractive index $n = 1.6$
Angle of deviation $\delta = 3.6^{\circ}$
Substituting the values in the formula:
$3.6^{\circ} = A(1.6 - 1)$
$3.6^{\circ} = A(0.6)$
$A = \frac{3.6^{\circ}}{0.6}$
$A = 6^{\circ}$
Therefore,the angle of the prism is $6^{\circ}$.

(D) The refractive index $n$ of a prism is given by the formula: $n = \frac{\sin((A + D_m)/2)}{\sin(A/2)}$.
Given that the angle of minimum deviation $D_m$ is equal to the angle of the prism $A$,we substitute $D_m = A$ into the formula:
$1.5 = \frac{\sin((A + A)/2)}{\sin(A/2)}$
$1.5 = \frac{\sin(A)}{\sin(A/2)}$
Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$,we get:
$1.5 = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)}$
$1.5 = 2 \cos(A/2)$
$\cos(A/2) = 1.5 / 2 = 0.75$
Since $\sin(48^{\circ} 36^{\prime}) = 0.75$,we know that $\cos(90^{\circ} - 48^{\circ} 36^{\prime}) = 0.75$,which is $\cos(41^{\circ} 24^{\prime}) = 0.75$.
Therefore,$A/2 = 41^{\circ} 24^{\prime} = 41.4^{\circ}$.
Thus,$A = 2 \times 41.4^{\circ} = 82.8^{\circ}$.

(D) For a prism,the sum of the angles of refraction at the two faces is equal to the angle of the prism $(A)$:
$r + r^1 = A$
Given that the angle of the prism $A = 50^{\circ}$ and $r = 10^{\circ} + t^2$,we can substitute these values into the equation:
$10^{\circ} + t^2 + r^1 = 50^{\circ}$
Solving for $r^1$:
$r^1 = 50^{\circ} - 10^{\circ} - t^2$
$r^1 = 40^{\circ} - t^2$

(C) Given,the prism is equilateral,so the angle of the prism $A = 60^{\circ}$.
Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.
The refractive index $\mu$ of the prism is given by the formula:
$\mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Substituting the values:
$\mu = \frac{\sin \left(\frac{60^{\circ} + 60^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
We know that the refractive index $\mu$ is related to the speed of light in vacuum $c$ and the speed of light in the medium $v$ as $\mu = \frac{c}{v}$.
Therefore,$v = \frac{c}{\mu} = \frac{3 \times 10^8 \ m/s}{\sqrt{3}} = \sqrt{3} \times 10^8 \ m/s$.

(C) For an equilateral glass prism,the angle of the prism is $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and each is $\frac{3}{4}$ of the angle of the prism:
$i = e = \frac{3}{4} A = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The formula for the angle of deviation $\delta$ is given by:
$\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ} = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Thus,the angle of deviation is $30^{\circ}$.

(D) Given,angle of prism $= A$.
Refractive index of prism,$\mu = \cot \frac{A}{2}$.
We know the formula for refractive index in terms of minimum deviation $\delta_m$ is:
$\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Substituting the given value of $\mu$:
$\cot \left(\frac{A}{2}\right) = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Using $\cot \theta = \frac{\cos \theta}{\sin \theta}$:
$\frac{\cos (A/2)}{\sin (A/2)} = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A/2)}$
$\cos (A/2) = \sin \left(\frac{A+\delta_m}{2}\right)$
Since $\cos \theta = \sin (90^{\circ} - \theta)$,we have:
$\sin (90^{\circ} - A/2) = \sin \left(\frac{A+\delta_m}{2}\right)$
Equating the angles:
$90^{\circ} - \frac{A}{2} = \frac{A+\delta_m}{2}$
$180^{\circ} - A = A + \delta_m$
$\delta_m = 180^{\circ} - 2A$.

(A) For a prism,the deviation $D$ is given by $D = (i_1 + i_2) - A$,where $i_1$ and $i_2$ are the angles of incidence and emergence,and $A$ is the prism angle.
Given $D = 44^{\circ}$ for $i_1 = 42^{\circ}$ and $i_2 = 62^{\circ}$ (or vice versa),we have:
$44^{\circ} = (42^{\circ} + 62^{\circ}) - A$
$44^{\circ} = 104^{\circ} - A$
$A = 104^{\circ} - 44^{\circ} = 60^{\circ}$.
At minimum deviation $(D_m = 38^{\circ})$,the angle of incidence $i$ is given by $i = \frac{A + D_m}{2}$.
$i = \frac{60^{\circ} + 38^{\circ}}{2} = \frac{98^{\circ}}{2} = 49^{\circ}$.

(B) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
At the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1$ is equal to $r_2 = r$.
From the relation $A = r_1 + r_2$,we have $A = 2r$,so $r = A/2 = 60^{\circ}/2 = 30^{\circ}$.
According to Snell's law,$\mu = \frac{\sin i}{\sin r}$.
Given $\mu = \sqrt{2}$ and $r = 30^{\circ}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.
$\sin i = \sqrt{2} \times \sin 30^{\circ} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$.