Refraction Through Prism Questions in English

(B) For an equilateral prism, the angle of the prism is $A = 60^\circ$.
Let the light ray be incident normally on the first face, so the angle of incidence $i_1 = 0^\circ$ and the angle of refraction $r_1 = 0^\circ$.
Inside the prism, the angle of incidence at the second face (the painted face) is $r_2 = A - r_1 = 60^\circ - 0^\circ = 60^\circ$.
For total internal reflection $(TIR)$ to occur at the painted face, the angle of incidence $r_2$ must be greater than or equal to the critical angle $C$ for the interface between the prism and the painted material.
Thus, $r_2 \geq C$, which implies $\sin(r_2) \geq \sin(C)$.
Given $\sin(C) = \frac{n_2}{\mu_{\text{prism}}}$, we have $\sin(60^\circ) \geq \frac{n_2}{1.6}$.
Substituting $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, we get $\frac{\sqrt{3}}{2} \geq \frac{n_2}{1.6}$.
Solving for $n_2$, we get $n_2 \leq 1.6 \times \frac{\sqrt{3}}{2} = 0.8\sqrt{3}$.
Since we need the minimum value of $n_2$ for $TIR$, and the condition is $n_2 \leq 0.8\sqrt{3}$, the maximum possible value for $n_2$ is $0.8\sqrt{3}$. However, the question asks for the minimum value of $n_2$ required for $TIR$. In this specific configuration, the condition for $TIR$ is satisfied if $n_2$ is less than or equal to $0.8\sqrt{3}$. The value $0.8\sqrt{3}$ is the threshold.

(B) First, calculate the refractive index $\mu$ of the prism material using the formula $\mu = \frac{c}{v}$, where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light in a vacuum and $v = 2.12 \times 10^8 \text{ m/s}$ is the speed of light in the prism.
$\mu = \frac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 = \sqrt{2}$.
For an equilateral prism, the angle of the prism $A = 60^\circ$. The formula for the refractive index in terms of the minimum angle of deviation $\delta_m$ is $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.
Substituting the values: $\sqrt{2} = \frac{\sin((60^\circ + \delta_m)/2)}{\sin(30^\circ)}$.
Since $\sin(30^\circ) = 0.5$, we have $\sqrt{2} = \frac{\sin(30^\circ + \delta_m/2)}{0.5}$.
$\sin(30^\circ + \delta_m/2) = 0.5 \times \sqrt{2} = \frac{1}{\sqrt{2}}$.
This implies $30^\circ + \delta_m/2 = 45^\circ$.
$\delta_m/2 = 15^\circ$, therefore $\delta_m = 30^\circ$.

(C) For an equilateral prism, the prism angle $A = 60^\circ$.
Given that the angle of minimum deviation $\delta_m = A/2 = 60^\circ / 2 = 30^\circ$.
The formula for the refractive index $\mu$ is given by $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.
Substituting the values, we get $\mu = \frac{\sin((60^\circ + 30^\circ)/2)}{\sin(60^\circ/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$.
Since $\sin(45^\circ) = 1/\sqrt{2}$ and $\sin(30^\circ) = 1/2$, we have $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(D) In an equilateral prism, the angle of the prism $A = 60^\circ$.
When the refracted ray is parallel to the base, the prism is in the state of minimum deviation.
In this condition, the angle of incidence $i$ is equal to the angle of emergence $e$.
Given, $i = 50^\circ$, therefore $e = 50^\circ$.
The angle of deviation $\delta$ is given by the formula: $\delta = i + e - A$.
Substituting the values: $\delta = 50^\circ + 50^\circ - 60^\circ$.
$\delta = 100^\circ - 60^\circ = 40^\circ$.

(C) For a thin prism, the angle of minimum deviation $\delta$ is given by the formula: $\delta = (n - 1)A$, where $n$ is the refractive index of the prism material relative to the surrounding medium and $A$ is the prism angle.
In air, $\delta_1 = (^a n_g - 1)A = (\frac{3}{2} - 1)A = \frac{1}{2}A$.
When the prism is immersed in liquid, the refractive index of the prism relative to the liquid is $^l n_g = \frac{^a n_g}{^a n_l} = \frac{3/2}{4/3} = \frac{9}{8}$.
The new angle of minimum deviation is $\delta_2 = (^l n_g - 1)A = (\frac{9}{8} - 1)A = \frac{1}{8}A$.
Comparing $\delta_1$ and $\delta_2$: $\frac{\delta_2}{\delta_1} = \frac{1/8 A}{1/2 A} = \frac{2}{8} = \frac{1}{4}$.
Therefore, $\delta_2 = \frac{1}{4} \delta_1$.

(C) For a thin prism, the angle of minimum deviation $\delta$ 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 angle of the prism.
In air, the refractive index of the prism relative to air is $\mu_a = \frac{\mu_g}{\mu_{air}} = \frac{3/2}{1} = 3/2$.
Thus, $\delta = (3/2 - 1)A = \frac{1}{2}A$.
When the prism is immersed in water, the refractive index of the prism relative to water is $\mu_w = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
The new angle of minimum deviation $\delta'$ is given by: $\delta' = (\mu_w - 1)A = (9/8 - 1)A = \frac{1}{8}A$.
Comparing $\delta'$ with $\delta$: $\frac{\delta'}{\delta} = \frac{(1/8)A}{(1/2)A} = \frac{1}{4}$.
Therefore, $\delta' = \frac{\delta}{4}$.

(A) For an equilateral prism, the angle of the prism $A = 60^{\circ}$.
Given refractive index $\mu = \sqrt{2}$.
The emergent ray grazes the adjacent face, which means the angle of emergence $e = 90^{\circ}$.
At the second face, by Snell's law: $\mu \sin r_2 = 1 \cdot \sin e$.
$\sqrt{2} \sin r_2 = \sin 90^{\circ} = 1 \implies \sin r_2 = 1/\sqrt{2} \implies r_2 = 45^{\circ}$.
Since $r_1 + r_2 = A$, we have $r_1 + 45^{\circ} = 60^{\circ} \implies r_1 = 15^{\circ}$.
Applying Snell's law at the first face: $1 \cdot \sin i = \mu \sin r_1$.
$\sin i = \sqrt{2} \sin 15^{\circ} \implies i = \sin^{-1} (\sqrt{2} \sin 15^{\circ})$.

(B) Let the angle of incidence at the first surface be $i = 2A$ and the angle of refraction be $r_1$.
According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin 2A}{\sin r_1}$.
For a prism, the refracting angle $A = r_1 + r_2$.
Since the light ray returns back along the same path after reflection at the second surface, it must strike the second surface normally.
Therefore, the angle of incidence at the second surface is $0$, which implies the angle of refraction $r_2 = 0$.
Substituting $r_2 = 0$ into the prism equation: $A = r_1 + 0$, so $r_1 = A$.
Now, substitute $r_1 = A$ into the Snell's Law equation: $\mu = \frac{\sin 2A}{\sin A}$.
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 $2 \cos A$.

(C) For a thin prism, the angle of minimum deviation $(\delta_m)$ is given by the formula: $\delta_m = (\mu - 1)A$, where $\mu$ is the refractive index and $A$ is the angle of the prism.
Given $\mu = 1.5$, we have $\delta_m = (1.5 - 1)A = 0.5A = \frac{A}{2}$.
For a thin prism, the angle of refraction $r$ is related to the prism angle $A$ by the relation $A = 2r$ (since at minimum deviation, $r_1 = r_2 = r$, and $A = r_1 + r_2$).
Substituting $A = 2r$ into the equation for $\delta_m$, we get $\delta_m = \frac{2r}{2} = r$.
Therefore, the correct relation is $\delta_m = r$.

(D) For an equilateral prism, the angle of the prism is $A = 60^\circ$.
From the given graph, the minimum angle of deviation is $\delta_m = 30^\circ$ at an angle of incidence $i = 45^\circ$ (Note: The graph shows $\delta_m = 30^\circ$ at $i = 45^\circ$ based on standard prism geometry for this specific graph).
The refractive index $\mu$ is given by the formula:
$\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$
Substituting the values:
$\mu = \frac{\sin(\frac{60^\circ + 30^\circ}{2})}{\sin(\frac{60^\circ}{2})} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Thus, the correct option is $D$.