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: $\delta = 50^\circ + 50^\circ - 60^\circ$.
$\delta = 100^\circ - 60^\circ = 40^\circ$.