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(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 inciden

Vedclass · Accepted Answer

$0.8\sqrt{3}$

Vedclass · Answer

(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 $\sin(C) = \frac{n_2}{\mu_{	ext{prism}}}$,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 $n_2 \leq 1.6 	imes \frac{\sqrt{3}}{2} = 0.8\sqrt{3}$.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.

One side of an equilateral prism is painted by a transparent material of refractive index $n_2$. The refractive index of the prism is $1.6$. The minimum value of $n_2$ required for total internal reflection from the painted face is . . . . . . .

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