Question

For the prism shown in the figure, the angle of incidence is adjusted such that the emergent ray suffers minimum deviation, $\delta_m=60^\circ$. The relative refractive index with respect to the medium in which it is kept is $\mu$. Choose the correct option(s).

• A. $i_m=60^\circ$
• B. $r_m=45^\circ$
• C. $\mu=\frac{\sqrt{3}+1}{2}$
• D. $\mu=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Answer 1

Answer: (B), (C)

$r_m=45^\circ$, $\mu=\frac{\sqrt{3}+1}{2}$

Answer 2

The prism angle is $A = 90^\circ$. For minimum deviation, $A = 2r_m$, so $90^\circ = 2r_m$, which gives $r_m = 45^\circ$. The minimum deviation formula is $\delta_m = 2i_m - A$. Given $\delta_m = 60^\circ$ and $A = 90^\circ$, we have $60^\circ = 2i_m - 90^\circ$, so $2i_m = 150^\circ$, and $i_m = 75^\circ$. Using Snell's Law, $\mu = \frac{\sin i_m}{\sin r_m} = \frac{\sin 75^\circ}{\sin 45^\circ}$. We know that $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$. And $\sin 45^\circ = \frac{1}{\sqrt{2}}$. Therefore, $\mu = \frac{\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{3}+1}{2}$.