Question

The light ray is incident at angle of $60^{\circ }$ on a prism of angle $45^{\circ }$. When the light rays falls on the other surface at $90^{\circ }$, the refractive index of the material of prism mu and the angle of deviation $\delta$ are given by

• A. $\mu =\sqrt{2},\delta =30^{\circ }$
• B. $\mu =1.5,\delta =15^{\circ }$
• C. $\mu =\frac{\sqrt{3}}{2},\delta =30^{\circ }$
• D. $\mu =\sqrt{\frac{3}{2}},\delta =15^{\circ }$

Answer 1

Answer: (D)

From the problem it is clear that $\angle e=\angle r_2=0$
From $A=r_1+r_2$
$\Rightarrow r_1=A=45^{\circ }$
$\therefore \mu =\frac{\sin i}{\sin r_1}=\frac{\sin 60^{\circ }}{\sin 45^{\circ }}=\sqrt{\frac{3}{2}}$
Also, from $i+e=A+\delta$
$\Rightarrow 60+0=45+\delta$
$\Rightarrow \delta =15^{\circ }$