Question

The light ray is incident at an angle of ${60}^{\circ }$ on a prism of angle ${45}^{\circ }$. When the light ray 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. $\mu =\frac{\sqrt{2}}{2},\delta ={30}^{\circ }$
• D. $\mu =\sqrt{\frac{3}{2}},\delta ={15}^{\circ }$

Answer 1

The correct option is D $\mu =\sqrt{\frac{3}{2}},\delta ={15}^{\circ }$

Drawing the ray diagram for the light ray,



From the property of triangle,
Angle of refraction at first surface $={45}^{\circ }$
Using Snell’s law, $1\times \sin {60}^{\circ }=\mu \times \sin {45}^{\circ }$
$\Rightarrow \frac{\sqrt{3}}{2}=\mu \times \frac{1}{\sqrt{2}}$
$\Rightarrow \mu =\sqrt{\frac{3}{2}}$

Net deviation $\delta ={\delta }_1+{\delta }_2$
At the second surface, since angle of incidence is ${90}^{\circ }$, the ray will not suffer refraction. i.e ${\delta }_2=0$


Therefore $\delta ={\delta }_1={60}^{\circ }-{45}^{\circ }={15}^{\circ }$