Question

The light ray is incidence at angle of ${60}^0$ on a prism of angle ${45}^0.$ When the light ray falls on the other surface at ${90}^0$, the refractive index of the material of prism $\mu$ and the angle of deviation $\delta$ are given by

• A. $\mu =\sqrt{2},\delta ={30}^0$
• B. $\mu =1.5,\delta ={15}^0$
• C. $\mu =\frac{\sqrt{3}}{2},\delta ={30}^0$
• D. $\mu =\sqrt{\frac{3}{2}},\delta ={15}^0$

Answer 1

The correct option is D $\mu =\sqrt{\frac{3}{2}},\delta ={15}^0$

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