Q149 / 170 · Manipal (MET) 2018 · Physics
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$\mu=\frac{\sqrt{3}}{2}, \delta=30^{\circ}$
- Correct answerD$\mu=\sqrt{\frac{3}{2}}, \delta=15^{\circ}$
Solution
From the problem it is clear that $\angle e=\angle r_2=0$
From $A=r_1+r_2$ $\begin{array}{lc} \Rightarrow & r_1=A=45^{\circ} \\ \therefore & \mu=\frac{\sin i}{\sin r_1}=\frac{\sin 60^{\circ}}{\sin 45^{\circ}}=\sqrt{\frac{3}{2}} \end{array}$
Also, from $i+e=A+\delta$ $\begin{array}{ccc} \Rightarrow & 60+0 & =45+\delta \\ \Rightarrow & \delta & =15^{\circ} \end{array}$
From $A=r_1+r_2$ $\begin{array}{lc} \Rightarrow & r_1=A=45^{\circ} \\ \therefore & \mu=\frac{\sin i}{\sin r_1}=\frac{\sin 60^{\circ}}{\sin 45^{\circ}}=\sqrt{\frac{3}{2}} \end{array}$
Also, from $i+e=A+\delta$ $\begin{array}{ccc} \Rightarrow & 60+0 & =45+\delta \\ \Rightarrow & \delta & =15^{\circ} \end{array}$
Hence, the correct answer is option: 'D'.