Question

In minimum deviation position of a prism made of refractive index \(\sqrt3\), the angle of deviation of the light ray is \(\frac{\pi}{6}\). If a light ray incidents normally on the first face of the prism, then the angle of incidence of the light ray on the second face of the prism is:

• A. \(\frac{\pi}{3}\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi}{6}\)
• D. \(\frac{\pi}{4}\)

Hint:

When a ray enters normally into the first face of a prism, \(r_1=0\), hence the entire prism angle becomes the incidence angle at the second face.

Answer 1

The Correct Option is A

Concept: For minimum deviation, \[ \mu=\frac{\sin\left(\frac{A+\delta_m}{2}\right)} {\sin\left(\frac{A}{2}\right)} \] where \(A\) is prism angle.

Step 1: Determine prism angle. Given \[ \mu=\sqrt3 \] \[ \delta_m=\frac{\pi}{6} \] Substituting, \[ \sqrt3= \frac{\sin\left(\frac{A+\pi/6}{2}\right)} {\sin(A/2)} \] Solving, \[ A=\frac{\pi}{3} \]

Step 2: Find incidence on second face. The ray falls normally on the first face. Hence, \[ r_1=0 \] For a prism, \[ r_1+r_2=A \] Therefore, \[ r_2=A \] \[ r_2=\frac{\pi}{3} \] This is the angle of incidence on the second face. \[ \boxed{\frac{\pi}{3}} \]