Solveeit Logo
Login

Question

Physics Question on Optics

The refractive index of a prism with apex angle AA is cot⁡A2\cot \frac{A}{2}. The angle of minimum deviation is:

A

δm=180∘−A\delta_m = 180^\circ - A

B

δm=180∘−3A\delta_m = 180^\circ - 3A

C

δm=180∘−4A\delta_m = 180^\circ - 4A

D

δm=180∘−2A\delta_m = 180^\circ - 2A

Answer

δm=180∘−2A\delta_m = 180^\circ - 2A

Explanation

Solution

Given:

μ=sin⁡(A+δm2)sin⁡A2\mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}}

Using the relation:

cos⁡A2=sin⁡(A+δm2)\cos \frac{A}{2} = \sin \left( \frac{A + \delta_m}{2} \right)

We get:

δm=π−2A\delta_m = \pi - 2A

Therefore:

δm=180∘−2A\delta_m = 180^\circ - 2A