Question

There is minimum deviation of light ray on passing through a prism (n = \( \sqrt{3} \)). If the angle of incidence is twice the angle of refraction for this ray, then what will be the angle of prism and the angle of refraction?

Hint:

For minimum deviation, the angle of incidence is equal to the angle of refraction in a prism.

Answer 1

Let the angle of incidence be \( i \) and the angle of refraction be \( r \). We are given that: \[ i = 2r. \] Let the angle of the prism be \( A \), the refractive index \( n = \sqrt{3} \), and the minimum deviation angle be \( D \). For minimum deviation in a prism, the angle of incidence \( i \) and the angle of refraction \( r \) are equal. Using the following relation for minimum deviation: \[ D = i + r - A. \] Since at minimum deviation \( i = r \), we have: \[ D = 2r - A. \] The refractive index \( n \) is related to the prism angle \( A \) and the minimum deviation \( D \) by: \[ n = \frac{\sin\left( \frac{A+D}{2} \right)}{\sin\left( \frac{A}{2} \right)}. \] Substituting \( n = \sqrt{3} \) and solving for \( A \) and \( D \), we get: \[ A = 60^\circ \quad \text{and} \quad r = 15^\circ. \]