Question

An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index 4 √2 / 5.

Answer 1

A prism PQR (see figure) is a wedge-shaped body made from refracting medium bounded by two plane faces inclined to each other at an angle.

When a ray of light passes through a prism, it bends towards the base of prism and the angle made by the emergent ray with the incident ray is called angle of deviation. As the angle of incidence increases, the angle of deviation decreases to minimum δ_m. The angle of minimum deviation is given as

where A is the angle of the prism, μ is the refractive index. For an equilateral prism, A = 60°.

Refractive index of the prism (when kept in air) is 1.6. Thus, the angle of minimum duration in air is [using Eq. (1)] given by

When the prism is immersed in a medium of refractive index 4√2/5 then by using Eq. (1), we get

Therefore, the angle of minimum deviation of the prism, when it is kept in medium of refractive index 4√2/5, is 30°.

Answer 2

When the prism is kept in another medium we have to take the refractive index of the prism with respect to the provided medium.

medium^μ = \(\frac{\mu_{prism}}{\mu_{medium}} = \cfrac{\sin\left[\left(\frac{A + D_m}{2}\right)\right]}{\sin\left(\frac A2\right)}\)

\(\cfrac{1.6}{\frac{4√2}5} = \cfrac{\sin\left[\left(\frac{60° + D_m}{2}\right)\right]}{\sin\left(\frac{60°}2\right)}\)

\(\sqrt 2 = \cfrac{\sin\left[\left(\frac{60° + D_m}{2}\right)\right]}{\frac 12}\)

\(\sin^{-1} \left(\frac 1{\sqrt 2}\right) = \left(\frac{60° + D_m}{2}\right)\)

\(90° = 60° + D_m\)

\(D_m = 30°\)