Question

A ray of light is incident on a refracting face AB of a prism ABC at an angle of 45°. The ray emerges from face AC and the angle of deviation is 15°. The angle of prism is 30°. Show that the emergent ray is normal to the face AC from which it emerges out. Find the refraction index of the material of the prism.

Answer 1

Given: Angle of incidence (i) = 45°

Angle of deviation (δ) = 15°

Angle of prism (A) = 30°

Since, by the prism formula:

δ = i + e − A

15° = 45° + e − 30°

e = 15° + 30° − 45°

= 45° − 45°

= 0°

Since the angle of emergence (e) is zero, the emergent ray is normal to the face AC.

For the refractive index by using Snell’s law at face AB:

n₁sin i = n₂sin r

Since the incident ray is in air, n₁ = 1, and let n₂ = n be the refractive index of the prism material.

The angle of refraction at face AB is denoted by r.

sin 45° = n sin r

`sqrt 2/2` = n sin r    ...(i)

Using the prism angle formula:

r + r' = 30°

Since the emergent ray is normal to AC,

r' = 0

So, r = 30°

Now substituting r = 30° in equation (i),

`sqrt 2/2` = n sin 30°

`sqrt 2/2 = n xx 1/2`

n = `sqrt 2/2 xx 2`

n = `sqrt 2`

n ≈ 1.414