In the post dated 3^{rd} May 2007, four questions on Optics which appeared in Kerala Engineering Entrance 2007 test paper were discussed. Here is an additional question which appeared in the same paper:

**If the polarising angle of a piece of glass for green light is 54.74º, then the angle of minimum deviation for an equilateral prism made of same glass is (given tan 54.74º = 1.414)**

**(a) 45º (b) 54.74º (c) 60º (d) 90º (e) 30º **

Brewster’s law gives you the refractive index ‘n’ in terms of the polarising angle ‘θ’:

n = tan θ

Therefore, the refractive index of the given equilateral prism is 1.414, which is √2

Since n = sin[(A+D)/2] /sin(A/2) where ‘A’ is the angle of the prism and ‘D’ is the minimum deviation, we have

√2 = sin[(60º +D)/2]/sin 30º

This gives sin(60º +D)/2 = 1/√2 so that D = 30º

Now that you have worked out a problem on polarisation, note the following points:

(i) Transverse waves only can be polarized (Longitudinal waves cannot be polarised).

(ii) The proof for the transverse nature of light is the exhibition of polarisation by light.

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