Sunday, July 15, 2007

KEAM (Engineering) 2007 Question on Polarisation

In the post dated 3rd 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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