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A circular beam of light of diameter d = 2 cm falls on a plane surface of glass. The angle of incidence is $60^{0}$ and refractive index of glass is $μ = 3 / 2$ . The diameter of the refracted beam is

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2.52 cm

3.26 cm

3.0 cm

4.00 cm

answer is C.

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Detailed Solution

Let $d'$ be the diameter of refracted beam.
Then, $d = P Q \cos (60^{0})$
And $d' = P Q \cos (r)$

$i . e ., \frac{d'}{d} = \frac{\cos (r)}{\cos (60^{0})} = 2 \cos (r)$
Or $d' = 2 d \cos r$

$\sin (r) = \frac{\sin (i)}{μ} = \frac{(\sqrt{3} / 2)}{(3 / 2)} = \frac{1}{\sqrt{3}}$

$∴ \cos r = \sqrt{1 - \sin^{2} r} = \sqrt{\frac{2}{3}}$

$∴ d' = (2) (2) \sqrt{\frac{2}{3}} = 4 \sqrt{\frac{2}{3}} c m ≃ 3.26 c m$

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