First slide

Refraction of light

Question

A circular beam of light of diameter d = 2cm 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

Moderate

A

4.00 cm
B

3.0 cm
C

3.26 cm
D

2.52 cm

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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