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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

2.52 cm

3.26 cm

3.0 cm

4.00 cm

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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 $d' = 2 d \cos r$

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

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

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

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE