A circular beam of light of diameter d=2 cm falls on a plane surface of glass. The angle of incidence is 600 and refractive index of glass is μ=3/2. The diameter of the refracted beam is

d'd=cos rcos(600)d'=2dcosrsin⁡r =sin⁡iμsinr=3232=13cos⁡r=1−sin2⁡r=23d′=2(2)23=4⋅23=3.26cm

A circular beam of light of diameter d=2 cm falls on a plane surface of glass. The angle of incidence is 600 and refractive index of glass is μ=3/2. The diameter of the refracted beam is

d'd=cos rcos(600)d'=2dcosrsin⁡r =sin⁡iμsinr=3232=13cos⁡r=1−sin2⁡r=23d′=2(2)23=4⋅23=3.26cm

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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 60⁰ and refractive index of glass is $μ = 3 / 2$ . The diameter of the refracted beam is

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3.26 cm

4.00 cm

3.0 cm

2.52 cm

answer is C.

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Detailed Solution

$\frac{d'}{d} = \frac{\cos r}{\cos (60^{0})}$

$d' = 2 d \cos r$

$\sin r$ = $\frac{\sin i}{μ}$

$\sin r = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}$ = $\frac{1}{\sqrt{3}}$

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

$d^{'} = 2 (2) \sqrt{\frac{2}{3}} = 4 \cdot \sqrt{\frac{2}{3}} = 3.26 cm$

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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 600 and refractive index of glass is μ=3/2. The diameter of the refracted beam is

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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 60⁰ and refractive index of glass is $μ = 3 / 2$ . The diameter of the refracted beam is