First slide

Ray optics

Question

Refractive index of the material of a sphere of radius 10 cm is $\frac{2}{\sqrt{3}}$ . A point source of light P is placed just inside the surface of the sphere. Then area of the surface of the sphere through which light energy can pass into surrounding air, is

Difficult

A

200 $π {cm}^{2}$
B

100 $π {cm}^{2}$
C

50 $π {cm}^{2}$
D

60 $π {cm}^{2}$

Solution

From the ray diagram, $∠ PAC = ∠ PBC = θ_{C} = \sin^{- 1} \frac{1}{(\frac{2}{\sqrt{3}})} = 60^{o}$
$∴ ∠ ACB = 4 θ_{C} = 240^{o} = \frac{4 π}{3}$
Therefore area of surface through area of surface through which rays of light can pass into surrounding area = Area AQB
Solid angle $ACB = ϕ = 2 π (1 - \cos \frac{4 θ_{C}}{2})$
$\Rightarrow ϕ = 2 π (1 - \cos 2 \times \frac{π}{3}) = 2 π (1 + \frac{1}{2}) = 3 π radian$
$∴$ Area $AQB = 3 π \times R^{2} = 2 π \times {(10)}^{2} {cm}^{2} = 200 π {cm}^{2}$

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