Refractive index of the material of a sphere of radius $$10$$ cm is $$23$$. 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

Question

Refractive index of the material of a sphere of radius $$10$$ cm is $$23$$. 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

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From the ray diagram, ∠$$PAC=$$∠$$PBC=$$θ$$C=$$$$sin$$−$$1123=60o$$∴ ∠$$ACB=4$$θ$$C=240o=4$$$$π$$$$3Therefore$$ area of surface through area of surface through which rays of light can pass into surrounding area $$=$$ Area AQB Solid angle $$ACB=$$ϕ$$=2$$$$π$$$$1$$−$$cos4$$θ$$C2$$⇒ϕ$$=2$$$$π$$$$1$$−$$cos2$$×π$$3=2$$$$π$$$$1+12=3$$π radian∴ Area $$AQB=3$$π×$$R2=2$$π×$$102cm2=200$$π $$cm2$$

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

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Refractive index of the material of a sphere of radius 10 cm is 23. 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

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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