A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24   W . The radius of curvature of hemisphere is 10   cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is…

Question

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24 &#160; W . The radius of curvature of hemisphere is 10 &#160; cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ______ &#215; 10 - 8 &#160; N .

Aicharya · Accepted Answer

For completely reflecting surface, force, $F = \frac{2 I A}{c}$ , where, $I = \frac{P}{Surface area}$ is intensity of radiation, $A$ is surface area and $c$ is speed of light.

$F = \frac{2 I}{c} π R^{2} = 2 \frac{P_{0}}{4 π R^{2}} \frac{π R^{2}}{c}$

$= \frac{P_{0}}{2 c} = \frac{24}{2 \times 3 \times 10^{8}} = 4 \times 10^{- 8} N$

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a…

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