Dual Nature of Matter

JEE Main level, numerical answer, medium

136 of 200

Question 136 of 200

A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of 24W24W. The radius of curvature of hemisphere is 10cm10cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ______ ×10−8N\times {10}^{-8}N.

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Answer: 4


For completely reflecting surface, force,
F=2IAcF=\dfrac{2IA}{c}, where, I=PSurfaceareaI=\dfrac{P}{Surfacearea} is intensity of radiation, AA is surface area and cc is speed of light.
F=2IcπR2=2P04πR2πR2cF=\dfrac{2I}{c}\pi {R}^{2} =2\dfrac{{P}_{0}}{4\pi {R}^{2}}\dfrac{\pi {R}^{2}}{c}
=P02c=242×3×108=4×10−8N=\dfrac{{P}_{0}}{2c} =\dfrac{24}{2\times 3\times {10}^{8}} =4\times {10}^{-8}N

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