Electromagnetic Waves

JEE Main level, single correct, medium

104 of 165

Question 104 of 165

Sunlight of intensity 50Wm−250W{m}^{-2}is incident normally on the surface of a solar panel. Some part of incident energy (2525%) is reflected from the surface and the rest is absorbed. The force exerted on 1m21{m}^{2} surface area will be close to (c=3×108ms−1)\left(c=3\times {10}^{8}m{s}^{-1}\right)

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Answer: (B) 20×10−8N20\times {10}^{-8} N

The pressure exerted by the photons
P=0.75×Ic+0.25×2IcP=0.75\times \dfrac{I}{c}+0.25\times \dfrac{2I}{c} ( PP = pressure, cc = speed of light, II = intensity )

P=1.25×IcP=1.25\times \dfrac{I}{c}

We know the force can be computed using

F=P×AF=P\times A
( FF = force, PP = pressure, AA = area )

F=1.25×(50×13×108)F=1.25\times \left(\dfrac{50\times 1}{3\times 1{0}^{8}}\right)

F=20.83×10−8NF=20.83\times 1{0}^{-8}N

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