A 50 W m - 2 energy density of sunlight is incident normally on the surface of a solar panel. Some part of incident energy ( 25 % ) is reflected from the surface and the rest is absorbed. The force exerted on 1 m 2 surface area will be close to c = 3 × 10 8 m s - 1

Question

A 50 W m - 2 energy density of sunlight is incident normally on the surface of a solar panel. Some part of incident energy ( 25 % ) is reflected from the surface and the rest is absorbed. The force exerted on 1 m 2 surface area will be close to c = 3 × 10 8 m s - 1, 15 × 10 - 8 N, 20 × 10 - 8 N, 10 × 10 - 8 N, 35 × 10 - 8 N

MARKS App · Accepted Answer

The pressure exerted by the photons P = 0.75 × I c + 0.25 × 2 I c ( P = pressure, c = speed of light, I = intensity ) P = 1.25 × I c We know the force can be computed using F = P × A ( F = force, P = pressure, A = area ) F = 1.25 × 50 × 1 3 × 1 0 8 F = 20.83 × 1 0 - 8 N

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