Sunlight of intensity 50 W m^(-2) is incident normally on the surface of a solar panel. Some part of incident energy (25%) is reflected from the…

Question

Sunlight of intensity 50 W m^(-2) 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)

MARKS Content Team · Accepted Answer

Correct answer: (B) 20×10^(-8) N. The pressure exerted by the photons P=0.75×I/c +0.25×2I/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)/(3×10^8) F=20.83×10^(-8) N

Sunlight of intensity $50 W m^{- 2}$ 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 $(c = 3 \times 10^{8} m s^{- 1})$

Solution

Relevant Questions

Sunlight of intensity 50 W m-2 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 m2 surface area will be close to c=3×108 m s-1

Solution

Relevant Questions

Sunlight of intensity $50 W m^{- 2}$ 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 $(c = 3 \times 10^{8} m s^{- 1})$