Question 134: EM waves

Question

50 W/m
2
energy density of sunlight is normally
incident 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 1m
2
surface area will be close to
(c = 3 × 108 m/s) :-

Accepted Answer

Radiation pressure for 100% reflection = $${{2I} \over C}$$
Radiation pressure for 0% reflection = $${I \over C}$$
Hence, in given case, radiation pressure
= $$\left( {0.25} ight)\left( {{{2I} \over C}} ight) + \left( {0.75} ight)\left( {{I \over C}} ight)$$
$$\left( {1.25} ight)\left( {{I \over C}} ight)$$
$$ 	herefore $$ Force = P ÃÂ (Area) = 20.83 ÃÂ 10
Ã¢ÂÂ8
N

Answer

20 × 10
–8
N

Answer

35 × 10
–8
N

Answer

10 × 10–8 N

Answer

15 × 10
–8
N

EM waves

50 W/m 2 energy density of sunlight is normally incident 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 1m 2 surface area will be close to (c = 3 × 108 m/s) :-

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