JEE Main · 2019 · 09 Apr 2019 · Shift 2 · 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

1 m^{2}

surface area will be close to (

c = 3 \times 1 0^{8} m / s

)

Answer: (D)

Solution

Sunlight carries momentum, and the push on the panel comes from the momentum it delivers each second. The absorbed fraction (

75%

) gives up its momentum at the rate

\frac{I}{c}

per unit area, while the reflected fraction (

25%

) turns its momentum around and so contributes

\frac{2 I}{c}

. For

1 m^{2}

F = \frac{I}{c} (0.75 + 2 \times 0.25) = \frac{1.25 I}{c} = \frac{1.25 \times 50}{3 \times 1 0 ^{8}}

, which is about

20 \times 1 0^{- 8}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option