Question

50 W/m² 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 1m² surface area will be close to (c = 3 × 10⁸ m/s):
1. 15 × 10^-8 N
2. 20 × 10 ^-8 N
3. 10 × 10^-8 N
4. 35 × 10^-8 N

Answer 1

Correct Answer - Option 2 : 20 × 10 ^-8 N

Concept:

Force exerted on the surface is given by the following formula

\(F = \left( {1 + r} \right)\frac{{IA}}{C}\)

where

r is the Reflected radiation

A is the Surface area

c is the speed of the sound

Calculation:

Given,

Energy density, I = 50 W/m²

Speed of the sound, c = 3 × 10⁸ m/s

Surface area, A = 1 m²

Reflected radiation, r = 25% = 25/100 = 0.25

Force on the surface (25% is reflecting back and the remaining 75% has been absorbed from the surface)

Force exerted on the surface is given by,

\(F = \left( {1 + r} \right)\frac{{IA}}{C}\)

\(= \frac{{\left( {1 + 0.25} \right) \times 50 \times 1}}{{3 \times {{10}^8}}}\)

\(= 1.25 \times \frac{{50}}{{3 \times {{10}^8}}}\)

\(= \frac{{125}}{{100}} \times \frac{{50}}{{3 \times {{10}^8}}}\)

\(= \frac{{125}}{6} \times {10^{ - 8}}\)

≃ 20 × 10^-8 N

Thus, the force exerted on 1m² surface area will be close to 20 × 10^-8 N