Modern Physics 7 Question 61
9. $50 Q / m^{2}$ 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 $\left.=3 \times 10^{8} m / s\right)$
(Main 2019, 9 April II)
(a) $20 \times 10^{-8} N$
(b) $35 \times 10^{-8} N$
(c) $15 \times 10^{-8} N$
(d) $10 \times 10^{-8} N$
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Solution:
- Radiation pressure or momentum imparted per second per unit area when light falls is
where, $I$ is the intensity of the light.
In given case, there is $25 %$ reflection and $75 %$ absorption, so radiation pressure $=$ force per unit area
$$ \begin{aligned} & =\frac{25}{100} \times \frac{2 I}{c}+\frac{75}{100} \times \frac{I}{c} \\ & =\frac{1}{2} \times \frac{I}{c}+\frac{3}{4} \times \frac{I}{c}=\frac{5}{4} \times \frac{I}{c}=\frac{5}{4} \times \frac{50}{3 \times 10^{8}} \\ & =20.83 \times 10^{-8} N / m^{2} \approx 20 \times 10^{-8} N / m^{2} \end{aligned} $$
