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 $(c = 3 \times 10^8 \; m/s)$ :

• A. $15 \times 10^{-8} \; N$
• B. $35 \times 10^{-8} \; N$
• C. $10 \times 10^{-8} \; N$
• D. $20 \times 10^{-8} \; N$

Answer 1

Answer: (D)

Force on the surface (25% reflecting and rest absorbing)
$F = \frac{25}{100} \left(\frac{2I}{C} \right)+ \frac{75}{100} \left(\frac{I}{C}\right) = \frac{125}{100}\left(\frac{I}{C}\right) $
$ =\frac{125}{100} \times\left(\frac{50}{3\times10^{8}} \right) = 20.83 \times10^{-8} N $