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A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is

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$3.2 \times 10^{- 8} N$

$3.2 \times 10^{- 7} N$

$5.12 \times 10^{- 7} N$

$5.12 \times 10^{- 8}$

answer is B.

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Detailed Solution

$F = 40 % \frac{P}{C} + 60 % \frac{2 P}{C}$
$\begin{array}{l} = \frac{40}{100} \times \frac{60}{3 \times 10^{2}} + \frac{60}{100} \times \frac{2 \times 60}{3 \times 10^{8}} N \\ = 3.2 \times 10^{- 7} N \end{array}$

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