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A perfectly absorbing surface intercepts a parallel beam of monochromatic light of power 10 W ( $λ$ = 500 nm) incident on it normally the force exerted by light beam on the surface is

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$1 \times 10^{- 7} N$

$\frac{1}{4} \times 10^{- 7} N$

$\frac{1}{3} \times 10^{- 7} N$

$\frac{1}{2} \times 10^{- 7} N$

answer is B.

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Detailed Solution

Power = 10W
$\Rightarrow Force = \frac{P}{C} (forperfectlyabsorbingsurface)$
$F = \frac{10}{3 \times 10^{8} m / s} = \frac{1}{3} \times 10^{- 7}$ N

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