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A perfectly reflecting mirror has an area of 1 cm². Light energy is allowed to fall on it for 1 h at the rate of 1 $10 \frac{W}{c m^{2}}$ . The force that acts on the mirror is

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$1.34 \times 10^{- 7} N$

$2.4 \times 10^{- 4} N$

$6.67 \times 10^{- 8} N$

$3.35 \times 10^{- 8} N$

answer is B.

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Detailed Solution

Let I = energy falling on the surface per second = 10 W/cm² , area = 1 sq-cm

Momentum of photons, $p = \frac{E}{c} = \frac{I A}{c}$

On reflection, change in momentum per second

$= 2 p = \frac{2 I A}{c} = \frac{2 \times 10 \times 1}{3 \times 10^{8}} = 6.67 \times 10^{- 8} N$

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