The momentum transferred by the wave to the mirror each second will be

6.4 \times 10^{- 7} k g / m / s

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4.8 \times 10^{- 8} k g / m / s

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3.2 \times 10 k g / m / s

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1.6 \times 10^{- 10} k g / m / s

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Solution

The correct option is D $1.6 \times 10^{- 10} k g / m / s$
Given: $I = 6 W / m^{2} A = 40 c m^{2} = 40 \times 10^{- 4} m^{2}$
Radiant momentum per sec, $p = \frac{2 I A}{c}$
$p = \frac{2 \times 6 \times 40 \times 10^{- 4}}{3 \times 10^{8}} = 1.6 \times 10^{- 10} k g / m / s$

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