Light is incident normally on a completely absorbing surface with an energy flux of $25 W {cm}^{- 2}$ . If the surface has an area of $25 {cm}^{2}$ , the momentum transferred to the surface in $40 \min$ time duration will be:

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$3.5 \times 10^{- 6} Ns$

$6.3 \times 10^{- 4} Ns$

$1.4 \times 10^{- 6} Ns$

$5.0 \times 10^{- 3} Ns$

answer is B.

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Detailed Solution

For completely absorbing surface, momentum transferred is given by:

$p = \frac{Energy transferred}{speed of light}$

$p = \frac{25 \times 25 \times 40 \times 60}{3 \times 10^{8}} = 5 \times 10^{- 3} Ns$

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