Light with energy flux of $18 W / c m^{2}$ falls on a non reflecting surface at normal incidence. If the surface has an area of $20 c m^{2}$ . The average force exerted on the surface during a 30 minute time span is about.

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$1.2 \times 10^{- 5} N$

$1.2 \times 10^{- 6} N$

$1.2 \times 10^{- 7} N$

$1.2 \times 10^{- 8} N$

answer is C.

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Detailed Solution

The toal energy falling on the surface is
$U = (18 w / {cm}^{2}) \times (20 {cm}^{2}) \times (30 \times 60) = 6.4 \times 10^{5} J$

Total momentum delivered (for complete absorption) is $\frac{6.4 \times 10^{5}}{3 \times 10^{8}} = 2.16 \times 10^{- 3} k g m s^{- 1}$

$F = \frac{M o m e n t u m d e l i v e r e}{t i m e} = \frac{2.16 \times 10^{- 3}}{30 \times 60}$

$= 1.2 \times 10^{- 6} N$

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