A light of intensity 12   W   m - 2 incidents on a black surface of area 4   cm 2 . The…

A light of intensity

12 W m^{- 2}

incidents on a black surface of area

4 {cm}^{2}

. The radiation pressure on the surface is

$1 \times 10^{- 8} Pa$
$4 \times 10^{- 8} Pa$
$1.6 \times 10^{- 7} Pa$
$4.8 \times 10^{- 7} Pa$

Solution

Black surface will be considered as a perfectly absorbing surface. In this case, radiation pressure will be given by,
$P = \frac{I}{c} = \frac{12 W m^{- 2}}{3 \times 10^{8} m s^{- 1}} = 4 \times 10^{- 8} Pa$

Asked in: AP EAMCET 2022 (04 Jul Shift 2)

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