Question

The incident intensity on a horizontal surface at sea level from the sun is about $1kW{m}^{-2}$. Assuming that 50 per cent of this intensity is reflected and 50 per cent is absorbed, determine the radiation pressure on this horizontal surface (in pascals).

• A. $8.2\times {10}^{-2}$
• B. $5\times {10}^{-6}$
• C. $3\times {10}^{-5}$
• D. $6\times {10}^{-5}$

Answer 1

The correct option is B $5\times {10}^{-6}$

Given: $I=1000$ $W{m}^{-2}$

As 50 percent of light is reflected, thus $e=0.5$

Radiation pressure, $P=\frac{(1+e)I}{c}$

$\therefore$ $P=\frac{(1+0.5)\times 1000}{3\times {10}^8}=5\times {10}^{-6}$ Pa