Question

A source of power $600kW$ emits photon which incident on a surface $2.5m$ away out of which $50\%$ is reflected back. What is the radiation pressure on the surface :

• A. $3.9\times {10}^{-5}Pa$
• B. $7.8\times {10}^{-5}Pa$
• C. $11.7\times {10}^{-5}Pa$
• D. $15.6\times {10}^{-5}Pa$

Answer 1

Answer: (A)

$3.9\times {10}^{-5}Pa$

Given,

$p=600kW=6\times {10}^{5}W$

$r=2.5m$

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

Intensity, $I=\frac{p}{4\pi r^2}=\frac{6\times {10}^5}{4\times 3.14\times 2.5\times 2.5}$

$I=7.64\times {10}^{4}W/m^2$

$c=3\times {10}^{8}m/s$

$r=50$%$=0.5$ reflectivity

From equation (1)

$P=\frac{7.64\times {10}^4}{3\times {10}^8}(1+0.5)$

$P=4\times {10}^{-5}Pa$