Light with an energy flux of $25 imes 10^4$ Wm$^{-2}$ falls on a perfectly reflecting surface at normal inci

Question

Light with an energy flux of $25 	imes 10^4$ Wm$^{-2}$ falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm$^2$, then the average force exerted on the surface is:

Accepted Answer

The average force ($F$) exerted by a light beam on a perfectly reflecting surface is given by the change in momentum per unit time. For a perfectly reflecting surface, the force is $F = \frac{2P}{c}$, where $P$ is the power of the incident light and $c$ is the speed of light.

First, calculate the Power ($P$):
$P = 	ext{Energy Flux} 	imes 	ext{Area}$
Given Energy Flux ($I$) = $25 	imes 10^4$ W/m$^2$
Area ($A$) = $15 	ext{ cm}^2 = 15 	imes 10^{-4} 	ext{ m}^2$
$P = (25 	imes 10^4) 	imes (15 	imes 10^{-4}) = 375$ Watts

Now, calculate the Force ($F$):
$F = \frac{2 	imes 375}{3 	imes 10^8} = \frac{750}{3 	imes 10^8} = 250 	imes 10^{-8} 	ext{ N}$
$F = 2.5 	imes 10^{-6} 	ext{ N}$

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