Question

Light with an energy flux of $25\times {10}^4{\text{W/m}}^2$ falls on a perfectly reflecting surface at normal incidence. If the surface area is $15{\text{cm}}^2,$ the average force exerted on the surface is

• A. $1.25\times {10}^{-6}\text{N}$
• B. $2.50\times {10}^{-6}\text{N}$
• C. $1.15\times {10}^{-6}\text{N}$
• D. $3.50\times {10}^{-6}\text{N}$

Answer 1

The correct option is B $2.50\times {10}^{-6}\text{N}$

Given:-
Intensity of light, $I=25\times {10}^4{\text{W/m}}^2$

Also, $I=\frac{U}{A\cdot \Delta t}..........\left(1\right)$

Now, rate of change in momentum of light after being reflected from a perfectly reflecting surface is

$\frac{\Delta p}{\Delta t}=\frac{2U}{c\cdot \Delta t}$

Average force exerted on the surface is given by

$F=\frac{\Delta p}{\Delta t}=\frac{2U}{c\cdot \Delta t}=\frac{2IA\Delta t}{c\Delta t}$

$\therefore F=\frac{2IA}{c}$

$F=\frac{2\times 25\times {10}^4\times 15\times {10}^{-4}}{3\times {10}^8}$

$F=250\times {10}^{-8}\text{N}$

$\therefore F=2.5\times {10}^{-6}\text{N}$

Hence, option $\left(\text{B}\right)$ is correct.