Question

Choose the correct answer from the alternatives given.
Light with an energy flux of $18W/c{m}^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20c{m}^2$, the average force exerted on the surface during a $30minute$ time span is:

• A. $2.1\times {10}^{-6}N$
• B. $1.2\times {10}^{-6}N$
• C. $1.2\times {10}^{6}N$
• D. $2.1\times {10}^{6}N$

Answer 1

The correct option is B $1.2\times {10}^{-6}N$

Given: The energy flux falling on the non-reflecting surface is $18W/c{m}^2$.

The surface area of the surface is $20c{m}^2$.

To find: The average force exerted on the surface in $30minute$.

The total energy falling on the surface is:

$U=I\times A\times t=18W/c{m}^2\times 20c{m}^2\times (30\times 60)=6.48\times {10}^{5}J$

So total momentum delivered is,

$p=\frac{U}{c}\Rightarrow \frac{6.48\times {10}^5}{3\times {10}^{8}m/s}$

The average force exerted on surface is ,

$F=\frac{p}{t}\Rightarrow \frac{2.16\times {10}^{-3}}{0.18\times {10}^4}=1.2\times {10}^{-6}N$

Option $\left(B\right)$ is correct.