Physics - Modern Physics Question with Solution | TestHub

The momentum incident per second on the surface is given by $\frac{P}{c}$, where $P$ is the incident power and $c$ is the speed of light.

The surface absorbs $50\%$ of the light and reflects the remaining $50\%$.

The force exerted by the absorbed portion of the light is:

$F_a=\frac{0.5P}{c}$

The force exerted by the reflected portion of the light is:

$F_r=\frac{2\times 0.5P}{c}=\frac{P}{c}$

The total force exerted on the surface is the sum of these two forces:

$F=F_a+F_r=\frac{0.5P}{c}+\frac{P}{c}=\frac{1.5P}{c}$

Substituting $P=90$ W and $c=3\times 10^8$ m/s:

$F=\frac{1.5\times 90}{3\times 10^8}$

$F=\frac{135}{3\times 10^8}$

$F=45\times 10^{-8}$ N

$F=4.5\times 10^{-7}$ N

Answer: $4.5\times 10^{-7}\mathrm{N}$