Question

If a light of intensity $60\text{W}$ falls normally on an area of $1{\text{m}}^2$. If the reflectivity of the surface is $75\%$, find the force experienced by the surface.

• A. $3.5\times {10}^{-10}\text{N}$
• B. $3.5\times {10}^{-7}\text{N}$
• C. $3.5\mu \text{N}$
• D. $3.5\text{mN}$

Answer 1

The correct option is B $3.5\times {10}^{-7}\text{N}$

$I=60{\text{W/m}}^2$
$A=1{\text{m}}^2$
As, $r=0.75\Rightarrow 75\%$ photons are reflected while $25\%$ are absorbed.

Force due to absorbed photons, $F_a=0.25\times \frac{IA}{c}$
$=0.25\times \frac{60\times 1}{3\times {10}^8}=5\times {10}^{-8}\text{N}$

Force due to reflected photons $F_r=0.75\times 2\times \frac{IA}{c}$

$=0.75\times 2\times \frac{60\times 1}{3\times {10}^8}$

$=3\times {10}^{-7}\text{N}$

The total force experienced is, $F=F_a+F_r=5\times {10}^8+3\times {10}^{-7}$
$=3.5\times {10}^{-7}\text{N}$

Hence, $\left(B\right)$ is the correct answer.