Question

A plane electromagnetic wave of wave intensity $6W{m^{ - 2}}$ strikes a small mirror area $40c{m^2}$, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be:
(A) $6.4 \times {10^{ - 7}}kgm{s^{ - 2}}$
(B) $4.8 \times {10^{ - 8}}kgm{s^{ - 2}}$
(C) $3.2 \times {10^{ - 9}}kgm{s^{ - 2}}$
(D) $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$

Answer 1

Hint: In order to solve this question, we will use the general relation between intensity, and momentum and are to calculate the net transferred momentum by the electromagnetic wave to the mirror each second.

Formula used:
If I is the intensity of an electromagnetic wave, A is the area of the object on which its falling then net momentum transferred by the wave to the object is given by:
$p = 2IAc$
where, p is the momentum transferred $c = 3 \times {10^8}m{s^{ - 1}}$ is the speed of the light

Complete answer:
According to the question, we have given that,
The intensity of an electromagnetic wave is $I = 6W{m^{ - 2}}$
Area of the small mirror is $A = 40c{m^2} = 4 \times {10^{ - 3}}{m^2}$ and $c = 3 \times {10^8}m{s^{ - 1}}$

In order to calculate momentum transferred, put the values in the formula $p = 2IAc$ we get,
$p = 2(6)(4 \times {10^{ - 3}})(3 \times {10^8})$

on solving for p we get the value,
$p = 1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$

So, the net momentum transferred by the electromagnetic wave to the small mirror has the magnitude of $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$

Hence, the correct answer is option (D) $1.6 \times {10^{ - 10}}kgm{s^{ - 2}}$

Note: It should be remembered that electromagnetic waves are a form of energy but as light energy carries photon and photons has a mass, due to this particle nature of light, it carries a momentum along with it and the value of the speed of light is taken in free space as the speed of light varies in a different medium but it’s maximum in free space.