Question

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c is the velocity of light):
A. ${\displaystyle \frac{2E}{c}}$
B. ${\displaystyle \frac{2E}{c^2}}$
C. ${\displaystyle \frac{E}{c^2}}$
D. ${\displaystyle \frac{E}{c}}$

Answer 1

Hint: Recall the relation for the momentum possessed by a photon. You can derive this from the mass-energy equivalence principle, since we know that momentum is a product of a particle’s mass and its velocity. Then, using the fact that the momentum transferred to the reflecting surface by the photon is equivalent to the change in momentum of the photon over the course of its incidence and reflection, determine the appropriate relation for the same by accounting for both magnitude and directional changes in the momentum.

Formula used: Momentum of photonic radiation $p={\displaystyle \frac{E}{c}}$
Momentum transferred $\mathrm{\Delta }p=p_i-p_f$

Complete step by step answer:
From the mass-energy equivalence, we have the energy of the photon given as $E=m{c}^2$, where m is the energetic mass of the photon and c is the velocity of light.
We know that the momentum of the photon can be expressed as $p=mc$
$\Rightarrow E=(mc)c=pc$
This means that the momentum possessed by a photon with energy E is:
$p={\displaystyle \frac{E}{c}}$
There is a photon incident on a reflecting surface. The initial momentum of the photon is given as: $p_i={\displaystyle \frac{E}{c}}$
Now, the photon falls normally on a perfectly reflecting surface. This means that the photon is reflected back with the same magnitude of incident momentum but is directionally opposite.
The final momentum of the reflected photon is given as:
$p_f=-{\displaystyle \frac{E}{c}}$
We understand that the photon is able to reflect back after incidence because it imparts its momentum to the reflecting surface upon incidence. This momentum transferred between the photon and the reflecting surface will be equivalent to the change in momentum of the photon.
Therefore, the change in momentum of the photon is given as:
$\mathrm{\Delta }p=p_i-p_f={\displaystyle \frac{E}{c}}-(-{\displaystyle \frac{E}{c}})={\displaystyle \frac{E}{c}}+{\displaystyle \frac{E}{c}}$
$\Rightarrow \mathrm{\Delta }p={\displaystyle \frac{2E}{c}}$.
This is equivalent to the momentum transferred by the incident photon to the reflecting surface which consequently brings about perfect reflection.

So, the correct answer is “Option A”.

Note: Do not forget to account for the signs to indicate change in direction, since momentum is ultimately a vector quantity whose magnitude and direction are co-dependent. If the final momentum of a body is in a direction opposite to its initial motion, then the final momentum assumes a negative sign, irrespective of whether there is a complete transfer of the magnitude of momentum or not. Thus, it is always important to remember the directional nature of vector quantities such as momenta in order to maintain a numerical consistency in calculations pertaining to their orientation in space.