Question

Light with an energy flux of $18W/c{{m}^{2}}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20c{{m}^{2}}$, then the total momentum delivered on the surface during the span of 30 minutes is
$\begin{aligned} & \text{A}\text{. }1.2\times {{10}^{-6}}kgm{{s}^{-1}} \\\ & \text{B}\text{. }2.16\times {{10}^{-3}}kgm{{s}^{-1}} \\\ & \text{C}\text{. }1.18\times {{10}^{-3}}kgm{{s}^{-1}} \\\ & \text{D}\text{. }3.2\times {{10}^{-3}}kgm{{s}^{-1}} \\\ \end{aligned}$

Answer 1

Hint: With the help of given flux and the area, calculate the power incident. Further, applying the relationship between power and energy with time, find the energy that is delivered. Since, light is made up of massless photons, from the energy-momentum relation defined, find the value of the momentum.
Formula used: Formula for incident power:
$P=\phi A$
Formula for energy in terms of power and time:
$\text{Energy}=\text{Power}\times \text{Time}$
Energy-momentum relation for a massless object:
$~E\text{ }=\text{ }pc$

Complete step-by-step answer:
It is specified that the surface is non-reflecting. Therefore, all the light energy incidents are delivered in the form of momentum.
Flux for a quantity, here light, is defined as the amount of that quantity incident normally on a surface. Notice that in the question we are given an energy flux of $18W/c{{m}^{2}}$ which means for an area of unit centimetre square, 18 watts of power is incident. If A is the area and $\phi$ is the energy flux, the power incident on the surface is given by the following expression.
$\begin{aligned} & P=\phi A \\\ & \Rightarrow P=18W/c{{m}^{2}}\times 20c{{m}^{2}}=360W \\\ \end{aligned}$
For a given amount of power applied for a time interval T, the expression for energy is:
$\begin{aligned} & \text{Energy}=\text{Power}\times \text{Time} \\\ & \Rightarrow \text{Energy}=360\times 30\times 60=6.48\times {{10}^{5}}J \\\ \end{aligned}$
Since, the time the power is applied is given in minutes (T=30 minutes), it is converted into seconds in the equation above.
We have now calculated the amount of energy.
In physics, the energy–momentum relation also known as relativistic dispersion relation, is an equation which relates the object's rest mass, total energy, and momentum:
${{E}^{2}}={{(pc)}^{2}}+{{({{m}_{0}}{{c}^{2}})}^{2}}…... (1)$
Where, E is its energy, p is the momentum, ${{m}_{0}}$ is the rest mass and c is the speed of light.
$c=3\times {{10}^{8}}m{{s}^{-1}}$
If the body is a massless particle (${{m}_{0}}$ = 0), then equation (1) reduces to
$~E\text{ }=\text{ }pc..… (2)$
Light is made up of packets of photons having rest mass zero. Therefore, the momentum of delivered is calculated as:
$\begin{aligned} & p=\dfrac{E}{c} \\\ & \Rightarrow \dfrac{6.48\times {{10}^{5}}}{3\times {{10}^{8}}}kgm{{s}^{-1}} \\\ & \Rightarrow 2.16\times {{10}^{-3}}kgm{{s}^{-1}} \\\ \end{aligned}$
Therefore, the correct answer to this question is option B.

Note: An alternative way to find the expressions for some quantities, if one does not remember the formula is to observe the and derive a relation. For example, in this case, flux had the units $W/c{{m}^{2}}$ where the area had the units $c{{m}^{2}}$. Since, the unit of power is Watt is a known fact. One can guess that the power flux times the area.