Question

Light with an energy flux of $18W/c{{m}^{2}}$falls on a non-reflecting surface at normal incidence. The pressure exerted on the surface is:
$\begin{aligned} & \text{A}\text{. 2N/}{{\text{m}}^{2}}\text{ } \\\ & \text{B}\text{. 2}\times \text{1}{{\text{0}}^{-4}}\text{N/}{{\text{m}}^{2}} \\\ & \text{C}\text{. 6N/}{{\text{m}}^{2}} \\\ & \text{D}\text{. 6}\times \text{1}{{\text{0}}^{-4}}\text{N/}{{\text{m}}^{2}} \\\ \end{aligned}$

Answer 1

Hint: To approach this kind of question you need basic formulas of electrostatics and mechanics. First use the formula of pressure defined as force per unit area. Use Newton’s second law formula of force which gives relation with momentum. To put value in momentum, use Einstein's famous equation, $E=m{{c}^{2}}$. Then convert the formula of pressure in such a way that it will show the relation of flux and velocity of light only. S.I unit of pressure is $N/{{m}^{2}}$or V/m.it is equivalent to electric field intensity.

Complete step by step answer:
We know that pressure is defined as force per unit area.
Mathematically,
$P=\dfrac{F}{A}-----(1)$
Where,
P= pressure
F= force
A= area
Force can be defined as rate change of momentum per unit time.
I.e. $F=\dfrac{dp}{dt}-----(2)$
By Einstein equation, we know that $E=m{{c}^{2}}$
Where,
E= energy of incident light
c= speed of light
It is also can be written as $p=\dfrac{E}{c}------(3)$
(Since momentum is p=mv)
Therefore pressure exerted on the surface is given as,
From (1), (2) and (3)
Thus $P=\dfrac{F}{A}=\dfrac{\dfrac{dp}{dt}}{A}=\dfrac{1}{cA}\dfrac{dE}{dt}=\dfrac{flux}{c}=\dfrac{18\times {{10}^{4}}W/{{m}^{2}}}{3\times {{10}^{8}}m/s}=6\times {{10}^{-4}}N/{{m}^{2}}$
Thus, the pressure exerted on the surface is$6\times {{10}^{-4}}N/{{m}^{2}}$.
Hence option (D) is correct

Note:
1. Do you know that light has dual nature: wave nature and particle nature.
2. Electric flux and electric flux density are two different things. Electric flux passing through a surface is defined as an integral product of vector quantity electric field and vector quantity area.
3. Unit of electric flux is volt meter. Whereas electric flux density is defined as electric flux per unit area. Unit of electric flux density is weber per meter square or volt per meter.