Question

The pressure exerted by an electromagnetic wave of Intensity \[I(watt/{{m}^{2}})\] on a non-reflecting surface is: [ \[c\] is the velocity of light]
A) \[Ic\]
B) \[I{{c}^{2}}\]
C) \[I/c\]
D) \[I/{{c}^{2}}\]

Answer 1

Hint: In physics, the intensity of radiant energy is the power transferred per unit area, where the area is measured on the plane perpendicular to the direction of propagation of the energy. Intensity can be found by taking the energy density (energy per unit volume) at a point in space and multiplying it by the velocity at which the energy is moving. The resulting vector has the units of power divided by area (i.e., surface power density).

Formula Used:
\[\lambda =\dfrac{h}{p}\] , \[c=\lambda \times \eta \] , \[E=h\times \eta \] , \[I=\dfrac{E}{A}\]

Complete step by step solution:
From the expression for the de Broglie wavelength of a wave, we know that \[\lambda =\dfrac{h}{p}\] where \[\lambda \] is the wavelength, \[h\] is the Planck’s constant and \[p\] denotes the momentum of the wave
Momentum of the photon can hence be expressed as \[p=\dfrac{h}{\lambda }\]
We have an expression for the relation between the velocity, wavelength and the frequency of radiation as \[c=\lambda \times \eta \] where \[\eta \] is the frequency of the electromagnetic radiation, \[\lambda \] is the wavelength and \[c\] is the speed of light
Substituting the value of wavelength in the equation for the de Broglie wavelength, we get
\[\begin{align}
  & p=\dfrac{h}{{\dfrac{c}{\eta}}} \\
 & \Rightarrow p=\dfrac{h\times \eta }{c} \\
\end{align}\]
Energy carried by a wave can be expressed as a product of the Planck’s constant and the frequency, that is \[E=h\times \eta \]
We can now express the momentum of a wave (or electromagnetic radiation) as \[p=\dfrac{E}{c}\]
Momentum of the wave over unit area can be given as \[\dfrac{p}{A}=\dfrac{E}{Ac}\]
Intensity of the wave is the energy carried by the wave per unit area, that is \[I=\dfrac{E}{A}\] and hence the momentum of the wave over unit area can be given as \[\dfrac{I}{c}\]
Since the surface is mentioned to be non-reflecting, final momentum of the wave is zero.
Change in the momentum of the wave can thus be given as the difference between the initial and the final momentum which will eventually give us the value of the initial momentum as the final momentum is zero.
Force exerted by the wave over a unit area can be given as the change in momentum over that area, that is \[\dfrac{F}{A}=\dfrac{I}{c}\]
Force per unit area is the pressure exerted by the wave, and hence the pressure exerted by the wave can be given as \[P=\dfrac{I}{c}\]

Therefore, option (C) is the correct answer.

Note: We should be aware of the meaning of terms such as non-reflecting or fully reflecting. In the given question, we were told that the surface is non-reflecting which told us that the final momentum of the wave would be zero. Instead, if we were told that the surface is fully reflecting, the final momentum would be equal in magnitude and opposite in direction to the initial momentum. This will make the value of the change in momentum twice the initial momentum and would also increase the pressure exerted by the wave by a factor of two.