Prove that the force exerted on a sphere of radius $r$ by a light source of intensity $I$ is not affected even if the sphere is not perfectly absorbing.

Question 11,Photoelectric effect and wave-particle duality, Concepts of physics Volume-2 by Dr.HC Verma

The derivation for the force exerted on a perfectly reflecting sphere can be found here

A slight manipulation of eq-2.58 yields:


where $R$ and $a$ are the reflection and absorption coefficients of the sphere respectively.

Thus the final result would be:

$$F=\frac{I\pi r^2(2R+a)}{2c}$$

Its clear that for $R=1$ and $a=0$, the result is as expected for a perfectly reflecting sphere.

However, coming back to the initial question, the final expression clearly varies for different values of $R$.

The momentum/force, obviously varies for different absorbing coefficients of the sphere. This is contradicting the question. Where have I gone wrong? Or is the question incorrect?


Multiplying by (2R+a) does not work in Eq 2.58 because you need to consider the direction. Eq 2.58 describes the force along the normal. For a total reflection case, the force from the photons is directed normal to the sphere. For a total absorption case, all the force is directed horizontally regardless of where it hits. So absorption shouldn’t be mixed with eq 2.58.

You should not put the absorption term into the integral. Just add it after the integration for reflection is complete.

  • $\begingroup$ "For a total absorption case, all the force is directed horizontally regardless of where it hits." I dont understand why. Would you be kind enough to elucidate further(maybe with a diagram)? $\endgroup$ – newbie105 Nov 20 '20 at 12:52
  • $\begingroup$ F = dp/dt, so the direction of force on the photons is the same as the direction of the change in momentum of the photons. By Newton’s 3rd law, the force on the sphere by the photons is opposite the force on the photons. So we need to find the direction of change in momentum of the photons and reverse it to find the direction of force. If the photons are coming in from the left, they are moving rightwards. After absorption, their momentum will become zero. So the change of momentum is leftwards, force on sphere is rightwards. This is regardless of where the photon hits. $\endgroup$ – bluemystic Nov 20 '20 at 14:10
  • $\begingroup$ Sorry, I don’t know how to draw and upload images on my phone $\endgroup$ – bluemystic Nov 20 '20 at 14:19
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    $\begingroup$ So now let us consider 2 cases- one perfectly reflecting and the other perfectly absorbing. In the first case, considering the force to be $F=\int dF\cos\theta=2\int\frac{dp}{dt}\cos\theta$ while the second one must be $F=\int dF=\int\frac{dp}{dt}$(since it is already along the path of light). Am I right in saying this? If yes, the second integral gives us the force $F=\frac{2I\pi R^2}{c}\int_0^{\pi/2}{cos^2\theta\sin\theta d\theta}=\frac{2I\pi R^2}{3c}$ $\endgroup$ – newbie105 Nov 20 '20 at 16:13
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    $\begingroup$ Yes, but the final $F$ should be $F$ = $\frac{2IπR^2}{c} \int_0^{π/2} cosθsinθ \,dθ$ , Notice there is another cosine term gone, in the perfectly reflecting case $dF =2 dp (cosθ)$ - Eq 2.58, while in the perfectly absorbing case $dF = dp $ In the end it should give you $F = \frac{IπR^2}{c}$ In the perfectly reflecting case, one cosine was for taking the component of momentum along the normal to the sphere, the other was for taking the component of that horizontally (pointing to the right). In the perfectly absorbing case, these two are not needed, but it needs to be divided by 2. $\endgroup$ – bluemystic Nov 20 '20 at 16:56

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