I am asking myself why the lens must be pushed to the right in the following scenario: (image coming from Atoms and Sporks' nice video https://www.youtube.com/watch?v=UAmdoOX3870&t=327s)

enter image description here

This can be explained based on the fact that momentum changes after the collision with the lens, meaning that an external force is exerted on the lens (Newton's second Law for momentum):

$$\vec F = \frac{d \vec p}{dt}$$

Newton's second Law for momentum has to be enough to explain why the lens is pushed to the right in the above picture

NOTE: I am going to work out the most basic example I can think of before dealing with the original question:

Imagine the following inelastic collision:

enter image description here

Assume that the magnitude of the velocity of the ball after the collision is the same.

Question 1. What is the force exerted on the ball?

My whole work is based on the fact that an external force causes a change in momentum. This can be understood as:

$\vec F = \Delta \vec p = \vec p_f - \vec p_i$

enter image description here

(Word is not the best tool to build up diagrams I know XD)

Then, based on the diagram above, the force exerted on the ball by the box is:

$$-\vec F = -\vec p_f + \vec p_i$$

This makes sense because the force points to the left.

Question 2. What is the force exerted on the box?

By Newton's third law we know that the force exerted on the box by the ball is equal to $F$ but with opposite sign. This also makes sense; the box is pushed to the right.

If you apply this reasoning to the original question you can also explain why the lens moves to the right.

However, in the lens' case we are dealing with light. Am I somehow being sloppy with my explanation? How can I improve it?


  • 1
    $\begingroup$ As you point out, light does carry momentum, so it kind of behaves like that. Just one thing, the formula is $F\Delta t=\Delta p$, not just $F$. $\endgroup$
    – FGSUZ
    Apr 19, 2019 at 10:17
  • $\begingroup$ @FGSUZ Thanks for pointing $F\Delta t=\Delta p$ out. I am working with momentum per unit time though. $\endgroup$
    – JD_PM
    Apr 19, 2019 at 10:58

1 Answer 1


In order to speak of impulse of light, we need Einstein: According to Special Relativity, a photon of energy $E=h\cdot\nu$ has an impulse $p=\frac Ec$. In your image, it looks like the light is refracted in varying angles, let's say the typical angle is $30^\circ$, so inspired by $\sin30^\circ=\frac12$ we boldly take $\frac12\frac Ec$ as horizontal change of impulse. Summing over all photons in a time unit (and dividing by the time unit), we obtain a force $$ F\approx\frac12\frac Pc$$ where $P$ is the power of the light source.

  • 1
    $\begingroup$ In order to speak of impulse of light, we need Einstein Not true. Momentum of light directly follows from Maxwell's equations. According to Special Relativity, a photon of energy $E=h\nu$ ... Einstein never speaks of photons in his papers on relativity. $\endgroup$
    – Elio Fabri
    Apr 19, 2019 at 13:18

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