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When a beam of light passes through a prism parallel to the base of the prism, the light slows down and hence its wavelength decreases. This, in turn, increases its momentum.

Does this change in momentum exert a force on the prism, thereby increasing its weight?

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    $\begingroup$ Why would exerting a force increase an object's mass? $\endgroup$ – NDewolf Jan 6 '20 at 10:50
  • $\begingroup$ @NDewolf Sorry, I meant weight. $\endgroup$ – Physics Jan 6 '20 at 11:10
  • $\begingroup$ Why do you think the light's momentum increases? That the momentum of a single photon of the beam has increased does not mean the total momentum of the beam increases! $\endgroup$ – ACuriousMind Jan 6 '20 at 11:59
  • $\begingroup$ See physics.stackexchange.com/q/255340/123208 $\endgroup$ – PM 2Ring Jan 6 '20 at 12:02
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We could ask by how much the normal contact force changes if the light travels through the prism. Suppose for simplicity the light enters one side of the prism and is refracted to the horizontal (parallel to the base), before leaving the prism such that the setup is symmetrical.

If the light travels toward the prism at angle $\theta$ below the horizontal, and leaves the prism at $\theta$ below the horizontal, then the change in momentum of a single photon is $\Delta p = \frac{2E}{c}\sin{\theta}$. If $n$ photons are "deflected" per second, then our force equals $F = \frac{2nE}{c}\sin{\theta}$. In the case that the prism rests on a surface, this force would act upward and the result would be a reduction in the necessary normal reaction force!

To use this relation we'd also need to figure out what "$n$" is. If the power of the beam is $P$, then $n=\frac{P}{E}$. So we could actually simplify our answer to $F = \frac{2P}{c}\sin{\theta}$.

Top pan balances measure the weight of an object via the normal contact force. If this decreases, this reading will decrease (though please do note that the weight itself is not changing!). Perhaps this is what you meant?

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  • $\begingroup$ Thank you for responding to my question. However, according to your analysis, if the angle of incidence is zero, there would be no change in momentum, but this is not true. According to de Broglie, the initial momentum would be h/wavelength, which changes as the light enters into the prism. This change in momentum could then be multiplied by the speed of light to give the effective mass change. $\endgroup$ – Physics Jan 6 '20 at 12:00
  • $\begingroup$ @Physics An equal number of photons would also leave the block in the same interval (having their momenta increased by the same factor), and the net impulse on the beam would still equal zero in the horizontal direction. Please do forgive me if I am misunderstanding your question! $\endgroup$ – 13509 Jan 6 '20 at 12:05
  • $\begingroup$ But surely the force decreasing the beam's momentum as it leaves the block also acts downwards by the symmetry of the situation you described? I thought the momentum increases as the beam enters the block because its wavelength decreases. $\endgroup$ – Physics Jan 6 '20 at 12:15
  • $\begingroup$ @Physics We can think of the prism as a sort of "black box". Sure, when the photon enters the prism it receives an "impulse" with a horizontal and downward component, and its wavelength decreases accordingly. When it leaves, it receives another "impulse" with the same downward component yet an opposite horizontal component. The overall effect is that the photon's momentum has only changed in the vertical direction! $\endgroup$ – 13509 Jan 6 '20 at 12:17
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    $\begingroup$ There is also the light that gets reflected (at incidence and internally) that contributes to radiation pressure. $\endgroup$ – user137289 Jan 6 '20 at 13:24

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