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There are two perfect reflect mirrors facing opposite to each other... I place a lamp which is not lit in between the mirrors... I switch it on and switch it off and remove the lamp from between the two mirrors (assume this is instantaneous)... Now(assuming particle nature) the photons are emitted in all direction... The photons collide elastically with the mirror... As the collision is elastic the energy will be conserved and as the photon is reflected... Mirror will gain momentum too and hence kinetic energy.. But the total energy has to be constant so can we say that speed of light decreased then speed of light ? (If you can describe the situation considering wave nature then it will be helpful too)

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    $\begingroup$ This excellent answer here covers this physics.stackexchange.com/q/13937 . In a nut shell, the mass of the mirror is too large with respect to the energy carried by the photon and the effect of momentum and energy transfer to the mirrors will be very very small, not measureable $\endgroup$ – anna v Feb 1 '16 at 19:07
  • $\begingroup$ you could get a famous course and learn a lot about how physicists play with light between mirrors and quantum mechanics : on youtube search for Haroche ( in english ) $\endgroup$ – user46925 Feb 1 '16 at 19:08
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    $\begingroup$ If the mirrors are free to move, they will move and, as others have said, the light will be redshifted. The answer is the same in the photon as well as the wave picture. In the wave picture one has to use Maxwell's equations at the boundary and one can recover the usual formulas for light pressure and Doppler shift. No mystery. $\endgroup$ – CuriousOne Feb 1 '16 at 19:52
  • $\begingroup$ Do you all mean even a 100% perfect reflector can not reflect light with 100% energy. $\endgroup$ – Anubhav Goel Feb 2 '16 at 6:46
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What we call reflection is in reality a more complicated process than bouncing a ball to a wall. For the part of the electromagnetic radiation that we call visual light and for low densities of this light the surface electrons are responsible for the absorption and re-emission of this photons.

So yes, mirror will gain momentum and the photons will lose momentum. The photons get simply redshifted.

See this answer too, please.

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  • $\begingroup$ Can this even be explained by wave nature? $\endgroup$ – brainst Feb 1 '16 at 19:04
  • $\begingroup$ @HoglerFiedler The photon is red shifted... If I use a violet light then after sometime I'll see a red light?.... Will I even see it? Will it still be illuminated after removal of the lamp? $\endgroup$ – brainst Feb 1 '16 at 19:11
  • $\begingroup$ Not a question to me. I know that every photon of the involved EM radiation hits the surface and gets re-emitted. Since no one is able to visualize nor to measure the wave properties of a light beam I prefer to talk about EM radiation. The only EM waves I know are radio waves, there a wave generator modulate the EM radiation of an antenna rod. $\endgroup$ – HolgerFiedler Feb 1 '16 at 19:13
  • $\begingroup$ No,no. Redshift means that the outging light is shifted in the same (nearly the same) amount to a less frequency. $\endgroup$ – HolgerFiedler Feb 1 '16 at 19:15
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    $\begingroup$ @Anubhav Goel: Exactly. No energy conversion without loss. That is what thermodynamics say; without giving for this an explanation. Write a good paper about this and have your "track ... to have your name in history" :-) see your profile. $\endgroup$ – HolgerFiedler Feb 2 '16 at 6:48

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