# Energy conservation when light reflects from mirror [duplicate]

Suppose we have a perfectly reflecting mirror which reflects 100% of the light incident on it. When a light beam falls on the mirror normal to the surface it will reflect in opposite direction from where it came. Hence its momentum is changed and also the momentum of the mirror. The light reflected off with the same speed and so its energy will be equal to energy of the light before incidence. But the mirror now, has also gained kinetic energy (because it gained velocity) . So the final energy comes out to be greater than initial energy. From where does this extra energy came into the mirror

• – John Rennie Nov 16 '19 at 11:44
• – John Rennie Nov 16 '19 at 11:44
• @John Rennie - the third question suggested by you has an answer which also answers my question – Param_1729 Nov 16 '19 at 12:27

The frequency of the reflected light will be a tiny bit smaller than the incident light. So the energy (and also the momentum) gained by the recoiling mirror is taken from the light.

This can be understood by considering the reflection as a collision between the photons and the mirror. During reflection the photon's energy $$E=h\nu$$ gets a little bit smaller by reducing the frequency $$\nu$$. Also the photon's momentum $$p=h/\lambda$$ gets a little bit smaller by increasing the wavelength $$\lambda$$.