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If you had a light in a room which was entirely sealed with mirrors, then switched the light off, would the room stay lit?
Would the light just keep reflecting off the mirrors?
Would it slow down? Perhaps to the point where it is no longer visible?
Asked By: Mark H. from Melbourne, Australia
The light source produces a certain number of photons, and if the mirrors in this room are perfect, there is no way for them to be absorbed. In that sense, the room would stay lit.
Image Credit: Beth Rankin via flickr
Of course, you would have to be very clever in how you checked in on your room, because any normal way of looking at the photons (i.e. with your eye) will absorb some of the photons. Truly perfect mirrors not only will not absorb photons, they will scatter them elastically, i.e. with no loss of energy.
In that case, the photons will retain the same wavelength, and if they were in the visible spectrum initially, they will remain there. However, if we imagine mirrors that scatter the photons but at the same time absorb some of the photons' energy (these would be inelastic collisions), then the photons' wavelength becomes longer with each scattering. Eventually, the photons are degraded from the visible spectrum to the infrared spectrum, so, as you suggest, they are no longer visible.
But unlike what you suggest, the photons do not "slow down". Light always travels at the speed of light in a vacuum. Normal particles (like, say, molecules) slow down when they lose energy. Photons do not slow down, but their wavelength changes.Answered by:
Alan Chodos, PHD
Associate Executive Officer (retired)
American Physical Society