I have a rather basic question. Light that undergoes reflection also picks up a phase shift of $\pi$. Does it mean that light that is incident normally onto a perfect mirror simply undergoes perfect destructive interference with its own reflected wave?

This cannot be due to conservation of energy but I'm not sure how it can be resolved.


1 Answer 1


It only undergoes total destructive interference right at the point of reflection. This just means the amplitude is zero right at that point, not anywhere else.

The total phase shift is $\pi$ at that point, but at other points there is a path length difference that contributes an extra phase shift $\Delta\Phi=-{2\pi\Delta x\over\lambda}$. This gives constructive and destructive interference in different locations.

Also, as an aside, the phase shift is only $\pi$ when the wave is reflects off a medium of higher refractive index than the medium it is traveling through. Otherwise it's zero.

  • $\begingroup$ Two questions? Is this path difference also for normal incidence? And in the case of a mirror, does the phase shift exist since no light is transmitted and the index of refraction is not clear. $\endgroup$ Feb 21, 2018 at 3:54
  • 1
    $\begingroup$ @user1936752 Yes, the path difference is for normal incidence as well. It's twice the distance from the mirror in that case. Ordinarily there is a phase shift with just about any material you use for a mirror, since the index of refraction of air is so low. (Though note that pretty much all light you see on a day-to-day basis is incoherent, so you shouldn't expect to see interference patterns.) $\endgroup$
    – Chris
    Feb 21, 2018 at 3:59

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