0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user1936752 Feb 21 '18 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 '18 at 3:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.