0
$\begingroup$

In case of interference,we know,energy is neither destroyed,nor created; but only redistributed.But in the case of an extremely thin film,due to a reflection and hence a phase difference of pi, the film always appears dark due to destructive interference. So,where does the energy go?

$\endgroup$
  • $\begingroup$ A phase difference of $\pi$ is required for destructive interference, not $\frac{\pi}{2}$. $\endgroup$ – Chris Nov 23 '17 at 4:13
1
$\begingroup$

The logical answer is : into an increase in the motion of the atoms on which individual photons scattered off the thin film, i.e. heat.

This is a fascinating similar phenomenon with monochromatic laser light showing destructive interference. It is instructive to look, as it shows the quantum mechanical dependence of light.

$\endgroup$

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.