3
$\begingroup$

In my textbook it is given that in thin film interference a path difference of wavelength/2 is added up to the reflected light from the upper surface. But, how does that happen? No physics textbook of mine gave an answer on this. So, please help.

My thoughts: I know that reflection is actually electrons absorbing and re-emitting light. But, I'm not able to figure out how an extra path is added. The extra path added to the light reflecting from the bottom surface of the thin film is evidently due to the extra distance it has to travel. But, what about the upper one?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
3
$\begingroup$

This is due to the fact that for a reflection from optically denser medium, there is a phase reversal of 180 degree. Since 360 degree corresponds to wavelength $\lambda$ , 180 degree correspond to $\frac{\lambda}{2}$ . For an excellent explanation of why there is a 180 degree phase difference, look here .

Hope this will answer your question.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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