# Thin Film Phase Difference

## Homework Statement

Light of wavelength 526 nm is incident normally on a film of water 1.0 µm thick. The index of refraction of water is 1.33.
(c) What is the phase difference between the wave reflected from the top of the air-water interface and the one reflected from the bottom of the water-air interface in the region where the two reflected waves superpose?

## Homework Equations

phase difference=(delta(r)/lambda)(2*pi)

## The Attempt at a Solution

So I easily found the wavelength in the film for part a, which is 395.5 nm, and I easily found the number of wavelengths that are contained within the distance of the film, 2t, which was 5.06. The problem I am having is finding the phase difference. I tried doing (delta(r)/lambda)(2*pi)=(2t/lambda')(2*/pi)=(2.0µm/395.5 nm)(2*pi). I reduce to find the remaining fraction of wavelength, but no matter what combination I use, I can't get the right answer. And I can't derive a method off of the practice problems. Sample answer from a practice problem: Wavelength: 416; Film: 1.0 µm thick; Index of refraction: 1.33. Answer is 5.62 rad when I would get 2.45 rad, or some multiple of that.

## Answers and Replies

Doc Al
Mentor
Did you consider the phase shift on reflection, where applicable?

Thank you, I completely forgot to factor that in. That was what I was missing and the answer was 3.5 rad.