# Phase difference

1. Mar 23, 2016

### nmsurobert

1. The problem statement, all variables and given/known data
A plane-wave is incident on an air-glass interface at 30° as shown below. Calculate the phase difference between A and B. Assume λo=1μm.

attatched is an image that looks like this.
________________________

--------------------------- A

--------------------------- B

________________________

the angle is incident from above at 30°. the distance between A and B is 1μm.
2. Relevant equations

3. The attempt at a solution
i really just need a bit of guidance. I feel as if i should have a formula for this but i don't. the book and my class notes aren't being much help either.

2. Mar 23, 2016

### drvrm

what is the wavelength given -in air or glass-
if it is in air -one can calculate the wavelength in glass after refraction at 30 degree incidence and move ahead calculating the phase changes as it travels in glass.
where lies the actual hurdle?

3. Mar 23, 2016

### nmsurobert

How do I calculate the phase change? Like I said, I feel like I'm missing a formula.

4. Mar 23, 2016

### drvrm

Phase and path traversed is related by simple concept-
When a wave travels full wavelength the it returns to identical phase- so a translation of lambda leads to phase change of 2.Pi
therefore phase difference = (2.Pi/ Wavelength). path length

5. Mar 23, 2016

### nmsurobert

If that's the case then wouldn't the phase difference be 2pi since the wave length and path length are the same?

6. Mar 23, 2016

### drvrm

well one must consider the refraction and due to refraction the wave is inclined at certain angle -moreover the waves traverse the glass medium with changed wavelength which is related to refractive index of the medium -pick up a text book on refraction.