# Double-Slit Arrangement With Plastic Sheet

1. Nov 29, 2012

### Ellipses

1. The problem statement, all variables and given/known data
Consider the double-slit arrangement shown in the figure below, where the separation d is 0.280mm and the distance L is 1.16m. A sheet of transparent plastic (n = 1.50) 0.0513mm thick (about the thickness of a piece of paper) is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y'. What is y'?

2. Relevant equations
PD / d = xL
n = c/v

3. The attempt at a solution
I'm sorry. but I have absolutely no idea how to approach this. I've just been looking through my books and trying to find equations to plug the given information into, but no dice. Any help would be greatly, greatly appreciated. Thank you so much!

2. Nov 29, 2012

### Fightfish

The presence of the plastic sheet results in a additional phase shift of the light passing through the upper slit relative to the light passing through the lower slit.
This means that their phase difference has now changed.

3. Nov 29, 2012

### Ellipses

How would I go about calculating that phase difference? Should I use the general wave equation?

4. Nov 29, 2012

### haruspex

The plastic adds to the effective path length (refractive index of plastic - index of air)*thickness.

5. Nov 29, 2012

### Fightfish

When the light waves travel through an optically denser medium, they are slowed down - and so they travel more "cycles" (ie oscillate more cycles) in passing through that medium as compared to travelling the same distance in air.
Say I have two light waves initially in phase. One travels a distance L through air. The other travels the same distance L, but through a medium of refractive index n > 1. The effective optical length travelled by the second wave is in fact nL, and so there is a path difference of (nL-L) between the two waves.