Young's Double Slit with an Optical Flat over one slit

[truth is life]

Homework Statement

In a Young's double slit-type experiment using light of 600 nm with 0.5 mm between fringes, a thin plate of glass (d = 100 micrometers, n =1.5) is placed over one of the slits. What is the lateral fringe displacement on the screen?

2. The attempt at a solution

To be entirely honest, I really have no idea how to even start approaching this problem. I can see that obviously there will be *something* involving optical path length in there (the plate clearly increases it), but I have no real idea of what equations will be involved or how to go about doing anything.

 

[zzzoak]

Number of wavelength in glass of width L is
<br /> N=\frac{L}{\lambda (n)}<br />

where

<br /> \lambda (n)=\frac{\lambda}{n}<br />

where
<br /> \lambda \ is \ in \ vacuum.<br />

So we get
<br /> N=\frac{Ln}{\lambda }.<br />

One wavelength gives phase shift 2 \pi and N

<br /> \Phi=2 \pi N=2 \pi \frac{Ln}{\lambda }.<br />

At first a wave moves distance L in vacuum where n=1 and the phase shift is

<br /> \Phi_1=2 \pi \frac{L}{\lambda }.<br />

and in glass

<br /> \Phi_2=2 \pi \frac{Ln}{\lambda }.<br />

So the phase difference is

<br /> \Delta \Phi=\Phi_2-\Phi_1=2 \pi \frac{L(n-1)}{\lambda }.<br />

So a beam without a glass plate will have to go a longer distance to compensate this additional
phase shift.

 

[fluidistic]

Thanks zzzoak. So now we still have to calculate the optical path difference in the air between the fringes and the screen where the image is seen, right?
From what you got, we would have to add 2 \pi \frac{L(n-1)}{\lambda } to the optical path of waves leaving the glass, right?