# Young's Double Slit Experiment problem, regarding change in medium

In the question we have a simple YDSE setup , there's just the addition of a glass slab of thickness t and ,refractive index u, in front of the top slit ( S1). Now we submerge the entire setup into water (given refractive index u1). We now have to find the locations of the points, which are the points of constructive interference( max intensity).

I considered 3 approaches, first-

I calculated the optical path difference in water = t(u/u1 -1)- yd/D. where y is the reference coordinate of the maxima( any general position) from the original central maxima, d is the slit separation , and D is the distance from screen. For constructive interference path difference = n*wavelength( in water). Now wavelength in water= "wavelength in air"/u1 Therefore we get our answer.

Approach two:

I found out the path difference in water and multiplied it by u1 , to get the path difference in air , and I got the same result when I equated this path difference to- n*wavelength(in air).

Approach three- I found the path difference in air - t(u-1)-yd/D and divided this by u1, to get path difference in water, and equated this to n* wavelength (water). This time around , I did not get the answer.

Can anyone explain the difference between the second and third approaches. And also please do tell me, if I have committed some error.

• ... and YDSE is ... what? – garyp Nov 26 '15 at 4:15
• @garyp Young's double slit experiment. – Garvit Sharma Nov 26 '15 at 5:07