Homework - Young's Double Slit Experiment 
I have attached the question as a picture. I simply don't understand where to begin this question. It is given that $l \gg d$. Why shouldn't the phase difference at $O$ be zero? What exactly is different about this set-up as compared to the 'normal' set-up of Young's Double Slit experiment?
 A: The drawing is a little bit confusing - but the key is that the source $S$ is a relatively long way from the slits (compared to their spacing), and at a height of $\frac{d}{2}$ (from "directly behind $S_1$"). This means you can draw a diagram to calculate the relative path distance between S and each of the two slits. The difference in path length results in a phase difference before the light arrives at the slits - and that phase difference is maintained as you travel from slits to screen.
One way to look at it is to say that the "zero" of the diffraction pattern will be directly in line with the line connecting $S$ to the middle of the slits. If there was just a single large hole between the slits, you would have no problem seeing that this is the case:

When you add a refractive index, the wavelength of the light is shorter (by a factor $\frac{1}{n}$). It's interesting to note that you can find the "zero" of the diffraction pattern (regardless of whether the refractive medium is in front of the slits or behind) by repeating the above argument: pretend there are no slits but just a single hole, and imagine where the light would go. Snell's law will quickly give you the answer.

I will leave the rest up to you - our homework policy asks us not to give complete answers to this kind of question, but just to explain some of the principles.
A: *

*A phase difference may also be zero.

*Whenever the difference of the distances the beams travel is not a multiple of the wavelength, there is a non-zero phase difference - as simple as that. So if the source is not equidistant from both slits, and the point of measurement is equidistant from the slits, mean you already have a pretty good chance of non-zero phase difference.

A: The text states that the source $S$ is placed just behind the slit $S_1$, so it looks to me that the sources $S_1$ and $S_2$ already are out of phase by a factor $\exp(i k d)$. Thus the phase difference in $O$ is $\Phi_O = k d$.
This is different from the usual experiment where $S$ is placed at equal distance from $S_1$ and $S_2$.
