Why shift in fringe is expected in Michelson Morley Experiment? When I (beginner) learnt Michelson Morley experiment, it was expected that there should be a shift in the fringe pattern when the setup was rotated by 90 deg if the presence of ether is true. But why should there be a shift in fringe after the setup was rotated, assuming the presence of ether?
 A: One may calculate the time that light (electromagnetic waves) need to go through the arms of the interferometer. If the speed of light relatively to the aether is $c$, then $c$ is also the approximate speed through the arm that is approximately orthogonal to the velocity $\vec v$ of the Earth (or interferometer) relatively to the aether. Well, the speed is a bit smaller, $c/\gamma$, due to the transverse Doppler shift.
On the other hand, in the arm that is parallel to $\vec v$, the speed of the aether, the speeds with which the waves propagate are $c+v$ and $c-v$, respectively, depending on the direction (back and forth). We need to calculate the inverse speeds because we want to know the time it takes to get through the arm (back and forth). The inverse speeds are $1/(c\pm v)$ and their average is
$$ \frac 12 \left[ \frac{1}{c+v} + \frac{1}{c-v} \right] = \frac{c}{c^2-v^2}\approx \frac 1c (1+v^2/c^2) $$
The harmonic average speed is therefore the inverse of this or
$$ c (1-v^2/c^2) $$
up to subleading corrections. That's smaller than the transverse average speed that was
$$ c (1-v^2/2c^2) $$
So the difference between the "effective average speeds" through the arms is 
$$\Delta c = \frac{v^2}{2c} $$
where $v$ is the speed of the Earth relatively to the aether. If you divide the overall distance traveled by a photon (through one of the arms) by the speed, you get the time, and 
$$ \Delta t = \frac{s}{c} - \frac{s}{c-\Delta c} \approx \frac{s\cdot \Delta c}{c^2}\approx \frac{s v^2}{2 c^3} $$
So the two photons need a slightly different time to get through the arms (the total traveled distance is $s$ for a fixed arm). This $\Delta t$ will be translated into a phase shift $2\pi \cdot \Delta t  / t_{\rm period}$ is the angle that will decide about the phase shift that may be seen through the shifted interference pattern. ($2\pi$ corresponds to the first nonzero shift that is undetectable.)
Even though the prediction for the phase shift only appears in the second-order, it was already observable by the 19th century interferometers, and the result of the experiment was that this phase shift doesn't exist. So the hypothesis that the Earth is moving relatively to the "environment whose oscillations are perceived as light" was experimentally ruled out.
A: Consider paths A and B.
Orbital speed of Earth is v (30km/sec approx.), D is the length of both paths, P is the optical path length.
Supposing that the ether wind is along A, thus
$P_a - P_b = \frac{D v^2}{c^2}$.
Got some fringe pattern!
After rotating the apparatus by  $90^{\circ}$, B comes along the ether wind. So,
$P_b - P_a = \frac{D v^2}{c^2}$.
Got some fringe pattern!
Just draw the diagram, rotate it as it is(if you didn't get it) and think about this symmetrical result.
This means that a displacement in the fringe pattern equal to $\frac{2 D v^2}{\lambda c^2}$ should occur if Ether exists. The displacement as the original paper said was much less than the theoretical one which indicated some other complications like the movement of the solar system altogether which could have lowered v (the measure of relative movement of earth through ether) to display such small displacement in the fringes. So, the experiment was repeated at intervals of 3 months to avoid all ambiguity. The original paper is available online if you want to delve deeper.
