For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).
fringe = (delta t) / (λ/c)
We can find (delta t) with the equation:
delta t = [v^2(L1+L2)]/c^3
Derivation of this formula can be found in your physics text book. From here we find (delta t):
600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13
2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes
This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.
I believe that B is the answer.
