Calculating the time difference in the Michelson-Morley Experiment I'm having trouble understanding the explanation of the Michelson Morley experiment given in Physics for Scientists and Engineers by Raymond Serway.
Serway computes the time difference between the two beams of light arriving at the telescope as follows,

$$ \Delta t = \frac{2L}{c} \left[ \left(1-\frac{v^2}{c^2}\right)^{-1}-\left(1-\frac{v^2}{c^2}\right)^{-1/2}\right]$$

And using the binomial expansion $ (1-x)^n\approx 1-nx$ for $x<<1$
He gets

$$\Delta t \approx \frac{Lv^2}{c^3}$$

I understand up to this part just fine.
The next part I don't understand:

A shift in the interference pattern should be detected when the interferometer is rotated through 90° in a horizontal plane so that the two beams exchange
roles. This rotation results in a time difference twice that given by $\Delta t \approx \frac{Lv^2}{c^3}$.
Therefore, the path difference that corresponds to this time difference is $\Delta d = c(2 \Delta t) = \frac{2Lv^2}{c^2}$

Why is the time difference twice that given by $\Delta t \approx \frac{Lv^2}{c^3}$ when the interferometer is rotated through 90 degrees?
Wouldn't the time difference be the same? By rotating the apparatus by 90 degrees, the light beam that was parallel to the ether wind is now perpendicular to it, and vice versa. I do not understand why you need to double the time difference.
This doubling seems crucial because the formula for calculating the fringe shift is
Shift = $\frac{2Lv^2}{\lambda c^2}$
and there is that factor of 2 on the numerator. But I do not understand why.
 A: They've implicitly chosen a sign convention that makes $Δt>0$ in the original configuration. When you rotate the apparatus and the arms change places, the same convention gives $Δt=-\frac{Lv^2}{c^3}$. The difference between the $Δt$s is $\frac{2Lv^2}{c^3}$.
The original derivation assumed that the aether wind is parallel to one of the arms. Realistically, you wouldn't know that, so you would need to measure the fringe shift at many angles to find the amplitude of the sinusoid. (And at different times of day, if you can only rotate it in a plane.)
A: The reason for rotation of arms was that if even there is time difference for light in two arms, there might be no fringe shift seen because the arm lengths was not equal that could compensate for time difference. By rotating interferometer's arm, the advantage of unequal arm doubles the disadvantage for time difference.
The calculation for time difference was little erroneous, and I suppose they did it intentionally to propose a theory. Speed of light to aether is, $v_{la}=c$ and speed of aether to earth is, $v_{ae}=v$. Thus speed of light to earth is, $v_{le}=v_{la}+v_{ae}=c+v$.
The time taken for round trip of light in perpendicular direction to relative motion of aether for average length of interferometer's arm $l$ is given by,
$t_\bot=\dfrac{2l}{c/\cos \theta}=\dfrac{2l}{c\sqrt{1+\frac{v^2}{c^2}}}=\dfrac{2l}{c}\left(1-\frac{1}{2}\frac{v^2}{c^2}+\ldots\right)\tag{1}$
where relative speed of light to earth, $v_{le}$ due to sum of speeds in perpendicular direction is given by $c/\cos\theta$ and $\cos \theta=\frac{1}{\sqrt{1+v^2/c^2}}$
Now time taken for light for round trip in parallel direction to motion of aether is consists of two parts, one for along the motion of aether, $v$ and other is against the aether, $-v$ for average length $l$.
$\quad \quad t_{\parallel1}=\dfrac{l}{c-v}\quad$ & $\quad t_{\parallel2}=\dfrac{l}{c+v}$
where $t_{\parallel1}$ is time taken against the motion of aether and $t_{\parallel2}$ is time taken along the motion of aether. Thus, total time taken for round trip is sum of these two. Therefore,
$t_\parallel=\dfrac{l}{c-v}+\dfrac{l}{c+v}=\dfrac{2l}{c\left(1-\frac{v^2}{c^2}\right)}=\dfrac{2l}{c}\left(1+\frac{v^2}{c^2}+\ldots\right)\tag{2}$
Now time difference, $\Delta t$ between time taken by light to travel a round trip in parallel and perpendicular arms of interferometer is given by (1) & (2),
$\Delta t=t_\parallel-t_\bot=\dfrac{2l}{c}\left(\dfrac{3}{2}\dfrac{v^2}{c^2}\right)=\dfrac{3l}{c}\dfrac{v^2}{c^2}\tag{3}$
From here, one can calculate path difference, $d=c\Delta t$ and $n\lambda=d$, where $n$ is number of fringe or shift in fringe. So,
$n=\dfrac{3l}{\lambda}\dfrac{v^2}{c^2}\tag{4}$
From (4), it seems that whoever came to experimenter that fringe shift must this, they said that this was similar to their calculation which comes after rotation of interferometer otherwise, $n=\dfrac{l}{\lambda}\dfrac{v^2}{c^2}$.
