Galilean transformations - What am I messing up here?

This is a follow-up to a question I asked earlier. We have the following exercise:

We have two parallel mirrors, which are located at y=0 and y=l in the (x,y) plane. A photon is travelling between the mirrors, up and down along the y-axis. Consider an observer O at rest w.r.t. the mirrors.

1. What's the time (Δt) measure by O for the photon to make a full period.
Consider an observer O' which moves along the x-axis with a speed v, which is constant. Assume l' = l.

2. Using Galilean transformations from Newt. mech. , what's the time measured by O' for the photon to make a full period (draw a picture to illustrate the logic). Compare this with the time measured by O.

So for 1, the answer is $\Delta t= \dfrac{2l}{c}$. I tried to do 2 too, but I keep getting a somewhat incorrect answer. Here's what I do:

Draw an isosceles triangle $ABC$ with $|AB| = vt'$, and the height is l. I figured that $t' = \dfrac{2|AB|}{c'}$, where $c' = c-v$. So we try to solve the following equation: $(c't')^2 = 4l^2 + (vt')^2$, but if I solve for t', I get an almost correct answer, which of course is still incorrect. You get $t' = \dfrac{2l}{\sqrt{c^2-2cv}}$, and of course we want $t= \dfrac{2l}{c}$ because in newtonian mechanics t=t'. What have I done wrong?

If I'm not wrong, you've missed a factor 2 in t' (you gave the solution for just AC (or BC), and it should be both to consider the path that the photon would do for the O' observer. So it is, $t'=\frac{4l}{\sqrt{4c^2-v^2}}$, and when $v<<c$ (newtonian approach) then $t'=t=\frac{2l}{c}$.

If I'm not wrong, you've missed a factor 2 in t' (you gave the solution for just AC (or BC), and it should be both to consider the path that the photon would do for the O' observer. So it is, $t'=\frac{4l}{\sqrt{4c^2-v^2}}$, and when $v<<c$ (newtonian approach) then $t'=t=\frac{2l}{c}$.

If you draw a triangle with ABC where AB is vt', then wouldn't AC + BC be equal to simply c't'?

I'm not sure if I followed you in your definition $c'=c-v$ (remember Einstein's discovery about speed of light as the only constant element in his Theory of Relativity). Am I misunderstanding you?

I'm not sure if I followed you in your definition $c'=c-v$ (remember Einstein's discovery about speed of light as the only constant element in his Theory of Relativity). Am I misunderstanding you?
This question is a sort of intro question for a special relativity course, in the next question you have to assume c' = c. But in this question you have to assume that everything is Newtonian, i.e. you want to obtain that t=t'. But, as I say in the OP, I can't seem to figure out what I've done wrong.

Hope I'm getting it well (the axis and the mirror positions): I consider two parallel mirrors (y=0 and y=l) —parallel to the xz plane, although we don't need to consider z axis here. The observer O' travels perpendicular to these in x axis (AB in my triangle) —or equivalent, the mirrors move in x axis with constant speed v. In this scenario, O' looks how the photon doing a path which is not a straight line in y (case 1), but two diagonal lines (AC+CB). These three paths close the triangle of height l. Now, we know the two cathetus (AB/2 and the height, l) and the hypothenuse is half the distance that the photon travels for O' ($ct'/2$—which is what you write but with c' instead of c), and Pythagoras gives the answer for t' seen above.

Hope I'm getting it well (the axis and the mirror positions): I consider two parallel mirrors (y=0 and y=l) —parallel to the xz plane, although we don't need to consider z axis here. The observer O' travels perpendicular to these in x axis (AB in my triangle) —or equivalent, the mirrors move in x axis with constant speed v. In this scenario, O' looks how the photon doing a path which is not a straight line in y (case 1), but two diagonal lines (AC+CB). These three paths close the triangle of height l. Now, we know the two cathetus (AB/2 and the height, l) and the hypothenuse is half the distance that the photon travels for O' ($ct'/2$—which is what you write but with c' instead of c), and Pythagoras gives the answer for t' seen above.
Shouldn't it be c't'/2 = (c-v)t/2?

Keep in mind that the photon is moving in the y-axis, and the second observer is moving in the x-axis. The speed of the photon is always c, so the path would be AC+CB=t'c. If the photon moved in x axis, then we would face intuition against relativity, and the photon moves always at c, but this luxury costs something to the Universe: to deform space-time in order to make it happen.

vela
Staff Emeritus
Draw an isosceles triangle $ABC$ with $|AB| = vt'$, and the height is l. I figured that $t' = \dfrac{2|AB|}{c'}$, where $c' = c-v$.