0
$\begingroup$

so I'm working on the Michelson interferometer for a presentation, and I'm having trouble understanding how does the beamsplitter splits the incoming beam.

Let's consider the incoming bean as a monochromatic source represented by its electric field:

$\overrightarrow{E} = \overrightarrow{E_0}\cos(\varphi)$ where $\varphi$ depends on time $t$ and position $M$.

The light intensity is equal to: $ I_0 = 2\space\langle\overrightarrow{E}.\overrightarrow{E}\rangle_T = 2\space\langle {E_0}^2\cos^2(\varphi)\rangle_T = {E_0}^2 \space (V^2.m^{-2})$.

From what I've read on the first beam division, the beamsplitter splits it into two other beams of intensity $I'= \dfrac{I_0}{2}$.
If we call $E'$ to be the electric field amplitude of both the divided beams, we also have $I' = \dfrac{I_0}{2} = \dfrac{{E_0}^2}{2} = {E'}^2$ which means that $E' = \dfrac{E_0}{\sqrt2}$.

But this is were I don't really get it. If these two splitted beams ever reconstruct (not necessarily when the light comes back to beamsplitter a second time as the Michelson interferometer works), the resulting amplitude would be $\dfrac{2E_0}{\sqrt2} > E_0$.

How can the rebuilt beam has a bigger amplitude than the source?

I understand that when the light comes back to the beamsplitter from the fixed and moving mirror, it is refracted/reflected a second time, but I don't understand how it is possible to divide the beam intensity by two the first time.

Thanks!

$\endgroup$
-1
$\begingroup$

Sorry for the short answer, as I am on my phone right now. Probably other people can provide a more complete answer to your question later.

You ask what will happen if the two beams were ever to be "recombined", and you find a contradiction because you end up with more power than you injected. I would say that your calculation is correct, but unfortunately for you the important question here is precisely how you would "recombine" these two beams. I might be wrong, but I believe there are actually no way to recombine "perfectly" (in the exact same spatial mode and same polarization). Why is it important to recombine them "perfectly"? Because for instance if you add two light fields with orthogonal polarization, you will find that the amplitudes do not add up, but the intensities do. In your case, adding up two beams with intensity $I_0/2$ and orthogonal polarisation would just give you back $I_0$.

For the Michelson now, what happens? Let's say for simplification that the optical path on either arm is the same. It means that light travels the same distance on both paths. What you have to remember is that in order to recombine the two beams, they have to go through the same beamsplitter a second time. If it is balanced, it means that you would start from one beam with amplitude $E_0$, then split it into two beams with amplitude $E_0/ \sqrt{2}$ as you noted (one on each arm), but these two beams would again go through the beamsplitter and form $4$ beams with amplitude $E_0/2$, two of which would go back into the input port, and the other two going into the forth arm. If you look at one of these "output" arms now, and you suppose you add the two resulting beams constructively (meaning they do not pick up any phase difference going in two different arms), you would find that you need to add their amplitudes $E_0/2 + E_0/2 = E_0$, corresponding to an intensity of $I_0$.

Now everything is fine until you look at the other arm. Indeed if you apply the same reasoning on the second "output" arm, you would find another $I_0$, so the summed up intensity would be equal to $2 I_0$. However, there is a catch, as in this configuration, you also pick up a phase shift on some of the beams due to the beamsplitter itself. I will pass on the details but if one "output" arm is interfering fully constructively, the other one will necessary be interfering destructively, meaning that if you can sum up the amplitudes to $E_0$ on one output arm, the amplitudes have to sum to zero on the other one (the two electric fields having opposite amplitude).

Sorry again for not giving all the details. If you want to convince yourself on an example, like the Michelson, you have to really pay attention to the accumulated phase on each arm and ESPECIALLY the extra phase given by the beamsplitter.

In short, you cannot just "recombine" magically the two beams after the beamsplitter, as they are now separated spatially. In whatever situation you imagine, you will either lose some power trying to recombine them (for instance if you plan to use polarizing optics), end up in different polarizations (thus not being able to add the amplitudes in your calculation), or have an imperfect spatial overlap, resulting in additional phase shifts...

$\endgroup$
7
  • $\begingroup$ By the way, see this very nice answer which gives more details about the conservation of energy in a Michelson : physics.stackexchange.com/a/435894/235356 $\endgroup$ Dec 30 '19 at 9:50
  • $\begingroup$ See also this answer for more general considerations about energy conservation: physics.stackexchange.com/a/196919/235356 $\endgroup$ Dec 30 '19 at 9:57
  • $\begingroup$ Thanks a lot for this clear not so short anwser. Indeed, at ZPD, if we look at the phase shift created by the mirrors, the two output beams hitting the light sensor are in phase even if they undergo a phase shift of $270°$. But the two beams returning to the source are $180°$ out of phase. So the resulting electric field will be equal to 0, hence the intensity equals 0 too. If I understand right, we have to be careful to calculate the resulting electric field before applying the intensity formula. $\endgroup$
    – Vic
    Dec 30 '19 at 12:45
  • $\begingroup$ You have to be careful indeed. You can show under certain conditions (I do not remember exactly which ones, but I think it is true for a thin, non-absorbtive medium, with no magnetic field applied so time-reversal symmetry is true) that when reflecting on a beamsplitter, you will acquire an additional 180° phase when reflecting from one side and no additional phase when reflecting on the other side. It means that whatever actual path difference you acquire by propagating along each arm, you will need to add a 180° phase shift on one of the two beams for one of the output arms. $\endgroup$ Dec 30 '19 at 13:03
  • $\begingroup$ If you acquire a $\phi$ phase difference between the two arms, and you account for another 180° phase shift on one of the beams going into the 4th arm for instance, you will find on the input arm $E_0/2 \cos (\omega t) + E_0/2 \cos (\omega t + \phi)$ for the field and $E_0/2 \cos (\omega t) + E_0/2 \cos (\omega t + \phi + \pi)$ on the 4th arm. Using the intensity formula for both arms you find $I_0/2 (1 + \cos (\phi))$ on the input arm and $I_0/2 (1 + \cos (\phi + \pi)) = I_0/2 (1 - \cos (\phi))$ on the 4th arms. $\endgroup$ Dec 30 '19 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.