# Double-slit expirement fundamentals (half-silvered mirror version)

In the double-slit experiment variation in which 2 half-silvered mirrors and 2 mirrors are used to illustrate the interference of a stream of photons or single photons at a given time step, how is it when the experiment is comprised of all 4 mirrors that the photon always knows to go to Camera 1/A in every illustration of the experiment I've ever heard of? Why doesn't it ALWAYS go to Camera 2/B instead with this setup? Seems so arbitrary and I have never come across the reason.

Diagram to refresh you: (Assume all distances are equal in diagram.)

                          [CAM 1/A]  (beam)
|
regular mirror             |
//---------------------/     [CAM 2/B]  (no beam)
|                     | half-silvered mirror #2
|                     |
|                     |
|                     |
[XXX>--/---------------------//  regular mirror
^half-silvered mirror #1


[XXX> is the photon emitter/laser. The slash to the right of the photon emitter represents one of the two half-silvered mirrors that I've designated as half-silvered mirror #1.

So... why is it when interference occurs at half-silvered mirror #2 that the photon decides it prefers the camera designated CAM 1/A when it could easily just go to the other CAM every single time it interferes?

• Welcome to Physics SE! Why do you think interference is the cause? Is there a light path to CAM 2/B using geometric optics? Apr 20, 2014 at 12:55

the light from the two different paths are in phase going to cam1 and out of phase going to cam2. To see this lets look at each path going to cam1 and each path going to cam 2.

The important thing to know is that both "half-silvered mirrors" are in fact interfaces between optical media of two different indices of refraction. The interface goes from bottom left to top right in both cases, and in both cases the top left medium is has a lower index of refraction and the bottom right medium has a higher index of refraction. (Actually, they could be flipped, but they must be consistent with each other).

Let's start with cam1. And for cam1 lets start with the upper path. The light first strikes the mirror and accumulates $\pi$ of phase because it is reflected while moving from a low index of refraction to a high index of refraction interface. Then the light travels the vertical distance, acquiring some phase. Then the light reflects off the mirror, acquiring another $\pi$ phase. Then the light reflects off the other half-silvered mirror, again acquiring $\pi$ phase, then the light goes to cam 1. The total phase accumulated is distance + $3\pi$.

Now lets consider the other path to cam1 (the lower path). Here there is no phase at the mirror since it passes through. It accumulates phase from the horizontal distance and a $\pi$ of phase when it reflects off the mirror. It also accumulates the vertical distance phase. Lastly, notice that it doesn't accumulate phase as it passes through the mirror. Thus the amount of phase accumulated on this path to cam1 is distance+$\pi$.

Since the accumulated phases of the two paths differ by $2 \pi$, they add constructively and you do see a signal at cam1.

Now let's think about cam2. We expect the two paths to travel the same distance as before, so let's not consider the phase contributions from distance but only reflection. Let's start with the top path. Here, it accumulates $\pi$ of phase from reflecting off the first mirror, just as before. Also, it acquires $\pi$ of phase when reflecting off the second mirror. However, it does not acquire any phase when passing through the final mirror. Thus it has a total of $2 \pi$ of phase

Now let's consider the lower path to cam2. It does not acquire any phase when it passes through the first half-silvered mirror. It gets $\pi$ when reflecting off the regular mirror. Lastly, it does not acquire any phase when reflecting off the final mirror since it is being reflected while moving from a high index of refraction material to a low index of refraction material. Thus it has a accumulated total phase of $\pi$. This is different from the phase of the other path by $\pi$ and so these paths add destructively.

This is why you see signal at only one detector even though the set-up seems somewhat symmetric.

This half-silver based setup is known as Mach–Zehnder interferometer. you can see this behaviour when sending one photon in to this interferometer. The photon travels both upper and lower paths. The lower path assigns minus sign (phase shifting by 180 degree) to the photon so it is canceled out and reaching always CAM 1/A.

I tried at home with normal laser light. the beams are reaching both CAM 1/A and CAM 2/B. But I could see interference pattern at both beams.

The answer is quite simple. the reason is that the first half-silvered mirror reflects the light down and therefore there is no light in the path to Cam2

• Please do read the question more carefully. Jul 11, 2015 at 10:32
• I see the point here, it seems the half silvered mirrors (or at least the half silvered mirror #1) should be rotated 90 degrees. Sep 24, 2016 at 18:04