Optics - interferometers and interference I was watching a video of Sabine Hossenfelder, in which interferometers were a topic of discussion. Excuse me if the answer is obvious because I am not a physics pro by any means, just an interested reader.
She mentioned following construction:

Basically, a laser is getting shot at a beam splitter, which transmits and reflects 1/2 of the beam, respectively. Then the beams are getting reflected by a mirror on each side until they meet at the next beam splitter. Then both beams will recombine to the original beam which is measured by the detector D2.
However, my question is: Why does the first beam splitter reflect and transmit 1/2 of the beam respectively, while the second one does not? The beam on the bottom is only transmitted, the one on the right only reflected so that the whole original beam intensity is detected at D2.
She mentioned that is has something to do that the path of the two beams is the same length, but what is the underlying physical law here that it works like this and how is it calculated?
It is also mentioned that the second beam splitter indeed reflects and transmits if the path length for both is not the same. Why?

 A: The beam splitter both reflects and transmits every time, but you should keep in mind that it does this for each beam hitting it. So if two beams hit it (e.g. from above and from the left in your first diagram) then each is split and there are a total of four beams going out! But two of these travel to the right and two travel downwards, so it looks like two beams. And furthermore, when a pair of beams exactly overlap, sometimes they can cancel each other out. It is because they are waves. The wave motion can be pictured as a bit like water waves, only in a light wave it is the electric and magnetic field that is oscillating. If the electric field of one contributing beam is increasing while that of another is decreasing then we say the waves are "out of phase" and in this case the net result can be no change in the electric field at all. This results in no light, and it is what is happening in the downward path from the lower beam splitter in your first diagram.
In the second diagram the waves are no longer exactly out of phase as they emerge from the second beam splitter, so the light is not cancelled.
A: The beamsplitter introduces a phase shift for the reflected beam relative to the transmitted beam. This can, under ideal circumstances, give destructive interference at the one output port with the second beamsplitter. However, in a practical free-space setup, it extremely difficult to get such ideal destructive interference.
When the path are not the same the interference goes away, because the laser beam has a finite coherence length. This is the relative distance within which the beam is still in phase with itself. It has to do with the finite bandwidth of the laser.
We can model the process of a beamsplitter with a unitary matrix
$$ \left(\begin{array}{c} a \\ b \end{array}\right)_{out} 
= \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1  \end{array}\right)_{out} 
\left(\begin{array}{c} a \\ b \end{array}\right)_{in} , $$
where $a$ and $b$ represent the beams at the two input ports. To represent the beamsplitter in a symmetric way, I use $i=\exp(i\pi/2)$, which shows that the phase change for the reflection is $\pi/2$.
For the first beamsplitter, we have $b=0$. So, if the paths are equal, input to the second beamsplitter is
$$ \frac{1}{\sqrt{2}} \left(\begin{array}{c} a \\ ia \end{array}\right)_{in} . $$
Then the second beamsplitter produces
$$ \frac{1}{2}\left(\begin{array}{cc} 1 & i \\ i & 1  \end{array}\right)_{out} 
\left(\begin{array}{c} a \\ ia \end{array}\right)_{in} =
\left(\begin{array}{c} 0 \\ ia \end{array}\right)_{out} . $$
So you see that all the light comes out of just one port.
