I have drawn diagram so not to confuse.
So far, I've heard that in Mach-Zehnder interferometer, two output should have one constructive interference, and one destructive interference for other.
But, what I had calculated above for the phase shift, doesn't fit with the argument above.
What am I missing? Where is the constructive interference?
When you use a real beamsplitter, it has a finite thickness. When such a splitter is placed at the second position in a particular way, the green light beam going to screen 1 will be suffering Zero phase shift because it suffers a reflection coming from the denser medium and going back into the denser medium. The red beam going to screen 2 will suffer a phase shift of $\Pi$ radians because it is getting reflected coming from lighter medium and going back into lighter medium. This will be reversed if you change the position of the splitter but essentially, only one screen will have a constructive interference.
Use actual beamsplitter, from wikipedia
https://en.wikipedia.org/wiki/Beam_splitter
This is the solution!!!!
https://www.cs.princeton.edu/courses/archive/fall06/cos576/papers/zetie_et_al_mach_zehnder00.pdf
Forget about most of the instrument, just think about the final beam recombiner. You have two mutually coherent beams of some arbitrary phase $e^{\pm i\,\phi}$ at the two inputs to the final beamspliter (without loss of generalness, subtract out the common mode phase, so we can represent the two beams as $a_{\pm}\,e^{\pm i\,\phi}$ where $a_\pm$ are the real-valued amplitudes).
Now if the beamsplitter reflects negligible light back whence it came (which is a pretty good approximation with appropriately coated optics), and it is lossless then the sum of the powers of the two outputs has to be constant and the two variations of output power from each output as functions of the input phase difference $\phi$ must be two antiphase functions of $\phi$ so that their sum is the total input power, simply by conservation of energy.
To explore this idea in more detail, the two outputs $(y_1,\,y_2)^T$ written as a column vector are related, in the linear system case, to the two inputs $(a_+\,e^{i\,\phi},\,a_-\,e^{-i\,\phi})^T$ by a homogeneous, linear, unitary relationship:
$$\left(\begin{array}{c}y_1\\y_2\end{array}\right) = e^{i\,\chi}\left(\begin{array}{cc}\alpha&\beta\\-\beta^*&\alpha^*\end{array}\right)\left(\begin{array}{c}a_+\,e^{i\,\phi}\\a_-\,e^{-i\,\phi}\end{array}\right)$$
where $\chi\in\mathbb{R}$ and $\alpha,\,\beta$ are any two complex number fulfilling $|\alpha|^2+|\beta|^2=1$. Any $2\times 2$ unitary relationship can be written in the above form. Try working out the powers $|y_1|^2,\,|y_2|^2$ as functions of all the parameters $\chi,\,\alpha,\,\beta$: set it up in Mathematica or Maple or the like. You'll find that no matter what you put for these parameters, the powers $|y_1|^2,\,|y_2|^2$ will be antiphase, sinusoidal functions of the relative input phase $\phi$.
