I'd like to add to Peter Shor's comments. An ideal partly silvered mirror comes very close to conserving energy. This actually implies quite a lot about the phase shift imparted by both mirrors, particularly if the mirrors are symmetric in their action. The total power of the two outputs must sum to a constant and so, as the relative phase of the interferometer's arms changes, the two output powers both vary sinusoidally with the relative phase and these two sinusoids are exactly out of phase. Sometimes in practice one finds that they are not exactly out of phase and this means that energy is "leaking" somewhere: something is absorbing or (if we think of an optical fibre interferometer as below) the optical couplers can be wasting some light by coupling it out of the fibres to the radiation field.
However, it might help to derive the defining equations for this system in the case of symmetric mirrors. A good way to understand your problem is to imagine one optical mode. This kind of working is realized in the single mode optical fibre Mach-Zehnder interferometer. The single-modedness means we don't have the complicated beam structure as shown in Anupam's answer. This structure can certainly be present in multi-moded beans in interferometers, so for now simpler things to one optical mode, as happens in a single mode fibre Mach-Zehnder interferometer (the mirror's are replaced by $2\times2$ optical fibre couplers in that case). If the mirror is working in its linear regime we can then characterise the incident and reflected fields on both sides of such a mirror by a $2\times2$ scattering matrix:
$$\left(\begin{array}{c}b_1\\b_2\end{array}\right) = \left(\begin{array}{cc}s_{11}&s_{12}\\s_{21}&s_{22}\end{array}\right) \left(\begin{array}{c}a_1\\a_2\end{array}\right)$$
or, as simple matrices and column vectors:
$$b = S a$$
where we define the waves by single, scalar complex numbers (recall we assume everything is working one-modedly) as in the following diagram:
Energy is conserved, thus $b^\dagger b = a^\dagger a$ or, otherwise written, $a^\dagger (S^\dagger S - I) a = 0;\;\forall a\in \mathbb{C}^2$. By choosing arbitrary vectors in $\mathbb{C}^2$, one can show that this statement is the same as $S^\dagger S = I = S S^\dagger$ i.e. the scattering matrix is unitary.
Now let's assume that a mirror is symmetric in its action, i.e. its action is the same if the beams number one and two were to swap roles (or if it is flipped over), i.e:
$$\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\left(\begin{array}{cc}s_{11}&s_{12}\\s_{21}&s_{22}\end{array}\right)\left(\begin{array}{cc}0&1\\1&0\end{array}\right) = \left(\begin{array}{cc}s_{11}&s_{12}\\s_{21}&s_{22}\end{array}\right) = \left(\begin{array}{cc}s_{22}&s_{21}\\s_{12}&s_{11}\end{array}\right)$$
so that the most general form the scattering matrix can have is:
$$\left(\begin{array}{cc}\alpha e^{i\gamma}&\beta e^{i\delta}\\\beta e^{i\delta}&\alpha e^{i\gamma}\end{array}\right);\quad \alpha,\beta,\gamma,\delta\in\mathbb{R}$$
with the unitaryhood condition expressed by $\alpha^2+\beta^2 = 1$ and $\cos(\gamma-\delta)=0$. This means that the most general form the scattering matrix can have is:
$$S = e^{i\,\theta}\,\left(\begin{array}{cc}\cos\frac{\phi}{2}&i\,\sin\frac{\phi}{2}\\i\,\sin\frac{\phi}{2}&\cos\frac{\phi}{2}\end{array}\right);\quad \phi,\theta\in\mathbb{R}$$
which, if the mirrors are exactly half silvered ones (so that they split powers equally so that $\cos(\phi/2)^2 = \sin(\phi/2)^2 = 1/2$:
$$S = \frac{e^{i\,\theta}}{\sqrt{2}}\,\left(\begin{array}{cc}1&\pm i\\\pm i&1\end{array}\right)$$
The phase relationships between the matrix elements are what enforce the conservation of energy.
The action of the interferometer's two arms is described by the simple phase delays:
$$D = e^{i\xi}\left(\begin{array}{cc}e^{i\,\zeta/2}&0\\0&e^{-i\,\zeta/2}\end{array}\right)$$
where we can, without loss of generality, we can suck the phase $\xi$ common to both arms as a scalar forefactor. Therefore, the interferometer's two output amplitudes are:
$$S D S \left(\begin{array}{c}1\\0\end{array}\right) = e^{i(\xi+\theta)}\left(\begin{array}{c}\sin\frac{\zeta}{2}\\i\,\cos\frac{\zeta}{2}\end{array}\right) $$
Here we define the system "input" to be the column vector $\left(\begin{array}{c}1\\0\end{array}\right)$; note that the system input is such a vector because we could also drive the interferometer system with a beam at right angles to the input beam you have drawn from your source at the left of your diagram. In this case, one of the column vector's entries is nought because this input is not driven in your diagram.
We see that the output square magnitudes $\sin( \zeta/2)^2$ and $\cos( \zeta/2)^2$clearly vary sinusoidally exactly out of phase so that the total power sums up to one unit.
With general, unitary scattering matrices $S$ standing for general asymmetric mirrors, the equations are more complicated, but the results are the same. Unitaryhood enforces phase conditions at the mirrors and always leads to sinusoidally varying out-of-phase outputs. In the more general case, the "visibility" of the fringes is less as the sinsusoids in general have offsets that mean that neither output, notwithstanding the sinusoidal variation with phase, can ever be "nulled" i.e. can never output exactly zero power. Only the special, symmetric case considered above leads to full fringe visibility.
I think the opposite will happen here. The energy will be absorbed by whatever optical diode (e.g. Faraday rotator) is fitted to the laser that prevents reflection back into the laser (which causes it to misbehave)
– lionelbrits Nov 30 '13 at 01:50My guess is that you think this because you're just looking at the path lengths.
– MauranKilom Nov 30 '13 at 14:32