Interferometer consists of two Y-branches and two waveguides of different or equivalent lengths for imbalanced and balanced types respectively. 
Lets talk about Y splitter first. \(I_i\)\(\) is input intensity and \(E_i\) in input electric field. Light splits equally in two branches, so at the outputs we see \(I_1=I_2=I_i/2\), and electric field, proportional to the square root of intencity is \(E_1=E_2=Ei/\sqrt2\). When light goes backwards, we call this Y combiner. But when Y-branch is combiner we can not think of it as if it is a three-port device, it's a four-port device. If the light is passing through one branch, input splits equally between fundamental mode and all other modes plus radiation. So we can find electric field by following this equasions: \(E_{o1}=E_1/\sqrt2\) , \(E_{o2}=E_2/\sqrt2\). When we have two inputs, the output field is a vector summation of the input fields, divided by square root of 2, \(E_0=(E_1+E_2)/\sqrt{2}\).
Now lets assume, that waveguide lengths is \(L_1\) and \(L_2\)\(\Delta L=\mid L_1+L_2\mid\), propagation constant is \(\beta=2\pi n/\lambda\), where \(n\) is a refraction index and \(\lambda\) is wavelength. Including propagation loss \(\alpha\) in our theory, electrical field can be described by \(E_0=\frac{1}{2}E_i(e^{-j\beta_1L_1-\frac{\alpha1}{2}L_1}+e^{-j\beta_2L_2-\frac{\alpha2}{2}L_2})\) and as we remember intencity proportional to square of electric field, we have \(I_o=\frac{1}{4}I_i\mid (e^{-j\beta_1L_1-\frac{\alpha1}{2}L_1}+e^{-j\beta_2L_2-\frac{\alpha2}{2}L_2})\mid^2\) this equation represents the general form of an unbalanced MZI transfer function. In our case \(\beta_1=\beta_2=\beta\), and with zero loss we can simplify Intensity to \(\frac{I_o}{I_i}=\frac{1}{2}[1+cos\beta\Delta L]\) where \(\delta = \beta\Delta L\) is a phase shift. 
Look at the example of imbalanced MZI transfer function: