The beamsplitter is at a 45 degree angle. (slope=+1) One beam comes from the top of the paper. One beam from the right. Two receivers: One on the left and the second at the bottom. The beams are mutually coherent plane waves. One finer point you might have missed is mentioned in the details of the previous post that one of the reflections gets a ## \pi ## phase shift, which is essentially a minus sign. This can be seen at normal incidence where ## \rho=(n_1-n_2)/(n_1+n_2) ## (n's are indices of refraction). ## \rho ## from one direction is positive and from the other it is negative. ## \rho=E_r/E_i ## is the fresnel reflection coefficient for the electric field amplitude. The negative ## \rho ## (the minus sign ) is the same as a ## \pi ## phase shift. The same thing happens at 45 degree angle of incidence (the minus sign for ## n_2>n_1 ##), that happens at normal incidence,although the expressions for ## \rho ## become more complex (at 45 degrees and at any other angles of incidence). The materials can be chosen to make the energy reflection coefficient ## R=1/2 ## at 45 degrees, so that ## \rho=+1/\sqrt{2} ## or ## \rho=-1/\sqrt{2} ##. It turns out, the composite transmission coefficient ##\tau=E_t/E_i=1/\sqrt{2} ## for both beams (when ## R=1/2 ## ) so that you can readily compute the emerging E fields. The two sources can be assumed to have some arbitrary phase difference ## \phi ## , although the calculation is simplest if they are either assumed to be in phase with each other or ## \pi ## out of phase. In one case all the energy goes to one receiver, and in the other case it all goes to the other receiver. You can assume the top surface has the AR coating (beamsplitter is finite width), so that the reflection of the beam incident from the right will have an extra ## \pi ## phase change. When the two sources start out in phase, this makes them destructively interfere on the path to the bottom receiver, so that all the energy goes to the left where the two sources are in phase. One thing I should add is to compute the intensity (power) ## I=n E^2 ## (i.e. it is proportional to ## n E^2 ##. (In optics, the calculations for intensity are often done in this fashion without the additional constants that are in the Poynting vector for power. And n=1 for air.) You can write the amplitudes for the different wave components and compute the energy at the two receivers. e.g. for the bottom receiver ## E_{Bottom}=(1/\sqrt{2}-1/\sqrt{2})E_o=0 ##(one component is transmitted from the top and the other reflected from the right). For the left receiver ## E_{Left}=(1/\sqrt{2}+1/\sqrt{2})E_0=\sqrt{2}E_0 ## (one component is transmitted from the right and the other reflected from the top) and we see our final intensity at the left receiver ## I_{Left}=(\sqrt{2}E_0)^2=2 E_0^2 ## which is the sum of the intensities of the two initial beams ## I_0=E_o^2 ## for each. (Note for the above energy reflection coefficient ## R=I_r/I_i=E_r^2/E_i^2=\rho^2 ##. Notice also when there are two coherent sources, the energy reflection coefficients ## R ## can not be used to compute the resulting energy distribution (this system is not linear in its energy properties, but is completely linear in regard to the E fields), but the ## \rho ## 's computed from the ## R ## are completely valid for doing the computation.)
@Chris Frisella It's quite a lot of detail, but hopefully you can at least follow some of it.
