Michelson interferometer - destructive interference due to reflection phase change

I saw several analysis of the Michelson interferometer, most of them ignore the reflections back to the light source that as far as I understand accounts for the "missing" energy in the case of destructive interference.

I want to verify that:

In the case of: 1) Equal optical length paths (including the compensator plate) 2) Half silvered mirror surface is on the side facing the light source 3) Collimated, monochromatic beam

The detector gets no signal because:

1) The beam that reflects from the half silvered mirror gain 2 reflections phase changes of 180 degrees each.
2) The beam that goes through the half silvered mirror gain only one reflection phase change, the second reflection from the half silvered mirror is from low refractive index so no 180 degrees phase change. 3) In that case all the energy goes back to the source.

When there is an optical path difference "p" I got the following formulas for the powers:

Power_that_goes_to_the_detector = P_0/2 * (1-cos(k*p)) Power_that_come_back_to_the_source = P_0/2 * (1+cos(k*p))

k = 2*pi/lambda, P_0 is the source power.