How does the Michelson Interferometer Affect Light Intensity?

[Synetos]

Homework Statement

So I'm studying for my optics exam and some stuff that I find is quite odd and I can't find an explenation in the book.

So the question states that you have a michelson interferometer with two arms, that splits the light at 60% (so the first mirror reflects 40%)
The other two mirrors at the end of the arm (M1 and M2) reflect 100%
The intensity of the light that enters is I0 and the wavelength is 550nm

Homework Equations

Now I need to express the max and min Intensity that hits the detector.

The Attempt at a Solution

the answer states:
E1 = sqr(0.6)*sqr(0.4)E0
I = (2E1)²I0 = 0.96I0
But why do you take the roots of those numbers and why do you square it afterwards?

 

[NateD]

I think you take the root of the numbers because this is based on the intensity of the light. The intensity of the light that hits the first mirror is 0.6, so the root of 0.6 would be the intensity of the light that is reflected back. Similarly, the root of 0.4 would be the intensity of the light that is transmitted. Squaring this number gives you the total intensity of the light that hits the detector.