# Homework Help: Nightmarish michelson interferometer question

1. May 31, 2006

### sachi

We have to "closely spaced" frequencies f1 and f2. "The moving mirror moves at a constant speed v. Explain how the Doppler effect on the light reflected from the moving mirror affects the recorded signal as a function of time. Estimate the value of v required to resolve the freq. difference of 500 MHz between two modes of a laswe within a time of 0.1s."

I'm very confused about this. Under normal circumstances I'm pretty sure there is no moving mirror! I can only assume that the "closely" spaced frequencies are so closely spaced that they can't be resolved by the instrument, therefore we have to artificially introduce a doppler freq. shift in one of the components.
I'm using the doppler shift formula from spec. rel. that shift in freq. = (v/c) * original frequency. Therefore surely v = 500MHz *c/f1
or we could have used f2, as they are almost equal. I can't see how the time 0.1s comes into it. Thanks very much

2. Jun 1, 2006

### Andrew Mason

Can you give us the whole problem? You are assuming we know what you are talking about here.

There is a doppler shift only if the mirror is moving relative to the source of the light. How do you get such motion in a Michelson interferometer? You need to clearly explain what it is you are talking about.

AM