Nightmarish michelson interferometer question

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SUMMARY

The discussion centers on the application of the Doppler effect in a Michelson interferometer with closely spaced frequencies f1 and f2. Participants explore how the moving mirror, which operates at a constant speed v, introduces a frequency shift that can resolve a 500 MHz difference within 0.1 seconds. The Doppler shift formula from special relativity is utilized, specifically the equation for frequency shift: Δf = (v/c) * f_original. Clarification is sought on the role of the moving mirror and its necessity in generating the Doppler effect in this context.

PREREQUISITES
  • Understanding of the Doppler effect in light waves
  • Familiarity with Michelson interferometer principles
  • Knowledge of special relativity and its frequency shift equations
  • Basic concepts of laser frequency modes
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  • Study the Doppler effect in detail, focusing on light and sound
  • Research the operational principles of Michelson interferometers
  • Explore the implications of special relativity on frequency measurements
  • Investigate methods for resolving closely spaced frequencies in optical systems
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Physicists, optical engineers, and students studying wave mechanics or interferometry who are interested in the practical applications of the Doppler effect in experimental setups.

sachi
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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
 
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sachi said:
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
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
 

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