Amplitudes in a Michelson interferometer

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
Decimal
Messages
73
Reaction score
7
Hello,

I am having a hard time understanding a result relating to a michelson interferometer. I always assumed that when the beam hits the wave splitter both resulting waves will have half the amplitude of the original wave. However using this assumption does not give the correct irradiance for fringes on a michelson interferometer. $$ I = 4 I_0 * cos^2(\frac {2{\pi}d} {\lambda} + \frac {\pi} {2}) $$ Here ##d## is the difference in length between the two arms of the interferometer. I can only arrive at this expression by assuming that the amplitude of both beams after the beam amplitude is still the same as the source amplitude. Is this true, and if so, why is this? Wouldn't you be creating energy in this way?

Thanks!
 
Physics news on Phys.org