Physics: The element sodium can emit light at two wavelengths, λ1 = 588.9950 nm and λ2 = 589.5924 nm. Light in sodium is being used in a Michelson interferometer. Through what distance must mirror M 2 be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?

The element sodium can emit light at two wavelengths,

richard376

30.01.2022

The distance is $d = 0.00029065 \ m$

Explanation:

From the question we are told that

The first wavelength is $\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m$

The second wavelength is $\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m$

The difference in the fringe pattern is n = 1.0

Generally the equation defining the effect of the movement of the mirror M 2 in a Michelson interferometer is mathematically represented as

$2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n$

Here d is the mirror M 2 must be moved

substituting values

$2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0$

$d = 0.00029065 \ m$

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