Asample that is 2 cm thick is inserted into one arm of a michelson interferometer that was adjusted to give a maximum reading. the maximum is reestablished when the mirror is pushed 1.9 cm closer to the beamsplitter. what is the index of refraction of the sample?

μ = 1.95

Explanation:

Given that

thickness ,t= 2 cm

Displacement ,x= 1.9 cm

Now from Michelson interferometer ,the relationship between  index of refraction , thickness and the displacement given as

Now by putting the values

μ-1 = 0.95

μ = 1+ 0.95

μ = 1.95

So the index of refraction μ will be 1.95.

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