Question

A sample 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?

Answer 1

Answer:

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

$t=\dfrac{x}{\mu -1}$

Now by putting the values

$t=\dfrac{x}{\mu -1}$

$2=\dfrac{1.9}{\mu -1}$

μ-1 = 0.95

μ = 1+ 0.95

μ = 1.95

So the index of refraction μ will be 1.95.

Answer 2

Answer:

1.95

Explanation:

r = Thickness of sample = 2 cm

x = Displacement = 1.9 cm

$\mu$ = Refractive index of sample

The Michelson interferometer is an interferometer that uses arms which are split into two light sources.

Constructive interference is given by

$l=m\lambda$

Destructive interference is given by

$l=(m+0.5)\lamda$

where,

m = 1, 2, 3.....

$\lambda$ = Wavelength

In a Michelson Interferometer experiment, thickness is given by

$t=\frac{x}{\mu-1}\\\Rightarrow \mu-1=\frac{x}{t}\\\Rightarrow \mu=\frac{x}{t}+1\\\Rightarrow \mu=\frac{1.9}{2}+1\\\Rightarrow \mu=1.95$

The refractive index of the sample is 1.95