Answer to Question #182702 in Optics for Khushi

Question

Explain how Michelson interferometer is used to determine the refractive index of a

thin plate.

Answer 1

Principle: Monochromatic fringes are used and plates ranging in thickness from a few micrometres to several millimetres may be measured.

Consider the plate in one arm of the Michelson interferometer. Let the ray which is normal to the mirror M make an angle of incidence $\phi_i$ with the plate. The change in phase of the ray as it passes through the plate may be determined from considering below figure

The phase change in the ray in going from P to Q is

$\dfrac{2\pi n\overline {PQ}}{\lambda}=\dfrac{2\pi nd}{\lambda cos\phi_r}$

where,

$\lambda:$ is the wavelength of the monochromatic light

$n:$ is the refractive index

$d:$ is the thickness

$\phi_r:$ is the angle of refraction

The equivalent ray in the other arm of the interferometer will pass through a corresponding air thickness $\overline{PS}$ and suffer a phase change

$\dfrac{2\pi\overline{PS}}{\lambda}=\dfrac{2\pi d\space cos(\phi_i-\phi_r)}{\lambda\space cos\phi_r}$

Thus the phase difference between the two rays introduced by the plate is

$\Delta=\dfrac{2\pi nd}{\lambda\space cos\phi_r}-\dfrac{2\pi d\space cos(\phi_i-\phi_r)}{\lambda\space cos\phi_r}$

The factor 2 is required because the ray passes through the plate twice

When the plate is normal to the ray, $(\phi_i=\phi_r=0)$