Question

Suppose a transparent vessel 25.0 cm long is placed in one arm of a Michelson interferometer. The vessel initially contains air at 0° C and 1.00 atm. With light of vacuum wavelength 630 nm, the mirrors are arranged so that a bright spot appears at the center of the screen. As air is slowly pumped out of the vessel, one of the mirrors is gradually moved to keep the center region of the screen bright. The distance the mirror moves is measured to determine the value of the index of refraction of air, n. Assume that, outside of the vessel, the light travels through vacuum, and that if none of the mirror are moved, the central region of the screen changes would from bright to dark and back to bright 233 times—that is, 233 bright fringes are counted (not including the initial bright fringe). Calculate the distance that the mirror would be moved as the container is emptied of air.

Answer 1

Removing the vessel with air of refractive index 'n' leads to a change in the path difference by the amount 2*(n-1)*d where 'd' is the length of the vessel. The factor 2 comes because the light traverses the vessel two times. If 'x' number of fringes disappear at the central position on removing the vessel, then

If we move one of the mirrors by the amount 'y', then the change in path difference due to that is 2y. The factor 2 again comes because light traverses a path twice. Hence if the central fringe does not change:

Now, x=233 , λ = 630nm

This gives y=73395 nm. This is the distance we will have to move the mirror when we empty the vessel.