number of fringe shifts

Question

One arm of a Michelson interferometer utilizing light with λ=500 nm contains a cell 7.50 cm long between plane parallel windows. How many fringe shifts would be observed if all the air were evacuated from the cell? The index of refraction of the air was initially 1.00029.

 

Answer is supposed to be 87, please show me how to calculate

Answer 1

Here’s the quick derivation.

Idea: Evacuating the cell changes the optical path length (OPL) in that arm. In a Michelson, light passes through the cell twice, so the OPL change is doubled. Each time the OPL changes by one wavelength, the interference pattern shifts by one fringe.

Given:

• Wavelength: λ = 500 nm 5.00 x 10 ^{– 7} m

• Cell length: L = 7.50 cm = 7.50 x 10 ^{– 2} m

• Initial air index: n_air = 1.00029

• Final index (vacuum): n_vac = 1.00000

• Index change: Δn = n_air – n_vac = 0.00029

Change in optical path in that arm:

Δ (OPL) = 2L Δn = 2 x (7.50 x 10 ^–2) x (2.9 x 10 ^{– 4}) = 0.15 x 2.9 x 10 ^{– 4} = 4

Number of fringe shifts:

M = Δ (OPL) = 4.35 x 10 ^{– 5} = 87

λ 5.00 x 10 ^{– 7}

Answer: 87 fringes