Question

g One arm of a Michelson interferometer has a section of length 2.1 cm where the air can be evacuated. The dielectric constant of air is 1.00059. Find the number of fringes which will shift in the interference pattern of the interferometer when the air is evacuated for an interferometer illuminated with light of 663 nm wavelength.

Answer 1

The answer is "12388.17"

Step-by-step explanation:

`l = 2.1 cm = 2.1 * 10^(-2)\ m\\\\k = 1.00059\\\\\eta = √(k)= √(1.00059)\\\\\lambda = 663 nm = 663 * 10^(-9)\ m`

Users now know that perhaps the number of fingers the shift is provided when a path difference
`\Delta d` are inserts between both the two arms

`N = (\Delta d)/(\lambda)`

The optical pull-up in the arm is initially given by

`d = 2\eta l`

Its new length of the different sense as the reflection coefficient adjustments between
`\eta` (air) and 1 so if we evacuate air from of the arm (vacuum).

The new length of a path is therefore

`d'' = 2l`

Therefore, the different path

`\Delta d=d-d''=2l(\eta -1)`

So, The fringe shifts number are

`N= (2l(\eta -1))/(\lambda)`

`= (2 * 2.1 * 10^(-2) (√(1.00059)-1))/(663 * 10^(-9))\\\\=12388.17`