Optics – Michelson Interferometer

A gas-filled cell of length 5 cm is inserted in one arm of a Michelson interferometer as shown in the figure. The interferometer is illuminated by light of wavelength 500 nanometers. As the gas is evacuated from the cell, 40 fringes cross a point in the field of view. The refractive index of this gas is most nearly?

A. 1.02
B. 1.002
C. 1.0002
D. 1.00002
E. 0.98

(GR9277 #96)

Solution:

Michelson interferometer, 2ΔL = Nλ
N = ^2ΔL/_λ

N_vac = ^2ΔL/_λvac
N_gas = ^2ΔL/_λgas

λ_gas = ^λ_vac/_n
N_gas = ^2ΔL/_λgas  = ^2ΔLn/_λvac

ΔN = N_gas − N_vac
= ^2ΔLn/_λvac − ^2ΔLn/_λvac
= ^2ΔL/_λvac (n − 1)

Given:
ΔL = 5 cm = 5 × 10⁻² m
ΔN = 40 fringes
λ_vac = 500 nm  = 5 × 10⁻⁷ m

40 = [2(5 × 10⁻²)/(5 × 10⁻⁷)](n − 1)
40 = (2 × 10⁵)(n − 1)
40/(2 × 10⁵) = n − 1
2 × 10⁻⁴ = n − 1
n = 0.0002 + 1 = 1.0002

Answer: C

Optics - Michelson Interferometer

A Michelson interferometer is configured as a wave-meter, as shown in the figure above, so that a ratio of fringe counts may be used to compare the wavelength of two lasers with high precision. When the mirror in the right arm of the interferometer is translated through a distance d, 100,000 interference fringes pass across the detector for green light and 85,865 fringes pass across the detector for red (λ = 632.82 nanometers) light. The wavelength of the green laser light is

A. 500.33 nm

B. 543.37 nm

C. 590.19 nm

D. 736.99 nm

E. 858.65 nm

(GR0177 #100)

Solution

λ_red = 632.82 nm

λ_red  λ_green → (D) and (E) are FALSE.

Constructive path difference = Nλ

N_gλ_g = N_rλ_r

λ_g = N_rλ_r / N_g

= (85,865 × 632.82) / 100,000

≈ (8.6 × 10⁴ × 6.3 × 10²) / 10⁵

= 541.8

Answer: B