Optics - Interference

In a double-slit interference experiment, d is the distance between the centers of the slits and w is the width of each slit, as shown in the figure above. For incident plane waves, an interference maximum on a distant screen will be “missing” when

A. d = √2 w
B. d = √3 w
C. 2d = w
D. 2d = 3w
E. 3d = 2w

(GR9277 #20)

Solution:

For double-slit interference, d is always bigger than w
→ (C) and (E) are FALSE

Constructive or destructive patterns of interference (single or double slits) only deals with integer and half integer factors.
→ (A) and (B) are FALSE

Answer: D


Calculation:

“missing” = destructive pattern (minimum intensity).
For double slit: d sin θ = (m + ¹/₂) λ; m = 0, 1, 2, 3, ...
For single slit: w sin θ = mλ; m = 1, 2, 3, ...

^d/_w =  (m + ¹/₂)/m
Take m = 1,
^d/_w = ³/₂
2d = 3w

Optics - Interference

Two coherent sources of visible monochromatic light form an interference pattern on a screen. If the relative phase of the source is varied from 0 to 2π at a frequency of 500 hertz, which of the following best describes the effect, if any, on the interference pattern?

1. It is unaffected because the frequency of the phase change is very small compared to the frequency of visible light.
2. It is unaffected because the frequency of the phase change is an integral multiple of π.
3. It is destroyed except when the phase difference is 0 to π.
4. It is destroyed for all phase differences because the monochromaticity of the sources is destroyed.
5. It is not destroyed but simply shifts positions at a rate too rapid to be detected by the eye.

(GR8677 #13)

Solution:

Interference pattern is observed only when the sources are coherent (both sources have identical λ, f, and phase relationship).

The problem states that the light sources are coherent, thus the interference patterns are observed, not destroyed → (C) and (D) are FALSE.

Phase change will affect the constructive and destructive patterns of interference → (A) and (B) are FALSE.

Note: 500 Hz or 500/second is too rapid for human eye which only can perceive 60 Hz to 80 Hz flickering light.

Answer: E

Optics - Thin Film

It is necessary to coat a glass lens with a non-reflecting layer. If the wavelength of the light in the coating is λ, the best choice is a layer of material having an index of refraction between those of glass and air and a thickness of

A. λ/4
B. λ/2
C. λ/√2
D. λ
E. 1.5λ

(GR8677 #73)

Solution:

Index of refraction: 

Air to Layer,
phase difference: Δ_a = λ/2

Layer to Glass,
phase difference: Δ_b = 2t + λ/2

The relative shift: Δ = Δ_b − Δ_a =  2t + λ/2 − λ/2 = 2t

Non-reflecting → destructive interference
Δ = (m +½)λ

m = 0 (the thinnest the better)
2t = (0 +½)λ
t = λ/4

Answer: A


Notes: Click HERE for more info on Thin Film.

Optics - Interference

Light from a laser falls on a pair of very narrow slits separated by 0.5 micrometer, and bright fringes separated by 1.0 millimeter are observed on a distant screen. If the frequency of the laser light is doubled, what will be the separation of bright fringes?

A. 0.25 mm
B. 0.5 mm
C. 1.0 mm
D. 2.0 mm
E. 2.5 mm

(GR0177 #70)

Solution:

Double-slit interference:  

with


 

Answer: B

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