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Chapter 37 Interference of Light

Waves

Interference and Diffraction – Important phenomena that distinguish waves from

particles

Diffraction – Bending of waves around corners

37.1 Conditions for Interference 1. The source must be coherent; that is, they must maintain a constant phase with

respect to each other.

2. The sources should be monochromatic; that is, they should be of a single

wavelength.

What is the coherence length of the light?

What is the coherent time of the light?

37.2 Young’s Double-Split Experiment

Waves with the same phase.

∆x

∆t

2

37.3 Light Waves in Interference

Phase Difference and Coherence

Waves: ( )δω +− tkxAcos

Two harmonic waves of the same frequency and wavelength, the resultant wave is a

harmonic wave: ( ) ( ) ( ) ( )2/sin2/cos2sinsin 111 δωδδωω +−=+−+− tkxAtkxAtkxA .

phase difference is zero --> in phase, interfere constructively

phase difference is 180o --> out of phase, interfere destructively

phase difference: δ , can be a path difference or a time delay

xkx ∆=∆=

λπδ 2 or t

T

t ∆=∆= ωπδ 2

Example: (a) What is the minimum path difference that will produce a phase

difference of 180o for light of wavelength 800 nm? (b) What phase difference will that

path difference produce in light of wavelength 700 nm?

(a) 8002

x∆=π

π --> 400=∆x nm

(b) 700

4002πδ =

Time Delay

Path Difference

3

Constructive: λθ md =sin , Destructive: λθ

+=21

sin md

The phase difference between lights from the two slits is: λ

θπδ sin2

d=

L

y=θtan --> d

mLLLy mmensityimum

λθθ ~sin~tanint_max =

Example: Two narrow slits separated by 1.5 mm are

illuminated by yellow light of wavelength 589 nm from a

sodium lamp. Find the spacing of the bright fringes

observed on a screen 3 m away.

d

mLLLym

λθθ == sin~tan --> 361 102.1

105.1589

3 −×=×

==d

Lyλ

m

Example: Measuring the wavelength of laser light

A laser is used to illuminate a double slit. The distance between the two slit is 0.03

mm. A viewing screen is separated from the double slits by 1.2 m. The second-order

brighter fringe m = 2 is 5.1 cm from the center line. (a) Determine the wavelength of

the laser light.

d

Lmybright

λ= & 2=m --> ( )

03.0102.1

2101.53 λ×=× --> 410375.6 −×=λ mm

37.4 Intensity Distribution of the

Double-Slit Interference Pattern ( )tkxA ωψ −= sin1 , ( )δωψ +−= tkxAsin2

( ) ( ) 22222210 sinsin AtkxAtkxAIII =+−+−=+= δωω

( ) ( ) ( ) ( )2/sin2/cos2sinsin' δωδδωωψ +−=+−+−= tkxAtkxAtkxA

+−

==2

sin2

cos4 2222 δωδψ tkxAI

( )L

ykdkdkdxk ==∆= θθδ tan~sin

4

-->

=

=L

dyA

L

kdyAI

λπ2222 cos2

21

2cos4 or

=2

cos2 20

δII

Lloyd’s mirror for producing a two-slit

interference pattern

Phasor for Wave Addition

tkx ωα −=

( ) ( )δαα ++=+= coscos 2121 AAEEE

( )δπ −−+= cos2 2122

21

2 AAAAA

( ) ( )( )δαα

δααδα++++=+

coscossinsin

'tan21

21

AA

AA

δδδ

cossin

'tan21

2

AA

A

+=

--> ( ) ( )'cos'cos δωδα +−=+= tkxAAE

Example: Use the phasor method to derive the superposition of two waves, ( )αsin0A

and ( )δα +sin0A , of the same amplitude.

2'

δδ =

==2

cos'cos2 00

δδ AAA

--> ( )

+

=+=2

sin2

cos2'sin 0

δαδδα AAE

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37.5 Change of Phase Due to Reflection If light traveling in one medium strikes the surface of a medium in which light travels

more slowly, there is a 180o phase change in the reflected light.

As light travel from one medium to another, its frequency does not change but its

wavelength does.

Relative Intensity of Reflected and Transmitted Light

)cos(),( txkAtxy ll ω−= , ll

l k

Tv

ωµ

== ,

)cos(),(' txkBtxy ll ω−−=

)cos(),( txkCtxy rr ω−= , rr

r k

Tv

ωµ

==

0=x , rll yyy =+ ' , ==> CBA =+

rll ydx

dy

dx

dy

dx

d =+ ' , ==> CkBAk rl =− )(

Ann

nnA

kk

kkB

rl

rl

rl

rl

+−=

+−= , A

nn

nA

kk

kC

rl

l

rl

l

+=

+= 22

rl nn > --> B > 0 (in phase) : rl nn < --> B < 0 (out of phase)

The reflected intensity can be 0

22

0 Inn

nn

A

BII

rl

rl

+−=

=

37.6 Interference in Thin Films

Coherent <-- The same source light

Light 1: 180o (π ) reflected wave

Light 2: 0o reflected wave

Constructive: πδ =

Destructive: πδ n2=

6

How do we estimate the phase difference between Light 1 and 2? λ

πλ

πδ tx 22~2

∆=

Light 1: 180o (π ) reflected wave

Light 2: 180o (π ) reflected wave

λπ

λπδ tx 2

2~2∆=

Constructive: πδ n2=

Destructive: πδ =

Newton’s rings: the thin film is air

Example: A wedge-shaped film of air is made by

placing a small slip of paper between the edges of two

flat pieces of glass. Light of wavelength 500 nm is

incident normally on the glass, and interference

fringes are observed by reflection. If the angle θ

made by the two plates is 4103 −× rad, how many

dark interference fringes per centimeter are observed?

x

t~~sin θθ ,

λt

n2= -->

( )12

105103222

7

4

=××=== −

−

λθ

λx

t

x

n cm-1

37.7 The Michelson Interferometer