chapter 37 interference of light wavesocw.nctu.edu.tw/course/ph012/chapter_37.pdf · chapter 37...
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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
5
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