Wave Interference and Diffraction
Part 1: Introduction, Double Slt
Paul Avery
University of Florida
http://www.phys.ufl.edu/~avery/
avery@phys.ufl.edu
PHY 2049
Physics 2 with Calculus
PHY 2049: Chapter 35
1
Need to Understand Light as Wave!
Î(You
already have read this material)
ÎIndex
of refraction
Speed
of EM wave in medium:
Wavelength of light:
ÎPropagation
cn = c / n
λn = λ / n
of light: Huygens principle (36-2)
Explains
reflection and refraction
Explains interference (from superposition)
Explains diffraction (spreading of light around barrier)
PHY 2049: Chapter 35
2
Interference as a Wave Phenomenon
ÎInterference
of light waves
Caused
by superposition of waves
Intensity can increase or decrease!
Contrast with particle model of light
ÎEffects
and applications
Double
slit
Single slit
Diffraction gratings
Anti-reflective coatings on lenses
Highly reflective coatings for mirrors
Iridescent coatings on insects
Colors on thin bubbles
Interferometry with multiple telescopes
PHY 2049: Chapter 35
3
Interference from Wave Superposition
Basic rule: Add displacement at every point
Wave 2
Wave 1
Sum
PHY 2049: Chapter 35
4
Constructive Interference
ÎSame
wavelength, phase difference = 0°
ÎAmplitude larger: Higher intensity
Sum
E ( x ) = sin(kx) + 0.5sin(kx) = 1.5sin(kx)
PHY 2049: Chapter 35
5
Destructive Interference
wavelength, phase difference = 180° (1/2 λ)
ÎAmplitude smaller: Lower intensity
ÎSame
Sum
E ( x ) = sin(kx) + 0.5sin( kx + π ) = 0.5sin( kx)
PHY 2049: Chapter 35
6
Examples
ÎTwo
waves, same λ, with amplitudes 2A and A
Initial
ÎNo
intensities 4I and I, respectively (I ∝ A2)
interference
Combined
ÎMaximum
intensity: Inew = 4I + I = 5I
constructive interference (φ = 0)
New
amplitude:
New intensity:
ÎMaximum
New
Anew = 2A + A = 3A
Inew = 9I
destructive interference (φ = π)
amplitude:
New intensity:
Anew = 2A – A = A
Inew = I
PHY 2049: Chapter 35
7
General Treatment of Interference
ÎMost
interference is partial
Amplitudes
for 2 waves are generally different
Phase difference : 0 < φ < 180°
E ( x, t ) = E1 cos(kx − ω t ) + E2 cos( kx − ω t + φ )
ÎAdditional
considerations
Wavelengths
can be different
Multiple waves may interfere (e.g., diffraction grating)
But easy to accommodate: just sum over all waves
E ( x, t ) = ∑ i Ei cos(ki x − ωi t + φi )
PHY 2049: Chapter 35
8
Interference and Path Length
Two sources, spaced 3 wavelengths apart, emit waves with
the same wavelength and phase. In how many places on
the circle will the net intensity be a relative maximum?
Answer = 12
Can you see why?
Hint: Start at far right and move
counterclockwise towards top,
noting path length changes.
Key idea: Path difference leads
to phase difference
PHY 2049: Chapter 35
9
Interference and Path Length
ÎTwo
sources, separated by 4λ, emit waves at same
wavelength and phase. Find relative minima on +x axis.
Solution:
path difference must be a half-multiple of λ
(
)
ΔL = x + ( 4λ ) − x = n + 12 λ
2
2
x=
4λ
x
(
16 − n +
)
1 2
2
2n + 1
λ
4 values
n=0
x = 15.8λ
ΔL = λ/2
n=1
x = 4.58λ
ΔL = 3λ/2
n=2
x = 1.95λ
ΔL = 5λ/2
n=3
x = 0.54λ
ΔL = 7λ/2
PHY 2049: Chapter 35
10
Double Slit Interference
ÎIncident
light
Light
waves strike 2 narrow slits close together
Light goes through both slits, diffracts in all directions
ÎInterference
At
certain angles, waves constructively interfere
At other angles, waves destructively interfere
⇒ brighter
⇒ darker
Interference
Slits
Diffracted light
Screen
Light waves
PHY 2049: Chapter 35
11
Basic Requirements for Two Slit Setup
ÎLight
beam strikes normal to slits
ÎLight
beam illuminates both slits equally
ÎLight
beam is in phase at both slits: coherent
Young
used small slit in front of 2 slits to get coherence
Modern versions use laser for coherence (much brighter)
PHY 2049: Chapter 35
12
Two Slit Analysis
d
θ
Path difference = d sinθ
d sin θ = mλ
(
Maximum
)
d sin θ = m + 12 λ
Minimum
PHY 2049: Chapter 35
13
Example 1: d = 5λ
Max sin θ = m ( λ / d ) = 0.2m
Min
(
)
(
sin θ = m + 12 ( λ / d ) = 0.2 m + 12
)
m
sinθmax θmax
sinθmin
θmin
0
0
0
±0.1
±5.7
±1
±0.2
±11.5
±0.3
±17.5
±2
±0.4
±23.6
±0.5
±30
±3
±0.6
±36.9
±0.7
±44.4
±4
±0.8
±53.1
±0.9
±64.2
±5
±1.0
±90
±1.1
--
PHY 2049: Chapter 35
14
Intensity vs Angle for d = 5λ
PHY 2049: Chapter 35
15
Example 2: d = 2.0 μm , λ = 550 nm
ÎHow
many bright fringes? Where are they?
sin θ = m ( λ / d ) = 0.275m
Îm
can equal 0, ±1, ±2, ±3 ⇒ 7 maxima
m=0
sinθ = 0
θ=0
m = ±1
sinθ = 0.275
θ = 16.0°
m = ±2
sinθ = 0.55
θ = 33.4°
m = ±3
sinθ = 0.825
θ = 55.6°
PHY 2049: Chapter 35
16
Intensity vs θ for d = 2.0 μm , λ = 550 nm
PHY 2049: Chapter 35
17
Calculating Double Slit Intensity
ÎAssumptions
Each
slit acts as a source of waves
Waves radiate equally in all directions
d
θ
Path difference = d sinθ
PHY 2049: Chapter 35
18
Double Slit Intensity (2)
ÎAdd
amplitudes, include phase difference
Assume
equal size slit widths
Phase difference from path difference: 2π × # wavelengths
We ignore x dependence here (analysis does not depend on it)
E ( t ) = E cos(ωt ) + E cos(ωt + φ )
φ=
2π d sin θ
λ
PHY 2049: Chapter 35
19
Double Slit Intensity (3)
ÎIntensity
is time average of amplitude squared
Consider
single wave of amplitude E = E0 cosωt
2 2
Intensity from time average of E2: I = K E
cos 2 ω t = 12 K 2 E 2
<…> is time average over a period
I tot = K
2
( E cos(ω t ) + E cos(ω t + φ ) )
2
= K 2 E 2 cos 2 ω t + K 2 E 2 cos 2 (ω t + φ )
+2 K 2 E 2 cos ω t cos (ω t + φ )
We work this out on next page
PHY 2049: Chapter 35
20
Double Slit Intensity (4)
1
cos ω t =
2
1
2
cos (ω t + φ ) =
2
1
cos ω t cos (ω t + φ ) = cos φ
2
2
φ=
I tot = K 2 E 2 (1 + cos φ ) = 2 I 0 (1 + cos φ )
=
4 I 0 cos 2 12 φ
φ=
2 ⎛ π d sin θ
= 4 I 0 cos ⎜
⎝
λ
⎞
⎟
⎠
PHY 2049: Chapter 35
2π d sin θ
λ
2π d sin θ
λ
21
Double Slit Intensity (5)
ÎSo
the intensity is I = 4 I 0 cos 2 (π d sin θ / λ )
ÎMaxima
occur when argument inside cos() is nπ
d sin θ = nλ
ÎMinima
occur when argument inside cos() is (n+1/2)π
(
)
d sin θ = n + 12 λ
PHY 2049: Chapter 35
22
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