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 36
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 36
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 36
3
Interference from Wave Superposition
Basic rule: Add displacement at every point
Wave 2
Wave 1
Sum
PHY 2049: Chapter 36
4
Constructive Interference
ÎSame
wavelength, phase difference = 0°
ÎAmplitude larger: Higher intensity
Sum
E ( x ) = cos(kx) + 0.5cos(kx) = 1.5cos(kx)
PHY 2049: Chapter 36
5
Destructive Interference
wavelength, phase difference = 180° (1/2 λ)
ÎAmplitude smaller: Lower intensity
ÎSame
Sum
E ( x ) = cos(kx) + 0.5cos(kx + π ) = 0.5cos(kx)
PHY 2049: Chapter 36
6
Examples
ÎTwo
waves, same λ, with amplitudes 2A and A
‹ Initial
ÎNo
intensities 4I and I, respectively (I ∝ A2)
interference
‹ New
intensity:
ÎMaximum
constructive interference (φ = 0)
‹ New
amplitude:
‹ New intensity:
ÎMaximum
‹ New
Inew = 4I + I = 5I
Anew = 3A
Inew = 9I
destructive interference (φ = π)
amplitude:
‹ New intensity:
Anew = A
Inew = I
PHY 2049: Chapter 36
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 36
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 36
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 36
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 36
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 36
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 36
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 36
14
Intensity vs Angle for d = 5λ
PHY 2049: Chapter 36
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 36
16
Intensity vs θ for d = 2.0 μm , λ = 550 nm
PHY 2049: Chapter 36
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 36
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(ω t ) + E cos(ω t + φ )
φ = 2π d sin θ / λ
PHY 2049: Chapter 36
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 36
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
I tot = K 2 E 2 (1 + cos φ ) = 2 I 0 (1 + cos φ )
= 4 I 0 cos 2 12 φ
= 4 I 0 cos 2 (π d sin θ / λ )
PHY 2049: Chapter 36
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 36
22