phy 102: waves & quanta topic 7 diffraction john cockburn (j.cockburn@... room e15)
TRANSCRIPT
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PHY 102: Waves & Quanta
Topic 7
Diffraction
John Cockburn (j.cockburn@... Room E15)
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•Interference re-cap
•Phasors
•Single slit diffraction
•Intensity distribution for single slit
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Electromagnetic Waves
)sin(0 tkxEEy
)sin(0 tkxBBz
Where E0 and B0 are related by: E0 = cB0
INTENSITY of an EM wave E02
NB. we will see later that EM radiation sometimes behaves like a stream of particles (Photons) rather than a wave………………
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Interference
First, consider case for sound waves, emitted by 2 loudspeakers:
Path difference =nλConstructive Interference
Path difference =(n+1/2)λDestructive Interference
(n = any integer, m = odd integer)
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Interference
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Young’s Double Slit Experiment
•Demonstrates wave nature of light
•Each slit S1 and S2 acts as a separate source of coherent light (like the loudspeakers for sound waves)
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Young’s Double Slit Experiment
Constructive interference:
Destructive interference:
nd sin
2
1sin nd
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Young’s Double Slit Experiment
Y-position of bright fringe on screen: ym = Rtanm
Small , ie r1, r2 ≈ R, so tan ≈ sin
So, get bright fringe when:
d
nRym
(small only)
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Young’s Double Slit Experiment:Intensity Distribution
For some general point P, the 2 arriving waves will have a path difference which is some fraction of a wavelength.
This corresponds to a difference in the phases of the electric field oscillations arriving at P:
tEE sin01
tEE sin02
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Young’s Double Slit Experiment:Intensity Distribution
Total Electric field at point P:
tEtEEEETOT sinsin 0021
Trig. Identity:
2
1sin
2
1cos2sinsin
With = (t + ), = t, get:
2sin
2cos2 0
tEETOT
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2sin
2cos2 0
tEETOT
So, ETOT has an “oscillating” amplitude:
2cos2 0
E
Since intensity is proportional to amplitude squared:
2cos4 22
0
EITOT
Or, since I0E02, and proportionality constant the same in both cases:
2cos4 2
0
IITOT
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differencepath
2
difference phase
sin
2
d
2cos4 2
0
IITOT
sin
cos4 20
dIITOT
For the case where y<<R, sin ≈ y/R:
R
dyIITOT
20 cos4
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Young’s Double Slit Experiment:Intensity Distribution
R
dyIITOT
20 cos
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2-slit intensity distribution: “phasor” treatment
•Remember from Lecture 1, harmonic oscillation with amplitude A and angular frequency can be represented as projection on x or y axis of a rotating vector (phasor) of magnitude (length) A rotating about origin.
Light•We can use this concept to add oscillations with the same frequency, but different phase constant by “freezing” this rotation in time and treating the 2 oscillations as fixed vectors……
•So called “phasor method”
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2-slit intensity distribution: “phasor” treatment
Use phasor diagram to do the addition E1 + E2
tEE cos01
tEE cos02
Using cosine rule:
cos2 20
20
20
2 EEEETOT
cos2 20
20
20 EEE
cos12 20 E
2cos4 22
02
EETOT
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Another way: Complex exponentials
)sin(cos iAAei
)sin(cos00 titEeE ti
))sin()(cos(0)(
0 titEeE ti
tieEtE 00 Re)cos(
)(00 Re)cos( tieEtE
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Another way: Complex exponentials
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Single Slit Diffraction
“geometrical” picture breaks down when slit width becomes comparablewith wavelength
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Single Slit Diffraction
observed for all types of wave motion
eg water waves in ripple tank
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Single Slit Diffraction
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Single Slit Diffraction
•Explain/analyse by treating the single slit as a linear array of coherent point sources that interfere with one another (Huygen’s principle)…………………….
All “straight ahead” wavelets in phase → central bright maximum
Destructive interference of light from sources within slit for certain angles
Fraunhofer (“far-field”)case
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Single Slit Diffraction
•From diagram, can see that for slit of width A, we will get destructive interference (dark band on screen) at angles which satisfy…..:
2sin
2
a
2sin
4
a
a
sina
2sin
Choice of a/2 and a/4 in diagram is entirely arbitrary, so in general we have a dark band whenever;
a
m sin (m=±1, ±2, ±3………..)
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Position of dark fringes in single-slit diffraction
a
m sin
If, like the 2-slit treatment we assume small angles, sin ≈ tan =ymin/R, then
a
Rmy
min
Positions of intensity MINIMA of diffraction pattern on screen, measured from central position.
Very similar to expression derived for 2-slit experiment:
d
nRym
But remember, in this case ym are positions of MAXIMA
In interference pattern
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Width of central maximum
•We can define the width of the central maximum to be the distance between the m = +1 minimum and the m=-1 minimum:
a
R
a
R
a
Ry
2
Ie, the narrower the slit, the more the diffraction pattern “spreads out”
image of diffraction pattern
Intensitydistribution
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Single-slit diffraction: intensity distribution
To calculate this, we treat the slit as a continuous array of infinitesimal sources:
Can be done algebraically, but more nicely with phasors………………..
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Single-slit diffraction: intensity distribution
E0 is E-field amplitude at central maximum
2
0 2/
)2/sin(
II
= total phase difference for “wavelets” from top and bottom of slit
2/
)2/sin(0
EETOT
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Single-slit diffraction: intensity distribution
2
0 2/
)2/sin(
II
How is related to our slit/screen setup?
Path difference between light rays from top and bottom of slit is
sinax
differencepath
2
difference phase
sin
2
d
From earlier (2-slit)
sin2 a
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Single-slit diffraction: intensity distribution
2
0 2/
)2/sin(
II
sin2 a
2
0 /)(sin
)/)(sinsin(
a
aII