Lecture 29: Chapter 36
Diffraction
Friday 24 March 2006
Diffraction by a single slit
• Today
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Single-slit diffraction
Diffraction by a circular aperture
Use of phasors in diffraction
Double-slit diffraction
Single slit: Pattern on screen
Single Slit:
finding the minima
Divide the single slit into
many tiny regions. If the top
ray and the middle ray
interfere destructively, then
every pair of rays will also.
Bright and dark fringes appear behind a single very thin slit.
As the slit is made narrower the pattern of fringes becomes wider.
Single Slit:
the first minimum
So we get destructive
interference if:
(a / 2)sinθ = (λ / 2)
a sin θ = λ
So each pair cancels each other
and we have total cancellation!
Dark Fringes
in Diffraction
Second dark fringe?
First dark fringe.
Note this mustn’t be confused for
the condition for constructive
interference in the two-slit case!
Also note as slit width gets smaller,
angle gets larger, and vice versa.
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Lecture 29: Chapter 36
Dark fringes
a sin θ = λ
First is at
As we move up on the screen the next
dark fringe occurs when the first ray
interferes destructively with the one
from one-fourth the way down the slit.
(a / 4)sinθ = λ / 2
a sin θ = 2λ
We can continue to one-sixth, one-eighth,
etc. to get all the dark fringes at
a sin θ = mλ
Friday 24 March 2006
Summary of single-slit diffraction
• Given light of wavelength λ passing through a slit of
width a.
• There are dark fringes (diffraction minima) at
angles θ given by a sin θ = mλ where m is an integer.
• Note this exactly the condition for constructive
interference between the rays from the top and
bottom of the slit.
• Also note the pattern gets wider as the slit gets
narrower.
• The bright fringes are roughly half-way between the
dark fringes. (Not exactly but close enough.)
Again note as slit gets narrower, pattern gets wider.
Example: Problem 37-2
Light of wavelength 441 nm is incident on a
narrow slit. On a screen 2 meters away, the
distance between the second diffraction minimum
and the central maximum is 1.5 cm.
(a) Calculate the angle of
diffraction θ of the second
minimum.
(b) Find the width of the slit.
Example: (cont’d)
(a) Calculate the angle of
diffraction θ of the second
minimum.
y 1.5 × 10 −2
=
D
2
= 7.5 × 10 − 3 rad
θ ≅ tan θ =
(b) Find the width of the slit.
a sin θ = mλ
θ ≅ sin θ = m
λ
a
λ
441 × 10 −9 m
a=m =2
= 117.6 µm
θ
7.5 × 10 − 3
Diffraction by a circular aperture
Consider a round hole of
diameter d. Same idea as a
long slit – only the geometry
is different.
The Rayleigh Criterion
Using a circular instrument (telescope, human eye),
when can we just resolve two distant objects?
Angle for first minimum:
Long slit:
a sin θ = λ
Circular hole:
d sin θ = 1.22 λ
When images are separated by distance to first
minimum:
θ R = 1.22 λ / d
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Lecture 29: Chapter 36
Friday 24 March 2006
Sample Problem 36-4
Divide slit into many tiny slits. Use a tiny phasor
for each. Add them together graphically.
Camera or eye:
Smallest resolvable angle
to two distant objects:
Single slit phasor diagram
θ R = 1.22 λ / d
Multiple-Slit Diffraction
Now we can finally put together our
interference and diffraction results to see
what really happens with two or more slits.
See we get destructive interference if first and
last phasors interfere constructively!
So the condition for
a sin θ = mλ
the mth dark fringe is:
Double-Slit
Diffraction
a = slit width
d = slit separation
RESULT: We get the two-slit (or multiple-slit)
pattern as in chapter 35, but modified by the
single-slit intensity as an envelope.
Instead of all peaks being of the same
height, they get weaker at larger angles.
Double-slit diffraction
2 slits of zero width
1 slit of width a = 5λ
2 slits of width a = 5λ
θ = angle on screen
Bright fringes due to 2-slit
interference: θ = mλ / d
Zero due to diffraction:
θ = λ / a, 2λ / a, "
Diffraction grating: Many Slits
Very sharp maximum
when all rays are in phase.
d sin θ = mλ
m = “order” = 1, 2, 3, …
First-order maximum:
sin θ = λ / d
Second-order maximum:
sin θ = 2λ / d
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Lecture 29: Chapter 36
Friday 24 March 2006
Diffraction grating bright lines
Diffraction grating recap
Position of lines is
determined by
separation of rulings.
d sin θ = mλ
d sin θ = mλ
Second-order maximum:
Sharpness of lines is
determined by number
of rulings.
First-order maximum:
∆θ = λ / Nd
Spectroscopy: separate lines of different wavelengths.
R = mN
Diffraction
Grating
Five Slits
Note we still have θ max = mλ / d
But more slits makes the peaks sharper.
Resolving power is
determined by number of
rulings and order of line.
ƒ
For many slits, we get a diffraction grating.
Wavelength Resolution of a Grating
Width of
sharp lines
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Lecture 29: Chapter 36
Friday 24 March 2006
Resolving Power
Example: Yellow sodium vapor lines
d sin θ = mλ
Positions of maximums:
Small angles: θ = mλ / d
∆ θ = m∆ λ / d
∆θ = λ / Nd
Wavelengths just resolved: m∆λ / d = λ / Nd
m∆ λ = λ / N
Widths of sharp maximums:
Resolving power definition:
λ
R=
∆λ
So we get:
R = mN
Problem 36-50
The strong yellow lines in the sodium spectrum are at
wavelengths 589.0 nm and 589.6 nm.
How many rulings are needed in a diffraction grating to
resolve these lines in second order?
We need
But
R=
R = mN
Interference with 3 Slits
Path difference
between rays
from adjacent
slits:
∆L = d sin θ
N=
so
1. Central maximum:
φ =0
2. First minimum:
Phase difference between
rays from adjacent slits:
λ
= 2π
2π
3
ET = 0
φ=
Get intensity from
phasor diagram:
dθ
3. Next maximum:
φ = 180°
λ
4. Next minimum:
φ = 240°
5. Next maximum:
IT = 9 I 0
d sin θ =
ET = E0
2
2
φ = 2π
d sin θ = λ
3
3
ET = 0 IT = 0
d sin θ = λ
ET = 3 E0
IT = 9 I 0
3
d sin θ = λ / 2
IT = I 0
Summary
φ = 2π
λ
IT = 0
φ =π
3 Slits Continued
φ = 360°
θ =0
ET = 3 E0
φ = 120°
∆L
R 982
=
= 491
2
m
3 Slits Continued
θ ≈ ∆L / d
φ = 2π
λ
589 nm
=
= 982
∆λ 0.6 nm
9I 0
I0
θ ≈ λ /d
θ ≈ ∆L / d
∆L =
0
λ
λ
3
2
2λ
3
λ
θ max ≈ mλ / d
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