光的衍射与光谱学

P
Incident Wave
(wavelength l )
y
L
a
Diffraction
&
Spectroscopy
Overview
•Multiple-slit Interference formula*•Diffraction Gratings
–Optical Spectroscopy
–Spectral Resolution
•Single-Slit Diffraction*•Interference + Diffraction
•X-ray Crystallography
*Derivations in Appendix
Last Lecture: General properties of N-Slit Interference
•The positions of the principal maxima of the intensity patterns always
occur at f = 0, ±2p , ±4p, ...[f is the phase between adjacent slits]
(i.e., dsin q = ±m l , m = 0, 1, 2,…).•The principal maxima become taller and narrower as N increases.
•The intensity of a principal maximum is equal to N 2times the maximum intensity from one slit. The width of a principal maximum goes as 1/N.•The # of zeroes between adjacent principal maxima is equal to N-1. The # of secondary maxima between adjacent principal maxima is N-2.
I 0
16I 1N=4
0-2p I 025I 1
N=5
02p
-2p I
9I 1
N=3
10
20h5()
x 1010x -l/d 0 l/d f q 10
0100
590g()x 10
10x f
q
-l/d 0 l/d 010
16
0h()
x 10
10x
-l/d 0 l/d f
q
How do we calculate these interference patterns …
illuminates a screen. Now we double the number of slits, keeping the spacing constant.
What happens to the net power I on the screen?
a. stays same (I)
b. doubles (2I)
c. increases x4 (4I)
illuminates a screen. Now we double the number of slits, keeping the spacing constant. What happens to the net power on the screen?a. stays the same b. doubles c. increases by 4If we double the number of slits, we expect the net power on the screen to double. How does it do this…
●The location and number of the principle maxima (which
have most of the power) does not change.
●The principle maxima become 4x brighter.●But they also become only half as wide. (A/2)
●
Therefore, the net power (integrating over all the peaks) increases two-fold, as we would expect. (P goes as IA)
We will soon see that we often use such an array of slits (also called a “diffraction grating”) to perform very precise metrology, e.g, spectroscopy, crystallography, etc.
N-Slit Interference –Summary (see Appendix for derivation)
•
The Intensity for N equally spaced slits is given by:L
y and d sin d ≈⋅≈==q l q l q l δp f 221)2/sin()2/sin(⎪
⎭
⎫
⎝⎛=f f N I I N **
Your calculator can probably plot this up. Give it a try.
** Note: You will not be able to use the small angle approximations if d ~ l.
y
L
d
q
f is the phase difference between adjacent slits.
•
As usual, to determine the pattern at the screen (detector
plane), we need to relate f to q or y = Ltan q :
**
0 q min ?
l /d
q
In an N-slit interference pattern, at what angle q min does the intensity first go to zero?(In terms of l , d and N.)
What does phasor sum look like here?
0 q min ?
l /d
q
In an N-slit interference pattern, at what angle q min does the intensity first go to zero?(In terms of l , d and N.)
But f = 2p(d sin q)/l ≈2p d q /l =2p /N. Therefore, q min ≈l /Nd.
As the illuminated number of slits increases, the peak widths decrease!
General feature :
Wider slit features narrower patterns in the ―far field‖.(obviously doesn‘t hold just past slits!)
2
1sin(/2)sin(/2)N N I I f f ⎛⎫= ⎪
⎝⎭
has a zero when N f min /2= p , or f min =2p /N.
Note:The simple calculations we have done only hold in the ―far-field‖(a.k.a. ―Fraunhofer‖limit), where L >> d2/l.
Intermediate cases (―Fresnel diffraction‖) can be much more complex…
near
far
Optical spectroscopy –
a major window on the world
•Quantum mechanics →definite energy levels, e.g., of electrons in atoms or molecules.
•When an atom transitions between energy levels →emits light of a very particular frequency.
•Every substance has its own “signature” of what colors it can emit.
•By measuring the colors, we can determine the substance, as well as things about its surroundings (e.g., temperature, magnetic fields), whether its moving (via the Doppler effect), etc.
Optical spectroscopy is invaluable in materials research, engineering, chemistry, biology, medicine…
Interference Gratings
--the basis for optical spectroscopy Interference gratings (usually called Diffraction gratings) allow us to resolve sharp spectral signals.
2/d sin q
N-slit Interference:
I N = N2I1
/d sin q l1
l2Shift of the peak:
Spectroscopy Demonstration
We have set up some discharge tubes with various gases. Notice that the colors of the various discharges are quite different. In fact the light emitted from the highly excited gases is composed of many ―discrete wavelengths.‖You can see this with the plastic ―grating‖we supplied.
Your eye
Hold grating less than 1 inch from your eye.
Light source
Your view through
the grating:
Put light source at left side of grating.View spectral lines by looking at about 45o. If you don’t see anything, rotate grating 90o.
Room lights must
be turned off!
Diffraction Gratings (1)
•Diffraction gratings rely on N-slit interference.•They simply consist of a large number of evenly spaced parallel slits.
•Recall that the intensity pattern produced by light of passing through N slits with spacing d is given by:2122⎭⎬⎫⎩⎨⎧=)/sin()/N sin(I I N f f where:
01020
h5()x 1010x f 0-2p 025I 1
N=5dsin 2q f p l =Consider very narrow slits (), so is roughly constant.
y L d
q Slits Demo 500/550 nm
•The position of the first principal maximum is given by (can ‘t assume small q !) →Different colors →different angles.
Diffraction Gratings (2)
•How effective are diffraction gratings at resolving light of different wavelengths (i.e. separating closely-spaced
‗spectral lines‘)?
–Concrete example:Na lamp has a spectrum with two yellow
―
12Dl =
nm)
–Are these two lines distinguishable using a particular grating?
I N = N2I1
/d sin q
2/d sin q
l1 l2
min min sin d d
d Nd l l q q l l q D =→D ≈D →D ≈=wavelength separation w
e can resolve Dl min ≡l 2-l 1occurs when the maximum o
f l 2overlaps with the first diffraction minimum of l 1.(Dq min =l /Nd)I N = N 2I 1N 1 m in =l l D Larger N Smaller Dl min
(Higher spectral resolution)
0 l 1/N d sin q
q l 1/d “Rayleigh Criterion”
N = number of illuminated lines in grating.Dq min
l 2/d
•We can squeeze more resolution out of a given grating by working in ―higher order ‖. Remember, the principal maxima occur at sin q = m l /d , where m = 1,2,3…designates the ―order ‖. (Dq min ≈ l /Nd still *)You can
easily show:Nm 1
m in =l l D q
First order Second order Third order m = 1m = 2m = 3
* To be precise: Dq = l /(Nd cos q ), (but Dl /l = 1/Nm is correct.)
Larger Nm Smaller Dl min
(Higher spectral resolution)
Example: Diffraction Gratings
•Angular splitting of the Sodium doublet:
Consider the two closely spaced spectral (yellow ) lines of sodium (Na ), l 1= 589 nm and l 2= 589.6 nm .If light from a sodium lamp fully illuminates a diffraction grating with 4000 slits/cm , what is the angular separation of these two lines in the second-order (m=2)spectrum?
Hint: First find the slit spacing d from the number of slits per centimeter.
Solution
•Angular splitting of the Sodium doublet:
Consider the two closely spaced spectral (yellow ) lines of sodium (Na ), l 1= 589 nm and l 2= 589.6 nm , mentioned earlier.If light from a sodium lamp illuminates a diffraction grating with 4000 slits/cm , what is the angular separation of these two lines in the second-order (m=2)spectrum?
41cm 2.510 2.54000
d cm m μ-==⨯=111sin = 28.112 m d l q -⎛⎫=︒ ⎪⎝⎭
Hint: First find the slit spacing d from the number of slits per centimeter.
1121sin sin 0.031m m d d l l q --⎛⎫⎛⎫D =-=︒ ⎪ ⎪⎝⎭⎝⎭
problem (d = 2.5 m), how big must it be to resolve the Na lines (589 nm, 589.6 nm)?
(a)0.13 mm(b)1.3 mm(c) 13 mm
2. How many interference orders can be seen with this grating?
(a)2(b)3(c) 4
3. Which reduces the maximum number of interference orders?
(a) Increase wavelength
(b) Increase slit spacing
(c) Increase number of slits
problem (d = 2.5 μm), how big must it be to resolve the Na lines (589 nm, 589.6 nm)?
(a)0.13 mm (b)1.3 mm (c) 13 mm
3. Which reduces the maximum number of interference orders?(a) Increase wavelength
(b) Increase slit spacing
(c) Increase number of slits We need enough lines to narrow the diffraction peak:1Nm
l l D =589nm 4912(0.6nm)N m l l ≥==D size = N d
≥490
(2.5 μm)≈1.2 mm 2. How many interference orders can be seen with this grating? The diffraction angle can never be more than 90˚: From sin q = m l /d , it must be that m l ≤ d , or m ≤ d/l = 2.5 μm/0.589μm = 4.2 m = 4
Increase l , or decrease d. Changing the number of slits does not affect the number of orders.(a)2(b)3
(c) 4
•So far in the N-slit problem we have assumed that each slit is a point source.
–Point sources radiate equally in all directions.
–Real slits have a non-zero extent –-a ―slit width ‖a . The transmission pattern depends on the ratio of a l . –In general, the smaller the slit width, the more the wave will diffract!
Laser Light (wavelength l )
Diffraction profile I 1
Small slit:Laser Light (wavelength l )
profile
I 1
Large slit:
Single-slit ―Diffraction ‖
Laser Demo
Single-Slit Diffraction •Slit of width a. Where are the minima?
–The first minimum is at an angle such that the light from
the top and the middle of the
slit destructively interfere:–The second minimum is at an angle such that the light from the top and a point at a/4
Location of nth-minimum:P
Incident Wave
(wavelength l )
y
L
a
q a/2
δ
min sin 22a l δq ==min sin a
l
q ⇒=
min,2sin 42a l
δq ==min,22sin l q ⇒=min,sin n n a l
q =q
a/4δ(n = 1, 2, …)
Diffraction: Example Problem
Suppose that when we pass red light (l= 600 nm) through a slit of width a, the width of the spot (the distance between the first zeros on each side of the bright peak) is W = 1 cm on a screen that is L = 2m behind the slit. How wide is the slit?
a 1 cm = W
2 m
Solution
Suppose that when we pass red light (l = 600 nm) through a slit of width a , the width of the spot (the distance between the first zeros on each side of the bright peak) is W = 1 cm on a screen that is L = 2m behind the slit. How wide is the slit?
Solution:
The angle to the first zero is: q = ±l /a
1 cm = W
L = 2 m
a
q
Solve for a: a =2L l /W = (4m)(6⨯10-7m) /(10-2m)
= 2.4⨯10-4m = 0.24 mm
W = 2 Ltan q ≅ 2 L q = 2L l /a
(use tan q ≅q)
Which of the following would broaden the diffraction peak?a. reduce the laser wavelength b. reduce the slit width
c. move the screen further away
2 m
a
1 cm = W
a 1 cm = W
2 m
Which of the following would broaden the diffraction peak?
a. reduce the laser wavelength
b. reduce the slit width
c. move the screen further away
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Single-slit Diffraction —Summary (see Appendix for derivation)
form:Single Slit Diffraction Features:•First zero: b/2= p ⇒q ≈l /a
(agrees with phasor analysis)
10010
0.5
1
10
dif f()x 12.56
12.56x b q
-4p -2p 0 2p 4p -l/a 0 l/a 0I I 0At P, the phase difference b between 1st and last source is given by:
2
01)/a sin(I )(I ⎪
⎫ ⎛=l q p q Screen
(far away)
δa = a sin q ≈a q q
a
L
q =l
q ≈l q =l δ=p b L y a sin a 2a Therefore,
P
2
012/)2/sin(I I ⎪
⎭
⎫ ⎝
⎛=b b b = angle between 1st and last phasor
Summary: Slit Interference + Diffraction
Combine:
Multi-slit Interference,
and
Single-slit Diffraction,
to obtain
Total Interference Pattern,
Remember:f/2p = δ/l = (d sin q)/l ≈d q /l
b/2p = δa /l = (a sin q)/l ≈a q/l
2
012/)2/sin(⎭
⎬⎫⎩⎨⎧=b b I I 2
1)2/sin()2/sin(⎪⎭
⎫ ⎝⎛=f f N I I N 2
2
)2/sin()2/N sin(2/)2/sin(I I ⎭
⎬⎫⎩⎨⎧f f ⎭
⎬
⎫⎩⎨⎧b b = f = angle between adjacent phasors b = angle between 1st and last phasor
(for plane-wave sources, I 1= constant)
Light of wavelength l is incident on an N -slit system with slit width a and slit spacing d .
1. The intensity I as a function of y at a viewing screen located a distance L from the slits is shown to the right. What is N ? (L >> d , y , a)
(a)N = 2
(b)N = 3(c)N =4
00
5
9
0intensity ()x 18.84
18.84
x
0I
max 0+6
-6
Y (cm)
2. Now the slit spacing d is halved , but the slit width a is kept constant . Which
of the graphs best represents the new intensity distribution?
(a)
5
9
intensity ()x I
I max (c)
5
9
intensity ()x I
I max (b)
5
9
intensity ()x I
I max
Light of wavelength l is incident on an N -slit system with slit width a and slit spacing d .
1. The intensity I as a function of y at a viewing
screen located a distance L from the slits is shown to the right. What is N ? (L >> d , y , a)(a)N = 2
(b)N = 3(c)N =4
00
5
9
0intensity ()x 18.84
18.84
x
0I
max 0+6
-6
Y (cm)
2. Now the slit spacing d is halved , but the slit width a is kept constant . Which of the graphs best represents the new intensity distribution?
(a)
5
9
intensity ()x I
I max (c)
5
9
intensity ()x I
I max (b)
5
9
intensity ()x I
I max N is determined from the number of minima between two principal maxima. Here, there are two minima between principal maxima. Therefore, N = 3 .
Lecture 3, ACT 4
00
5
9
0intensity ()x 18.84
18.84
x
0I
I max 0+6
-6
Y (cm)
2. Now the slit spacing d is halved , but the slit width a is kept constant .
Which of the graphs best represents the new intensity distribution?
(a)
5
9
0intensity ()x 18.84
18.84
x 0+6
-6
Y
(cm)0I
I max (c)
00
5
9
0intensity ()x 9.42
9.42
x 0+6
-6
Y (cm)
I
I max (b)
00
5
9
0intensity ()x 9.42
9.42
x 0+6
-6Y (cm)
0I
I max Decreasing d will increase spacing between maxima.The spacing between maxima is increased , but diffraction
profile shouldn ’t expand, as seen here.
This one does it all.
Increased spacing between maxima and constant diffraction.
Light of wavelength l is incident on an N -slit system with slit width a and slit spacing d .
1. The intensity I as a function of y at a viewing
screen located a distance L from the slits is shown to
the right. What is N ? (L >> d , y , a)(a)N = 2(b)N = 3(c)N =4
•Diffraction gratings are excellent tools for studying visible light because the slit spacing is on the order of the wavelength of the light (~ few tenths of microns)•Visible light is a very small part of the spectrum of electromagnetic waves. How can we study e-m waves with smaller wavelengths (e.g., x-rays with l~ 10-10
m)?
–We can‘t use a standard diffraction grating to do this.
Why?
Calculate q for first order peak for x-rays with l = 10-
10 m for a grating with d=1000 nm
•Solution? We need a ―diffraction ‖grating with a spacing d that is on the order of the wavelength of e-m waves we ‘d like to measure -i.e., for x-rays: l ~10-10m –Where do we find such a ―grating ‖? Nature to the rescue!!!
Crystalline solids (having regularly spaced atoms with d ~ 10-10m) naturally occur.
ion
a 0
a 0X-rays Diffraction: A Primer •Illuminate crystal with x-rays.•The X-rays are scattered by the ions.
•Enhanced scattering at certain scattering angles reveal
constructive interference between
X-ray Diffraction for Crystallography (FYI)
•If we know about the grating, we can use the diffraction pattern to learn about the light source.
•If instead we know about the source, we can use the diffraction pattern to learn about the ―grating ‖.
•For this to work, we need to have a source wavelength that is less than the grating spacing (otherwise, there are no orders of diffraction).
•Crystals consist of regularly spaced atoms →regular array of scattering centers. Typical lattice spacing is 5 angstroms = 5 x 10-10m = 0.5 nm.→use x-rays!
Bragg Law for
constructive interference:2d sin q = l
d = lattic
e spacing q = x-ray angle (with respect to plane o
f crystal)
l = x-ray wavelength
X-ray Crystallography, misc.
•could distinguish different cubic lattices •discovered the crystal structure of diamond •Lawrence Bragg was the youngest Laureate ever (25) to receive a Nobel Prize (shared with his father in 1915)
•now standardly used for all kinds of materials analysis, even biological samples!•The same multi-layer interference
phenomenon is now used to make highly wavelength-specific mirrors for lasers (―distributed Bragg feedback‖ [DBF])
The Braggs made so many discoveries that Lawrence described the first few years as ‗like looking for gold and finding nuggets lying around everywhere ‘:
•showed that the sodium and chloride ions were not bonded into molecules, but arranged in a lattice
Appendix: N-Slit Interference •Intensity for N equally spaced slits is easily found from phasor analysis:Draw normal lines bisecting the phasors. They intersect, defining R as shown:
2
21f sin R A =2
2f N sin R A N =)/sin()
/N sin(A A N 221
f f =Solve for R and plu
g in here:
Result:
2
122⎪
⎭
⎫ ⎝⎛=)/sin()/N sin(I I N
f f f
N f
A N
R A 1
R
N slits:
1f f
R
R
2
f
2
f
2
f
2
f
2
f N
Appendix: Single-slit Diffraction
•To analyze diffraction, we treat it as interference
of light from many sources (i.e. the Huygens
wavelets that originate from each point in the slit
opening).
•Model the single slit as M point sources with
spacing between the sources of a/M. We will let
M go to infinity on the next slide.
•The phase difference b between first and last
source is given by b/2p = δ
a
/l= a sin q / l ≈a q/l .
Destructive interference occurs when the
polygon is closed (b= 2p):
Slits Demo 10-slits f a
A1 (1 slit)
A a
(1 source)
a
M f
b≈
P
Screen
(far away)
δa= a sin q ≈a q
q
a
L >> a implies rays are
parallel.
L
a
,
small
For
sin
a
means
his
T
l
q
q
l
q
≈
=1
Destructive
Interference
Single-slit Diffraction —the math
•
We have turned the single-slit problem into the M-slit problem that we already solved in this lecture.
•However, as we let M →∞, the problem becomes much
simpler because the polygon becomes the arc of a circle.•The radius of the circle is determined by the relation
between angle and arc length: b = A o /R.2
012/)2/sin(I I ⎪
⎭
⎫ ⎝⎛=b b Graph this function
Intensity is related to amplitude: I = A 2. So, here ‘s the final answer:
Remember:b /2p = δa /l = (a sin q)/l ≈a q/l
b = angle between 1st and last phasor
A o
q
l
p
b a 2=
A 1b /2
R
Trigonometry: A 1/2 = R sin(b /2)
With R = A o /b , A 1= (2A o /b) sin(b /2)
2
20
1/)
/sin(A A b b =。