6: Lecture Waves Review, Crystallography, and Examples
Lecture 6, p. 1
Single-Slit Diffraction Example
Suppose that when we pass red light ( = 600 nm) through a slit of unknown 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 = 2 m behind the slit. How wide is the slit?
a
W = 1 cm
L=2m
Lecture 6, p. 2
Solution
Suppose that when we pass red light ( = 600 nm) through a slit of unknown 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 = 2 m behind the slit. How wide is the slit?
a
W = 1 cm
L=2m
The angle to the first zero is: = /a W = 2L tan 2L = 2L/a a = 2L/W = (4m)(610-7 m) /(10-2 m) = 2.410-4 m = 0.24 mm
Lecture 6, p. 3
Multiple Slit Interference (from L4)
9I1
N=3
16I1
N=4
25I1
20
N=5
I3
I4
I5
10
0
2 /d
0 0
2 /d
0
2 /d
0 0
2 /d
0
2 /d
0 0
2 /d
The positions of the principal maxima occur at = 0, 2, 4, ... where is the phase between adjacent slits. = 0, /d, 2/d, ... The intensity at the peak of a principal maximum goes as N2. 3 slits: Atot = 3A1 Itot = 9I1. N slits: IN = N2I1. Between two principal maxima there are N-1 zeros and N-2 secondary maxima The peak width 1/N. The total power in a principal maximum is proportional to N2(1/N) = N.
Lecture 6, p. 4
Multiple-slit Example
Three narrow slits with equal spacing d are at a distance L = 1.4 m away from a screen. The slits are illuminated at normal incidence with light of wavelength = 570 nm. The first principal maximum on the screen is at y = 2.0 mm. 1. What is the slit spacing, d?
y
d L
2. If the wavelength, , is increased, what happens to the width of the principal maxima?
Lecture 6, p. 5
Solution
Three narrow slits with equal spacing d are at a distance L = 1.4 m away from a screen. The slits are illuminated at normal incidence with light of wavelength = 570 nm. The first principal maximum on the screen is at y = 2.0 mm. 1. What is the slit spacing, d?
The first maximum occurs when the path difference between adjacent slits is . This happens at sin = /d. We are told that tan = y/L = 1.43X10-3, so the small angle approximation is OK. Therefore, d / = 0.40 mm.
y
d L
2. If the wavelength, , is increased, what happens to the width of the principal maxima?
The relation between and is /2 = / = d sin / . Therefore, for every feature that is described by (peaks, minima, etc.) sin is proportional to . The width increases.
Lecture 6, p. 6
ACT 1: Multiple Slits
I =?
1. In 2-slit interference, the first minimum corresponds to = . For 3-slits, we have a secondary maximum at = (see diagram). What is the intensity of this secondary maximum? a) I1 b) 1.5 I1 c) 3 I1
9I1
I
0
2
=?
0
2
2. What value of corresponds to the first zero of the 3-slit interference pattern? a) =/2 b) =2/3 c) =3/4
3. What value of corresponds to the first zero of the 4-slit interference pattern? a) =/2 b) =2/3 c) =3/4
Lecture 6, p. 7
Solution
I =?
1. In 2-slit interference, the first minimum corresponds to = . For 3-slits, we have a secondary maximum at = (see diagram). What is the intensity of this secondary maximum? a) I1 b) 1.5 I1 c) 3 I1
9I1
I
0
2
=?
0
2
Draw the phasor diagram : =
A1
Two phasors cancel, leaving only one
I1
2. What value of corresponds to the first zero of the 3-slit interference pattern? a) =/2 b) =2/3 c) =3/4
Lecture 6, p. 8
Solution
I =?
2. What value of corresponds to the first zero of the 3-slit interference pattern? a) =/2 b) =2/3 c) =3/4
9I1
I
0
2
=?
0
2
A =/2
=2/3
=3/4
No. A is not zero.
Yes! Equilateral triangle gives A = 0.
A No. Triangle does not close.
3. What value of corresponds to the first zero of the 4-slit interference pattern? a) =/2 b) =2/3 c) =3/4
Lecture 6, p. 9
Solution
I =?
3. What value of corresponds to the first zero of the 4-slit interference pattern? a) = /2 b) = 2/3 c) = 3/4
9I1
I
0
2
=?
0
2
= /2 Yes. The square gives A = 0.
= 2/3
A No. With 4 slits we wrap around and A = A1.
A
= 3/4
No. A 0
To get a zero, we need a closed figure. N must be a multiple of 2, so the first zero is at 2/N.
Lecture 6, p. 10
Interference & Diffraction Exercise
Light of wavelength 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. L >> d, y, a. What is N? a) N = 2 b) N = 3 c) N = 4
Imax
I
0 6 0 +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) Imax
I
0 6
b) Imax
I
0 6
c) Imax
I
0 6
y
0 (cm)
+6
y
0 (cm)
+6
y
0 (cm)
+6
Lecture 6, p. 11
Solution
Light of wavelength 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. L >> d, y, a. What is N? a) N = 2 b) N = 3 c) N = 4
Imax
I
0 6 0 +6
y (cm)
N is determined from the number of minima between two principal maxima. N = #minima+1 Therefore, N = 3 . 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) Imax
I
0 6
b) Imax
I
0 6
c) Imax
I
0 6
y
0 (cm)
+6
y
0 (cm)
+6
y
0 (cm)
+6
Lecture 6, p. 12
Solution
Light of wavelength 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. L >> d, y, a. What is N? a) N = 2 b) N = 3 c) N = 4
Imax
I
0 6 0 +6
y (cm)
N is determined from the number of minima between two principal maxima. N = #minima+1 Therefore, N = 3 . 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) Imax
I
0 6
b) Imax
I
0 6
c) Imax
I
0 6
0 +6 y (cm) Interference spacing should change.
0 +6 y (cm) Diffraction profile shouldnt change.
0 +6 y (cm) Interference spacing doubles. Diffraction profile is unchanged.
Lecture 6, p. 13
Crystal diffraction
How do we know the atomic scale structure of matter around us?
A crystal is a very large number of atoms or molecules arranged in a periodic fashion. It acts like a grating with an extremely large number (~Avagadros number) of units that diffract waves coherently. Every crystal has its own signature of the various spacings between atoms. By measuring the diffraction, we can determine the atomic scale structure. Typical distances between atoms are of order 0.1-0.3 nm (1-310-10 m). What characteristic wavelengths are needed to study crystals? We want: < spacing (so that we can get > ). not too small (so that isnt too small). That is: ~10-10m. X-rays!
Lecture 6, p. 14
FYI: Crystal diffraction
The structure of the crystal can be found using almost the same law we have for optical gratings! Bragg Law for constructive interference: = 2d sin = m d = lattice spacing, = x-ray wavelength = x-ray angle (with respect to plane of crystal)
d
Example of planes in a NaCl crystal
Each crystal has many values of d - the distances between different planes. For a known wavelength the observed angles can be used to determine the crystal structure.
Lecture 6, p. 15
FYI: Application: Structure of DNA
How do we know the structures of DNA, proteins and other biological molecules? X-ray diffraction! The molecules are crystallized to create a crystal in which the molecules are arranged in a periodic lattice. By using the sharp Bragg diffraction from many molecules, the structure of each molecule is determined - the positions of thousands of atoms The original diffraction image of DNA taken by Rosalind Franklin in 1953 Actually the DNA was not crystallized for the first images, DNA but the DNA was dehydrated in a fiber, and formed a quasi-crystalline structure that showed the helical structure. The dark bands arranged in a cross were the first evidence of the helical structure.
See figure and more details on this and other X-ray scattering in the text, sec. 36.6
Lecture 6, p. 16
FYI: Modern Applications in Biology
X-rays remain the primary methods for establishing the atomic scale structures of complex molecules.
Example: Rabbit liver carboxylesterase (one molecule showing atomic groups and attached large scale structures with atoms not shown).
Alternative strategies to improve the antitumor efficacy have concentrated upon the design of novel camptothecin analogs. To effect this, we have determined the x-ray crystal structures of the rabbit liver carboxylesterase
Source: St. Jude Children Research Hospital http://www.stjude.org/faculty/0,2512,407_2030_4278,00.html
Lecture 6, p. 17
Diffraction from gratings
Resolving power of an N-slit grating: The Rayleigh criterion The slit/line spacing determines the location of the peaks (and the angular dispersing power () of the grating. The number of slits/beam size determines the width of the peaks (narrower peaks easier to resolve).
Lecture 6, p. 18
Diffraction Gratings (1)
Diffraction gratings rely on N-slit interference. They consist of a large number of evenly spaced parallel slits.
An important question:
How effective are diffraction gratings at resolving light of different wavelengths (i.e. separating closely-spaced spectral lines)?
IN = N2I1
1
0 1/d
sin depends on .
sin
2
0 2/d sin
Example: Na has a spectrum with two yellow lines very close together: 1 = 589.0 nm, 2 = 589.6 nm ( = 0.6 nm) Are these two lines distinguishable using a particular grating? We need a resolution criterion.
Lecture 6, p. 19
Diffraction Gratings (2)
We use Rayleighs criterion: The minimum wavelength separation we can resolve occurs when the 2 peak coincides with the first zero of the 1 peak:
1/d 2/d
So, the Raleigh criterion is (sin)min = /Nd. However, the location of the peak is sin = m/d. Thus, ()min = (d/m)(sin)min = /mN:
min
1 = Nm
Nd
sin
Comments: It pays to use a grating that has a large number of lines, N.
However, one must illuminate them all to get this benefit.
It also pays to work at higher order (larger m): The widths of
the peaks dont depend on m, but they are farther apart at large m.
First order m=1 Second order m=2 Third order m=3
0
/d
2/d
3/d
sin
Lecture 4, p 20
ACT 2
1. Suppose we fully illuminate a grating for which d = 2.5 m. How big must it be to resolve the Na lines (589 nm, 589.6 nm), if we are operating at second order (m = 2)? a. 0.12 mm b. 1.2 mm c. 12 mm
2. How many interference orders can be seen with this grating? a. 2 b. 3 c. 4 3. Which will reduce the maximum number of interference orders? a. Increase b. Increase d c. Increase N
Lecture 6, p. 21
Solution
1. Suppose we fully illuminate a grating for which d = 2.5 m. How big must it be to resolve the Na lines (589 nm, 589.6 nm), if we are operating at second order (m = 2)? a. 0.12 mm b. 1.2 mm c. 12 mm
We need to make N large enough to satisfy Raleighs criterion.
1 = Nm
589nm So: N = = 491 m 2(0.6nm)
Size = Nd 4912.5 m 1.2 mm 2. How many interference orders can be seen with this grating? a. 2 b. 3 c. 4
Lecture 6, p. 22
Solution
1. Suppose we fully illuminate a grating for which d = 2.5 m. How big must it be to resolve the Na lines (589 nm, 589.6 nm), if we are operating at second order (m = 2)? a. 0.12 mm b. 1.2 mm c. 12 mm
2. How many interference orders can be seen with this grating? a. 2 b. 3 c. 4 The sin of the diffraction angle can never be larger than 1: sin 1. From sin = m/d, we obtain m d/ = 2.5 m/0.589 m = 4.2. So, m = 4. 3. Which will reduce the maximum number of interference orders? a. Increase b. Increase d c. Increase N
Lecture 6, p. 23
Solution
1. Assuming we fully illuminate the grating from the previous problem (d = 2.5 m), how big must it be to resolve the Na lines (589 nm, 589.6 nm)? a. 0.12 mm b. 1.2 mm c. 12 mm
2. How many interference orders can be seen with this grating? a. 2 b. 3 c. 4 3. Which will reduce the maximum number of interference orders? a. Increase b. Increase d c. Increase N m d/, so increase , or decrease d. Changing N does not affect the number of orders.
Lecture 6, p. 24
Angular Resolution (from L5)
Diffraction also limits our ability to resolve (i.e., distinguish) two point sources. Consider two point sources (e.g., stars) with angular separation viewed through a circular aperture or lens of diameter D.
Two point sources
D =c =c/3
I0 I
=2c
I0 I0.5 0 2I0 I0 I 0 2I0
Just as before, Rayleighs Criterion define the images to be resolved if the central maximum of one image falls on the first minimum of the second image.
0 2I0
c = 1.22
D
Sum
0
Sum y Two images resolvable
0
Sum y Diffraction limit of resolution
0
y Two images not resolvable
NOTE: No interference!! Why not?
Lecture 6, p. 25
Exercise: Angular resolution
Car headlights in the distance: What is the maximum distance L you can be from an oncoming car at night, and still distinguish its two headlights, which are separated by a distance d = 1.5 m? Assume that your pupils have a diameter D = 2 mm at night, and that the wavelength of light is = 550 nm.
Lecture 6, p. 26
Solution
Car headlights in the distance: What is the maximum distance L you can be from an oncoming car at night, and still distinguish its two headlights, which are separated by a distance d = 1.5 m? Assume that your pupils have a diameter D = 2 mm at night, and that the wavelength of light is = 550 nm. Use Rayleighs criterion: c = 1.22
D = 3.4 10 4 (radians)
Then, L d/c = 4500 m = 2.8 miles (assuming perfect eyes).
The small angle approximation is valid.
Lecture 6, p. 27
Michelson Interferometer (from Lect. 5)
The Michelson interferometer works by varying the phase difference between the two paths the light can take. One possibility is to vary the lengths L1 or L2. This makes possible very accurate measurements of displacements.
mirror Total path length L1 Path-length difference = L2 - L1 Total path length L2
Be careful !! Whats important is the total time to get back to the beam splitter...
2I1
Source
2I1 2I1
4I1
mirror
2I1
beam splitter
I1
I1
I = 4I1 cos2(/2), with = 2 /
Lecture 6, p. 28
Exercise: Michelson Interferometer
An interferometer is sensitive to the phase difference, , between the two paths. One way to vary is to change L1 or L2. Another way is to change the index of refraction of the medium in one arm. Suppose that in our interferometer, the mirrors are 1 m from the beam splitter (L1 = L2 = 2 m). Also, assume the index of refraction in arm 1 is n1 = 1.0. What is the smallest deviation of n2 from 1.0 that will give destructive interference when the beams are recombined? Assume that = 600 nm in vacuum.
L1
beam splitter mirrors
Source
L2
Remember: L1 and L2 are the round trip distances.
detector
Lecture 6, p. 29
Solution
An interferometer is sensitive to the phase difference, , between the two paths. One way to vary is to change L1 or L2. Another way is to change the index of refraction of the medium in one arm. Suppose that in our interferometer, the mirrors are 1 m from the beam splitter (L1 = L2 = 2 m). Also, assume the index of refraction in arm 1 is n1 = 1.0. What is the smallest deviation of n2 from 1.0 that will give destructive interference when the beams are recombined? Assume that = 600 nm in vacuum.
L1
beam splitter mirrors
Source
L2
Box of gas with n2.
detector
We want the number of wavelengths in arm 2 to differ from that in arm 1 by . The number in arm 1 is N1 = L1/ = 3333333.3, so we want N2 = 3333333.8. The ratio, N2/N1 = 1+1.510-7 is the ratio of wavelengths, which, by definition, is the index of refraction in the gas. This instrument is very sensitive to small changes in n. For comparison nair ~ 1+310-4.
Lecture 6, p. 30