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Final reviewInterference and diffraction
Phys 322Lecture 37
Reminder: Please complete the online course evaluation
Final examFinal Exam, Saturday December 14, 1:00-3:00 pm, in PHYS 223.Material covered: Hecht Chapters 2-10 (except 5.8, 5.9, 6.3-6.5, 8.9-8.13, 10.3-10.5).Please bring two formula sheets (letter-size paper), scientific calculator, ID, etc.
Notes:1. The final exam is comprehensive, but the emphasis will be on the last part of
the material (Ch. 9-10)2. Two crib sheets (8 ½" x 11") are permitted.3. There will be 4-5 problems, one with qualitative questions and the rest
quantitative problems.
In a double slit interference experiment, the distance D from the slits to a screen is 1 m, the slit separation is 1 mm, and the wavelength of the incident plane wave is 500 nm.
a) Find the difference in the positions of two consecutive maxima near the center of the interference pattern.
Da
y
Find the separation between the 6th and 8th maxima located on each side of the central fringe (the 6th is on the positive side and the 8th fringe is on the negative side).
Review problem 1
a
s
Interferometers9.3 Wavefront-splitting interferometers9.3.1 Young’s experiment (1801)Young’s originality: Obtaining spatially coherent light from a pinhole, and splitting the wavefront of the spatially coherent light with two pinholes.
Optical path length difference:r2 – r1 = a sin θ = ay/s ( If s>> a, s>> y )
Position of bright fringes:
m = 0, ±1, ±2, … is the order number.asmy
am
m
m
ma msin
Space between fringes:asy
Intensity distribution of the fringes:
sayIrrkIII 2
0212
02
0 cos42
)(cos42
cos4
ay/s
In a double slit interference experiment, the distance D from the slits to a screen is 1 m, the slit separation is 1 mm, and the wavelength of the incident plane wave is 500 nm.
a) Find the difference in the positions of two consecutive maxima near the center of the interference pattern.
93
small angle approximation
1.0 (500 10 ) 0.510
Dya
my m mmm
D = s = 1 ma
y
Find the separation between the 6th and 8th maxima located on each side of the central fringe (the 6th is on the positive side and the 8th fringe is on the negative side).
~ 6 ( 8) 14 ~ 7D Dy mma a
A soap film surrounded by air has an index of refraction of 1.34. If a region of the film appears bright red (0 = 633 nm) in normally reflected light, what is the minimum thickness of the film there?
Review problem 2
9.4 Amplitude-splitting interferometers9.4.1 Dielectric films – double-beam interferenceApplication: Optical coating reduces or enhances reflection of optical devices through thin-film interference effects.
Fringes of equal inclinationConsider the first two reflections (others are weak).Optical path length difference:
tft
tf
t
f
itt
fi
t
f
f
dndndn
dndn
ACndn
ADnBCABn
cos2cos
sin2cos2
sintan2cos2
sincos2
)(
2
11
1
dA
B
C
D
P
S
i
t
nf
n1
n2
Phase difference:Assume nf >n1, nf >n2. There is a phase shift of between external and internal reflections (when the incident angle is not large).
tf d
nk cos
4
Bright spot: 4
)12(cos ft md
Dark spot:
42cos f
t md
A soap film surrounded by air has an index of refraction of 1.34. If a region of the film appears bright red (0 = 633 nm) in normally reflected light, what is the minimum thickness of the film there?
0
cos (2 1) eqn. 9.364
ft
ff
d m
n
0 633118
4 4(1.34)f
nmd nm
n
A double-slit diffraction pattern is formed using a HeNe laser light at 632 nm. Each slit has a width of 0.080 mm. The pattern reveals that the third-order interference maxima are missing from the pattern.(a) What is the slit separation?(b) What is the irradiance of the first two interference fringes, relative to the zeroth-order
maximum?
Review problem 3
Coherent line source:y
x
P (x,y)
dy
r
-D/2
D/2dy
rikryxE
D
DL )exp(),(2/
2/
L is the source strength per unit length.This equation changes a diffraction problem into an integration (interference) problem.
10.2 Fraunhofer diffraction10.2.1 The single slit
y
x
P (x,y)
y
r
-D/2
D/2
R
2sin)0()(
II
sin)2/( ),exp(sinsin)2/(
]sin)2/sin[()exp(
)]sin(exp[
)exp(),(
2/
2/
2/
2/
kDikRRD
kDkDikR
RD
dyR
yRik
dyrikryxE
L
L
D
DL
D
DL
In amplitude, r is approximated by R
In phase, r is approximated by R-y sinFraunhofer condition
The phase is the same as a point source at the center of the slit
2sin)0()(
II
0.047 0.016
maxima ,...47.3 ,46.2 ,43.1 ,0 ,tanminima ...3 ,2 , ,0sin
0)sincos(sin2)0(
sin)0()(
3
2
IddI
II
10.2.2 The double slit
z
x
P (x,z)R-a sin
R
ab )]sin(exp[)exp(sin
)()(),(
)]sin(exp[)(2/
2/
2/
2/
aRikikRbC
dzzFCdzzFCzxE
zRikzFba
ba
b
b
2
2
0 cossin4)(
II
sin)2/(sin)2/(
kbka
The result is a rapidly varying double-slit interference pattern (cos2) modulated by a slowly varying single-slit diffraction pattern (sin2/2).
(©WIU OptoLab)
22
/sin)/sinsin(sincos)0()(
bbaII
single-slit diffractiontwo-slit interferenceEnvelopeFringes
Question:Which interference maximum coincides with the first diffraction minimum?
bam
bma
sinsin
“Half fringe” (split fringe) may occur there.
2
2
0 cossin4)(
II
sin)2/(sin)2/(
kbka
A double-slit diffraction pattern is formed using a HeNe laser light at 632 nm. Each slit has a width of 0.080 mm. The pattern reveals that the third-order interference maxima are missing from the pattern.(a) What is the slit separation?(b) What is the irradiance of the first two interference fringes, relative to the zeroth-order
maximum?
Review problem 3
For f = (sin )/
What is the relative irradiance of the subsidiary maxima in a three slit Fraunhofer diffraction pattern?
Review problem 4
10.2.3 Diffraction by many slits
)2exp(1)2exp(1)exp(sin
])1(2exp[...)4exp()2exp(1)exp(sin
)(...)(
)()(),(
)]sin(exp[)(
2/)1(
2/)1(
2/2
2/2
2/
2/
2/
2/
iNiikRbC
NiiiikRbC
dzzFCdzzFC
dzzFCdzzFCzxE
zRikzF
baN
baN
ba
ba
ba
ba
b
b
z
x
P (x,z)
R-a sin
R
ab
R-2a sin
22
20
sinsinsin)(
N
NII
Principle maxima:Minima (N-1):
Subsidiary maxima (N-2):
... ,2 , ,0
)1(... ,3 ,2 , N
NNNN
2
)32( ..., ,25 ,
23
NN
NN
64sin
sinsin 22
Nba
N
What is the relative irradiance of the subsidiary maxima in a three slit Fraunhofer diffraction pattern?
32 2N
2 2
2
(0) sin sin( )sin
I NIN
sinfor small , ~ 1
2
( ) 1 1~(0) 9
II N
You want to distinguish between =500.2 and 500.3 nm light with a grating at normal incidence to the light and with a maximum angle for third-order diffraction of 5. What is the minimum width of the grating?
Review problem 5
GratingsDiffractive grating: An optical device with regularly spaced array of diffracting elements.Transmission gratings and reflection gratings.
Grating equation: i
m
a
i
m
a
ma im )sin(sin
Blazed grating: Enhancing the energy of a certain order of diffraction.Blaze angle: Specular reflection: mir 2 i
r
0
a
specular
0th
Grating spectroscopy:N-slit interference:
cos2
2
)sin)(sin2/(sin
sin)(2
0
Na
N
ka
NII
i
Angular width due toinstrumental broadening
Angular dispersion:m
mim a
mddma
cos)sin(sin
dm
d
Limit of resolution:
)sin(sin
coscos
coscos2
minminmin
min
criterion sRayleigh'
im
mm
m
m NamN
am
am
dd
NaNa
min
Angular width
Angular dispersion
min
min
Resolving power
Free spectral range:
mmm
ma fsri
)sin(sin
sinm =1
m =3
m =2
fsr
Question: Why does the resolving power increase with increasing order number and with increasing number of illuminated slits?
You want to distinguish between =500.2 and 500.3 nm light with a grating at normal incidence to the light and with a maximum angle for third-order diffraction of 5. What is the minimum width of the grating?
Review problem 5
1. T or F: Zone plates are used to change the polarization of an electromagnetic wave.2. T or F: The blaze on a reflection grating can improve the amount of energy in a desired
order of diffraction.3. T or F: The resolving power of a grating spectrometer used in a particular diffraction
order depends only on the number of lines illuminated (not wavelength or grating period).
4. T or F: The central peak of the Fraunhofer diffraction from two narrow slits separated by spacing h has the same width as the central diffraction peak from a single slit with width x = h.
5. T or F: The Fraunhofer diffraction pattern appearing at the focus of a lens varies in angular width, depending on the focal length of the lens used.
6. T or F: Fraunhofer diffraction can be viewed as a spatial Fourier transform (or inverse transform if you prefer) on the field at the aperture.
7. T or F: It is always possible to completely eliminate reflections with a single-layer antireflection coating as long as the right thickness is chosen for a given real index.
8. T or F: A half-wave plate rotates light initially linearly polarized at an angle with respect to the fast axis through a total angle 2.
9. T or F: A Michelson interferometer can be used to measure the spectral intensity of light I ().
10. T or F: A Michelson interferometer can be used to measure the wavelength of light.
Sample qualitative questions
T or F: Vertically polarized light illuminates a Young’s double-slit setup and fringes are seen on a distant screen with good visibility. A half wave plate is placed in front of one of the slits so that the polarization for that slit becomes horizontally polarized. Here’s the statement: The fringes at the screen will shift position but maintain their good visibility.
More qualitative questions
Effects of finite coherent length:
• Light from each slit has a coherent length lc.For sunlight lc is about 3.
• The waves from two slits can only interfere ifr2 – r1 <lc.
• The contrast of the fringes degrades when the amount of the overlap between uncorrelated wavegroups increases.
PA B
B'A'
lc
Other Wavefront-splitting interferometers
Fresnel’s double mirror: Slits S1 and S2 act as virtual coherent sources. They are images of slit S in two mirrors.
= r2 – r1, y = s/a
PM1
M2
S1
S2
S
a
r1
r2
s
y
Shield
Fresnel double prism: Interference between light refracted from the upper and the lower prisms. Two virtual coherent source S1 and S2 are produced by the prisms.
Question: Where are S1 and S2?
S1
S2
Sa
Lloyd’s mirror: Interference between light from source S and its image S ' in the mirror. Glancing incidence causes a phase shift of . The fringes are complementary to Young’s.
sayII2
0 sin4
S
S'
as
y
Extended source: All rays inclined at the same angle arrive at the same point.Fringes of equal inclination: Arcs centered on the perpendicular from the eye to the film.
Haidinger fringes: The fringes of equal inclination viewed at nearly normal incidence. Concentric circular bands.
t
f dn
cos2
sin 2
Fringes of equal thicknessFizeau fringes: Contours from a non-uniform film when viewed at nearly normal incidence.
x
n1
n2
nf
Thickness: d = xInterference maximum: 2nf dm = (m + ½)
Fringe space: x = f /24
)12( ,4
)12( fm
fm mxmd
Newton’s rings: Interference pattern from air film between glass surfaces.
xd
Thickness:
Bright rings: 2nf dm = (m + ½)
Dark rings:
RxddRdx2
)2(2
2
Rmxmd fmf
m
21 ,
4)12(
Rmx fm
22
2sin x
Rn
Beam splitter
Movable mirror
(©WIU OptoLab)
S S1 S2
M1 M2
P
cos2d
cos4 d
Optical path length difference:
Phase difference:
Dark fringes: md m cos2
Multiple beam interference9.4.2 Mirrored interferometersMichelson interferometer1) Compensation plate: Negates dispersion from the beam splitter2) Collimated source: Fringes of equal thickness3) Point source: Interference of spherical waves4) Extended source: Fringes of equal inclination
9.6 Multiple beam interferenceInterference between multiple reflection and multiple refractions:
td
S
i
nf
0E
rE0 ''0 ttrE ''30 ttrE ''50 ttrE
'0ttE 20 ''rttE 4
0 ''rttE
2
212
rrFCoefficient of finesse:
Airy function:2sin
121
2sin
12
1
2sin
121
1
cos21)1(
11
1)1(
'1'
...)'''1('1' ,'
cos4
cos2
22
2
22
2
22
2
24
222
2
2
2
20
20
362420
2
rr
rr
II
II
rr
rrr
err
II
errE
erttE
erererttEErttrr
dn
k
dn
i
t
i
r
ii
t
ii
iiit
tf
tf
)2/(sin1)2/(sin
)2/(sin11
2
2
2
FF
II
FII
i
r
i
t
)2/(sin11 )( 2
F
A
)2/(sin11
12
2
2
2
FII
rrF
i
t
it II /
/
F = 0.2 (r2 = 0.046)F = 1 (r2 = 0.17)F = 200 (r2 = 0.87)
9.6.1 The Fabry-Perot interferometer1) High resolving power2) Prototype of laser cavity
S P
dFF
FA
21sin2
21
)2/(sin11 )(
12/1
2
Half-width of transmission:
Finesse:F42 2/1
22 Ff
Chapter 10 DiffractionFraunhofer diffraction10.1 Preliminary considerationsDiffraction: The deviation of light from rectilinear propagation.There is no significant physical distinction between interference and diffraction.Huygens-Fresnel Principle: Every unobstructed point of a wave front serves as a source of spherical wavelets. The amplitude of the optical field at any point beyond is the superposition of all these wavelets, considering their amplitudes and phases.
Fresnel (near field) diffraction: Fraunhofer (far field) diffraction: Both the incoming and outgoing waves approach being planar. R > a2/. R is the smaller of the two distances from the source to the aperture (with size a) and from the aperture the to observation point.Mathematical criteria: Path difference is a linear function of the aperture variables.
S
P
aR1
R2S
a
P
a sin
10.2.4 The rectangular apertureCoherent aperture:
Aperture
)exp( dSrikrE A
2
2
2222
222
222
/)(1
/)(21
/)(2/)(1
)()(
RZzYyR
RZzYyR
RZzYyRzyR
zZyYXr
ZYXR
dS=dydz P(Y,Z)rR
x
y
z
Y
Za
b
Aperture
/)(exp)exp(),( dSRZzYyikR
ikRZYE A
,'
'sin'
'sin)exp(
/)(exp)exp(),(2/
2/
2/
2/
RikRab
dydzRZzYyikR
ikRZYE
A
a
a
b
bA
Rectangular aperture:
RkbYRkaZ
2/',2/'
22
''sin
''sin)0(),(
IZYI
22
''sin
''sin)0(),(
IZYI
kbRm YRkbY 2 ... ,2 ,2/' Y minimum:
kaRm ZRkaZ 2 ... ,2 ,2/' Z minimum:
10.2.5 The circular apertureImportance in optical instrumentation: The image of a distant point source is not a point, but a diffraction pattern.
2
12
222
12
0 0
2
0 0
0
2
0 0
Aperture
/)/(4
)/(2)exp(
)/(2)exp(
cos/exp)exp(
)cos(/exp)exp(
/)(exp)exp(),(
)cos(coscossinsin
RkaqRkaqJ
RAE
RkaqJkaqRa
RikR
dRqkJR
ikR
ddRqikR
ikR
ddRqikR
ikR
dSRZzYyikR
ikRZYE
qqqZzYy
A
A
aA
aA
aA
A
21)(lim
)()]([
])cos(exp[2
)(
)cosexp(21)(
1
0
1
2
0
2
00
uuJ
uJuuJudud
dvvumviiuJ
dvviuuJ
u
mm
mm
m
m
Bessel functions:.
21
2
1
sin)sin(2)0(
/)/(2)0()(
kakaJI
RkaqRkaqJII
P(Y,Z)R
x
y
z
Y
Z
q
a
Circular aperture
21
sin)sin(2)0()(
kakaJII
J0(u)
J1(u)
u
)0(/)( II
sinka
3.83
0.018
q1
Df
aRq 22.1 2
22.11
Radius of Airy disk:
83.3/ 0)/( 11 RkaqRkaqJ
DP
f
10.2.6 Resolution of imaging systemsOverlap of two incoherent point sources:
D P1S1
S2
P2
Dfq 22.11
Rayleigh’s criterion for bare resolution:The center of one Airy disk falls on the first minimum of the other Airy disk.
Angular limit of resolution: D 22.1min
Our eyes: '1mm 2
nm 550min
Wavelength dependence: CD DVD
Question: Comparing the circular and the square aperture, why does the square aperture produce a smaller diffraction pattern? (/D vs. 1.22 /D)