Download The Physics 431 Final Exam Wed, DECEMBER 15, 2010

The Physics 431 Final Exam
Wed, DECEMBER 15, 2010
12:45 -- 2:45 p.m. 
BPS 1320 (not 1308!!!)
•
Calculators, 2 letter-size sheets “handwritten notes”
•
Graded lab reports ================ OK
•
Books, old HW, laptops
OK
NO
The exam includes topics covered throughout the semester
Greater emphasis will be placed on the 2nd half of the course
The exam consists of problems totaling 250 pts.
Show all work on exam pages — circle your answers
Grades will be posted at BPS 4238 by 12 pm Friday, December 17. Remember
your “pass code” from the final exam.
Check “Midterm Review Slides” for topics covered in Midterm I.
Review “Final Exam Topics” posted/handed out in class.
Telescope
• Object is at infinity so image is at f
• Measure angular magnification
• Length of telescope light path is sum of
focal lengths of objective and eyepiece
A Magnifying Lens
fo
M= −
fe
CA0 s θ '
= = = M.
CAe s ' θ
The exit pupil is the image of the aperture stop (AS).
Define CA0 = entrance pupil clear aperture
CAe= exit pupil clear aperture
From the diagram, it is clear that
Microscope
• x‘ is the tube length:
s2 − x '
h'
− =
M0 = =
standard x’ ranging160mm to 250mm
h
s1
f0
25
• Magnification is product of lateral
Me =
fe
magnification of objective and angular
magnification of eyepiece
− x ' 25
⋅
M total = M 0 × M e =
• Note: Image is viewed at infinity
f0 fe
Eye (Hecht 5.7.1 and Notes)
Topics/Keywords:
Eye model, Visual Acuity, Cones/Rods
accomodation, eyeglasses,
nearsightedness/myopia,
farsightedness/hyperopia
Human Eye – Gullstrand Model
Retina – Cones and Rods
Current understanding is that the 6 to 7 million cones can be
divided into "red" cones (64%), "green" cones (32%), and
"blue" cones (2%) based on measured response curves.
Numerical Aperture
Paraxial approximation
sin (θ ) ≈ tan (θ ) ≈ θ
→ NA=
D/2
1
=
f
2f /#
We will learn that
the spatial resolution limit due to diffraction ≈ 1.22×f λ /D=0.61×λ/NA [Rayleigh Criterion].
The Chief Ray
For an off-axis object, the chief ray (CR) is the ray that passes
through the center of the aperture stop. Rays that pass through
the edge of the aperture stop are marginal rays (MR).
Aperture Stop and Entrance & Exit Pupil
Aperture Stop (AS)
The aperture stop (AS) is defined to be the stop or lens ring, which physically limits
the solid angle of rays passing through the system from an on-axis object point. The
aperture stop limits the brightness of an image.
The entrance pupil of a system is the image of the aperture stop as seen from an axial
point on the object through those elements preceding the stop. (Hecht p. 171)
The exit pupil of a system is the image of the aperture stop as seen from an axial point
on the image plane through the interposed lenses, if there is any. (Hecht p. 172)
Monochromatic plane waves
Plane waves have straight wave fronts
– As opposed to spherical waves, etc.
– Suppose
E ( r ) = E 0 e ik  r

− iωt
E ( r, t ) = Re{E(r )e
λ
}
= Re{E0 eik  r e − iωt }
Ex
By
= Re{E0 ei ( k  r −ωt ) }
– E0 still contains: amplitude, polarization, phase
– Direction of propagation given by wavevector:
k = (k x , k y , k z ) where |k|=2π /λ =ω /c
– Can also define
E = ( Ex , E y , Ez )
– Plane wave propagating in z-direction
=
E ( z , t ) Re{=
E0 ei( kz −ωt ) }
1
2
{E0 ei( kz −ωt ) + E*0 e − i( kz −ωt ) }
Key words: energy, momentum, wavelength, frequency, phase, amplitude…
vz
Poynting vector & Intensity of Light S = E× H
•Poynting vector describes flows of E-M power
•Power flow is directed along this vector
•Usually parallel to k
•Intensity is equal to the magnitude of the time averaged Poyning vector: I=<S>
1239.85
cε 0 2 cε 0
2
2

ω
[
eV
]
=
S =
I ≡| E ( t ) × H ( t ) |= E =( Ex + E y )
λ[nm]
2
2
cε 0 ≈ 2.654 ×10−3 A / V
=
 1.05457266 × 10 Js
example E = 1V / m
−34
I = ? W / m2
Wave equations in a medium
The induced polarization in Maxwell’s Equations yields another term in
the wave equation:
2
2
∂
E
1
∂
E
∂ E
∂ E
− 2 2 =
0
−
=
µε
0
2
∂z
v ∂t
∂z 2
∂t 2
2
2
This is the Inhomogeneous Wave Equation.
The polarization is the driving term for a new solution to this equation.
∂2 E
∂2 E
∂2 E 1 ∂2 E
− µ 0ε 0 2 =
0
− 2 2 =
0
2
2
∂z
∂t
∂z
c ∂t
Homogeneous (Vacuum) Wave Equation
E ( z , t ) = Re{E0 ei( kz −ωt ) }
=
1
2
{E0 e
i ( kz −ωt )
+E e
*
0
− i ( kz −ωt )
= | E0 | cos ( kz − ωt )
}
c
=n
v
Phase velocity
*Phase velocity can exceed the speed of light in a
dispersive medium where the refractive index n is not
necessarily >1.
Spherical waves
A spherical wave is also a solution to Maxwell's equations and is a good model for the
light scattered by a molecule.
Note that k and r are
not vectors here!

E (r , t ) ∝ ( E0 / r ) Re{exp[i (kr − ω t )]}
•
•
where k is a scalar, and
r is the radial magnitude.
A spherical wave has spherical wave-fronts.
Unlike a plane wave, whose amplitude remains constant as it
propagates, a spherical wave weakens. Its irradiance goes as 1/r2.
Interference [Hecht 9.1-9.4, 9.7.2; Fowles 3.1-3.1; Notes]
E ( r ) = E0 e
Michelson Interferometer
ik  r
E ( r, t ) = Re{E(r )e − iωt }
= Re{E0 eik  r e − iωt }
= Re{E0 ei ( k  r −ωt ) }
Consider the Optical Path Difference (OPD)
Or simply the superposition of two plane waves
=
E ( r ) E1eik •r + E2 eik •r
1
1
2
2
I= | E ( r ) |=2 E × E*
Key words/Topics:
Michelson Interferometer, Dielectric thin film, Anti-reflection coating,
Fringes of equal thickness, Newton rings.
The Michelson Interferometer and Spatial Fringes
x
z
•
•
•
•
Input
beam
Mirror
Suppose we misalign the mirrors
so the beams cross at an angle
Beamwhen they recombine at the beam
splitter
splitter. And we won't scan the delay.
Fringes
Mirror
• If the input beam is a plane wave, the cross term becomes:
Re•{ E0 exp [i (ω t − kz cos θ − kx sin θ ] E0* exp [ −i (ω t − kz cos θ + kx sin θ ]}
∝ Re {exp [ −2ikx sin θ ]}
∝ cos(2kx sin θ )
Crossing beams maps delay
onto position.
Fringes (in position)
I
x
x
z
Input
beam
Mirror
•Suppose we change one
arm’s path length.
Beamsplitter
Fringes
Re { E0 exp [i (ω t − kz cos θ − kx sin θ + 2kd ] E0* exp [ −i (ω t − kz cos θ + kx sin θ ]}
Mirror
∝ Re {exp [ −2ikx sin θ + 2kd ]}
∝ cos(2kx sin θ + 2kd )
Fringes (in position)
I
The fringes will shift in
phase by 2kd.
x
The Unbalanced Michelson Interferometer
can sensitively measure phase vs. position.
See HW#8 Problem #1
Spatial fringes distorted by
a soldering iron tip in one
path
Placing an object in one arm of a misaligned
Michelson interferometer will distort the
spatial fringes.
Input
beam
θ
Mirror
Beamsplitter
Mirror
• Phase variations of a small fraction of a wavelength can be measured.
Michelson interferometers: the compensator plate
Input
beam
Beamsplitter
Output
beam
Mirror
So a compensator plate
(identical to the beam
splitter) is usually added
to equalize the path
length through glass.
If reflection occurs off the
front surface of beam
splitter, the transmitted
beam passes through
beam splitter three times;
the reflected beam passes
through only once.
Mirror
Interference Fringes and Newton Rings
Phase shift on reflection at an interface
Near-normal incidence
π phase shift if ni < nt
0 (or 2π phase shift) if ni > nt
Young’s double slit interference experiment
order m maxima occur at:
mλ ≈ a sin θ m ≈ a
ym
s
Diffraction
Diffraction Geometry
We wish to find the light electric field after a screen with a hole in it.
This is a very general problem with far-reaching applications.
y0
A(x0,y0)
y1
x0
P1
0
Incident
wave
x1
This region is assumed to be
much smaller than this one.
What is E(x1,y1) at a distance z from the plane of the aperture?
Diffraction Solution
The field in the observation plane, E(x1,y1), at a distance z from the aperture
plane is given by:
E ( x1 , y1 , z=
)
∫∫
h( x1 − x0 , y1 − y0 , z ) E ( x0 , y0 ) dx0 dy0
A (x0 , y0 )
where :
1 exp(ikr01 )
h( x1 − x0 , y1 − y0 , z ) =
iλ
r01
and :
r01 =
z 2 + ( x0 − x1 ) + ( y0 − y1 )
2
2
Spherical
wave
A very complicated result! And we cannot approximate r01 in the exp by z
because it gets multiplied by k, which is big, so relatively small changes in r01
can make a big difference!
Fraunhofer Diffraction: The Far Field
We can approximate r01 in the denominator by z, and if D is the size of the
aperture, D 2 ≥ x02 + y02, so when k D2/ 2z << 1, the quadratic terms << 1, so
we can neglect them:
r01 =
z 2 + ( x0 − x1 ) + ( y0 − y1 ) ≈ z 1 + ( x0 − x1 ) / 2 z 2 + ( y0 − y1 ) / 2 z 2 


2
2
2
2
kr01 ≈ kz + k ( x02 − 2 x0 x1 + x12 ) / 2 z + k ( y02 − 2 y0 y1 + y12 ) / 2 z
Small, so neglect
these terms.
E ( x1 , y1 )
 x12 + y12 
exp(ikz )
exp ik

2
iλ z
z


Independent of x0 and y0,
so factor these out.
∫∫
 ik

exp − ( x0 x1 + y0 y1 )  E ( x0 , y0 ) dx0 dy0
 z

A ( x0 , y0 )
This condition means going a distance away: z >> kD2/2 = πD2/λ
If D = 1 mm and λ = 1 micron, then z >> 3 m.
Fraunhofer Diffraction
We’ll neglect the phase factors, and we’ll explicitly write the aperture function
in the integral:
∞ ∞
E ( x1 , y1 ) ∝
∫∫
 ik

exp − ( x0 x1 + y0 y1 )  A( x0 , y0 ) E ( x0 , y0 ) dx0 dy0
 z

−∞ −∞
This is just a Fourier Transform!
E(x0,y0) = constant if a plane wave
Interestingly, it’s a Fourier Transform from position, x0, to another position
variable, x1 (in another plane). Usually, the Fourier “conjugate variables” have
reciprocal units (e.g., t & ω, or x & k). The conjugate variables here are really x0
and kx = kx1/z, which have reciprocal units.
So the far-field light field is the Fourier Transform of the apertured field!
Diffraction: single, double, multiple slits
Study Guide: Hecht Ch. 10.2.1-10.2.6 (detailed lengthy discussions),
Fowles Ch. 5 (short but clear presentation), or Lecture Notes
2
 sin β 
I ( β ) = I (0) 

 β 
kb
b
β =
sin θ π sin θ
=
λ
2
Java applet – Single Slit Diffraction
http://www.walter-fendt.de/ph14e/singleslit.htm
Diffraction: Double and Multiple Slits
2
2
1
1
 sin β   sin N γ 
I (θ ) = I ( 0 ) 
kb sin θ ; γ
ka sin θ
=
 =
 β
2
2
 β   N sin γ 
See also
http://demonstrations.wolfram.com/MultipleSlitDiffractionPattern/ and
http://wyant.optics.arizona.edu/multipleSlits/multipleSlits.htm
Fraunhofer diffraction from
two slits (Fourier Transform)
w
-a
w
0
a
x0
A(x0) = rect[(x0+a)/w] + rect[(x0-a)/w]
E ( x1 ) ∝ F { A( x0 )}
∝ sinc[w(kx1 / z ) / 2]exp[+ia(kx1 / z )] +
sinc[w(kx1 / z ) / 2]exp[−ia (kx1 / z )]
E ( x1 ) ∝ sinc( wkx1 / 2 z ) cos(akx1 / z )
kx1/z
Diffraction from one- and two-slit screens
Fraunhofer diffraction patterns
One slit
Two slits
Diffraction Gratings
•Scattering ideas explain what happens when
light impinges on a periodic array of grooves.
Constructive interference occurs if the delay
between adjacent beamlets is an integral
number, m, of wavelengths.
a
Scatterer
D
C
θm
θm
θi a
Path difference: AB – CD = mλ
a [sin(θ m ) − sin(θ i ) ] =
mλ
Incident wavefront
A
θi
Scatterer
where m is any integer.
A grating has solutions of zero, one, or many values of m, or orders.
Remember that m and θm can be negative, too.
B
Potential
diffracted wavefront
AB = a sin(θm)
CD = a sin(θi)
Diffraction orders
Because the diffraction angle depends on λ,
different wavelengths are separated in the
nonzero orders.
Diffraction angle, θm(λ)
Incidence angle,
θi
First order
Zeroth order
No wavelength
dependence
occurs in zero
order.
Minus first
order
The longer the wavelength, the larger its deflection in each nonzero order.
The Diffraction Grating
Hecht 10.2.8 or Fowles Ch. 5 p.123 (handout)
Grating Equation
(Optical Path Difference OPD= m λ)
a ( sin θ m − sin θi ) =
mλ
a sin θ m = mλ
Normal incidence θi =0
The chromatic/spectral resolving power of a grating
R≡
λ
=
mN
∆λ
m is the order number, and
N is the total number of gratings.
Uniform Rectangular Aperture
a
b
 sin α
I (θ ) = I ( 0 ) 
 α
2
1
1
  sin β 
ka sin θ ; β
kb sin θ
=

 α
=
2
2
  β 
2
Uniform Circular Aperture
R
R
 2 J1 ( ρ ) 
I (θ ) = I ( 0 ) 

ρ


ρ
kR
=
sin θ ; k
2π
A circular aperture
yields a diffracted
"Airy Pattern,"
which involves a
Bessel function.
λ
2
Diffraction from small and large circular apertures
Far-field intensity
pattern from a
small aperture
Recall the Scale Theorem!
This is the Uncertainty
Principle for diffraction.
Far-field
intensity pattern
from a large
aperture
Wave optics of a lens
The spot diameter is
λf
λ
=
d 1.22
= 1.22
θ
w
The resolution of the lens as defined by the “Rayleigh” criterion is
d / 2 = 0.61λ / θ
For a small angle θ,
λ / sin θ 0.61
=
d / 2 0.61
=
λ
NA
Gaussian Beam Optics (only eq. (4))
Basic Fourier Optics (~30-50 points)
Fourier Transform Notation
There are several ways to denote the Fourier transform of a function.
If the function is labeled by a lower-case letter, such as f,
we can write:
f(t) → F(ω)
If the function is already labeled by an upper-case letter, such as E, we
can write:
E (t ) → F {E (t )}
or:
∩
Sometimes, this symbol is
used instead of the arrow:
E (t ) → E (ω )
Example: the Fourier Transform of a rectangle function: rect(t)
F (ω ) =
1/ 2
∫
exp(−iωt )dt =
−1/ 2
1
[exp( −iωt )]1/−1/2 2
−iω
1
[exp(−iω / 2) − exp(iω/2)]
−iω
1 exp(iω / 2) − exp(−iω/2)
=
(ω/2)
2i
sin(ω/2)
=
(ω/2)
F=
(ω ) sinc(ω/2)
F(ω)
Imaginary
Component = 0
ω
The Fourier Transform of δ(t) is 1.
∞
∫ δ (t ) exp(−iω t ) dt =exp(−iω [0]) =1
−∞
δ(t)
0
1
ω
t
∞
And the Fourier Transform of 1 is 2πδ(ω):
∫ 1 exp(−iω t ) dt =
−∞
2πδ(ω)
1
t
0
ω
2π δ (ω )
The Fourier transform of exp(iω0 t)
∞
=
F {exp(iω0 t )}
∞
=
∫
∫
exp(iω0 t ) exp(−i ω t ) dt
−∞
exp(−i [ω − ω0 =
] t ) dt
2π δ (ω − ω0 )
−∞
exp(iω0t)
Im
0
Re
0
F {exp(iω0t)}
t
t
0
ω0
ω
The function exp(iω0t) is the essential component of Fourier analysis.
It is a pure frequency.
The Fourier transform of cos(ω0 t)
∞
=
F {cos(ω0t )}
1
2
=
=
1
2
∞
∞
∫
∫ cos(ω t ) exp(−i ω t ) dt
0
−∞
[exp(i ω0 t ) + exp(−i ω0 t )] exp(−i ω t ) dt
−∞
∫ exp(−i [ω − ω ]t ) dt
0
−∞
+
1
2
∞
∫ exp(−i [ω + ω ]t ) dt
0
−∞
= π δ (ω − ω0 ) + π δ (ω + ω0 )
F {cos(ω0t )}
cos(ω0t)
0
t
−ω0
0
+ω0
ω
Scale Theorem
F { f (at )} = F (ω /a ) / a
The Fourier transform
of a scaled function, f(at):
∞
Proof:
∫
=
F { f (at )}
f (at ) exp(−iω t ) dt
−∞
Assuming a > 0, change variables: u = at
∞
∫
F
=
{ f (at )}
f (u ) exp(−iω [ u /a ]) du / a
−∞
∞
=
∫ f (u) exp(−i [ω /a] u) du / a
−∞
= F (ω /a ) / a
If a < 0, the limits flip when we change variables, introducing a
minus sign, hence the absolute value.
F(ω)
f(t)
The Scale
Theorem
in action
The shorter
the pulse,
the broader
the spectrum!
This is the essence
of the Uncertainty
Principle!
Short
pulse
t
ω
t
ω
t
ω
Mediumlength
pulse
Long
pulse
The Fourier Transform of a sum of two functions
F(ω)
f(t)
ω
t
G(ω)
g(t)
F {a f (t ) + b g (t )} =
aF { f (t )} + bF {g (t )}
t
f(t)+g(t)
Also, constants factor out.
t
ω
F(ω) +
G(ω)
ω
Shift Theorem
The Fourier transform of a shifted function, f (t − a ) :
F
{ f (t − a)} =
exp(−iω a ) F (ω )
Proof :
∞
F
{ f ( t − a )}= ∫
f (t − a ) exp(−iωt )dt
−∞
Change variables : u = t − a
∞
∫
f (u ) exp(−iω[u + a ])du
−∞
∞
exp(−iω a ) ∫ f (u ) exp(−iωu )du
=
−∞
= exp(−iω a ) F (ω )
Fourier Transform with respect to space
If f(x) is a function of position,
=
F (k )
∫
∞
−∞
f ( x) exp(−ikx) dx
x
F {f(x)} = F(k)
We refer to k as the spatial frequency.
k
Everything we’ve said about Fourier transforms between the t and ω
domains also applies to the x and k domains.
The 2D Fourier Transform
F
=
(2){f(x,y)}
∫∫
f(x,y)
= F(kx,ky)
y
f(x,y) exp[-i(kxx+kyy)] dx dy
If f(x,y) = fx(x) fy(y),
then the 2D FT splits into two 1D FT's.
But this doesn’t always happen.
x
F
(2){f(x,y)}
Fibers (will not be covered in 2010)
and
.
NA
=
(
n 2f
− nc2
)
1/2
The number of modes in a stepped-index fiber is
N m ≈ 12 (π D× NA / λ0 )
2