The Physics 431 Final Exam Wed, DECEMBER 3, 2014
3:00 – 5:00 p.m. BPS 1425
•
Calculators, 4 leQer-­‐size sheets “handwriQen notes”
•
Graded lab reports and HW ===========è OK •
Textbook ================è OK •
HW soluYons and laptops OK NO The exam includes topics covered throughout the semester The exam consists of problems totaling 250 pts. Show all work on exam pages Grades will be posted at BPS 4238 & Angel by 12 pm Friday, December 12. Remember your “pass code” from the final exam if you intend to check your grades in my office. Review “Final Exam Topics” handed out in class. Index of refraction and ‘speed’ of light
The speed of light in vacuum is a physical constant.
c = 299 792 458 m/s (exact) ~ 3x108 m/s
In a medium, light generally propagates more slowly.
–  in air: v = c/1.0003
–  in water: v = c/1.33
–  in glass: v = c/1.52
nair = 1.00
nwater = 1.33
nglass = 1.52
In general:
v = c/n is the “phase velocity”
wavelength/n
frequency is the same (in linear optics)
n also depends on the wavelength è dispersion.
Snell’s Law Frequency is the same. 1.  Huygens’ principle
2.  Fermat’s principle
3.  Interference of all possible paths of light wave from source to observer
— it results in destructive interference everywhere except extrema of phase
(where interference is constructive)—which become actual paths.
4.  Application of the general boundary conditions of Maxwell equations for
electromagnetic radiation. è amplitude of reflected and refractive waves
[Chapter 23]
5. Conservation of momentum based on translation symmetry considerations
Derive Snell’s Law by Transla?on Symmetry A homogeneous surface can not change the transverse momentum.
The propagation vector is proportional to the photon's momentum.
The transverse wave number must remain the same.

'

k1 i x̂ = k i x̂ = k2 i x̂
k1 sin θ1 = k 'sin θ ' = k2 sin θ 2
2π
2π
2π
k1 = k ' =
=
=
n1 = k0 n1
λ1 λ0 / n1 λ0
2π 2π
k2 =
=
n2
λ2 λ0
n1k0sinθ1 = n2k0sinθ2
n1sinθ1 = n2sinθ2
E = pc
p = k
x̂
Reflec?on and Transmission (Fresnel’s equa?ons) Can be deduced from the applicaYon of boundary condiYons of EM waves. Reflec?on and Transmission of Energy @ dielectric interfaces Polariza?on by Reflec?on hMp://hyperphysics.phy-­‐astr.gsu.edu/hbase/phyopt/polar.html Polariza?on by scaMering (Rayleigh scaMering/Blue Sky) Ray Diagrams for Spherical Mirrors R<0
1 1 1
R
+ = , f =−
s s' f
2
s'
m=−
lateral magnification
s
1 1 1
R
+ = , f =−
so si f
2
hi
si
m= =−
ho
so
y
object origin= V x
image x'
y'
origin= V R>0
Ray Diagrams
Imaging Forma?on by a Mirror Refrac?on at a Spherical Surface y'
y
image x'
object x
origin= V R<0
R>0
n1 n2 n2 − n1
+ =
s s'
R
hi
n1s '
m= =−
ho
n2 s
When R → ∞ ( i.e. a plane surface )
⎛n ⎞
s' = −⎜ 2 ⎟ s
⎝ n1 ⎠
m = +1
EXAMPLE A goldfish in a bowl EXAMPLE A goldfish in a bowl Example of refrac?on by spherical surfaces Explain the imaging forma?on by a cylinder filled with water. Be specific (i.e. use ‘real’ values’), See Example 2-­‐2 (page 34) Summary of Image Forma?on: Spherical Mirrors & Thin Lenses y
y'
object image x
origin= V x'
y
y'
image object x
x'
origin= V y
y'
object image x
o x'
n1=1 (air) n2=n The lensmaker’s formula A “Thin” lensè d is negligible the lensmaker’s formula y
y'
image object x
x'
origin= V The refrac?ve power of a lens of focal length f D [diopters] =
1
f [in meters]
Newtonian Equa?on for the Thin Lens =si
=xi
=xo
1 1 1
= +
f s0 si
x0 xi = f 2
yi
si
ML ≡
=−
y0
so
lateral magnification
dxi
f2
MT ≡
=− 2
dx0
x0
transverse magnification
Major Rays Thin Lens Combina?on è Sequen?al Imaging Reading Assignments 1.  Example 2-­‐3 (page 38) 2. Java Applets hMp://silver.neep.wisc.edu/~shock/tools/ray.html (See also Lab #1 Appendix) Simple Magnifiers αM
h / s 25
=
=
α 0 h / 25 s
25
M=
[ cm ] image at infinity (s = f )
f
25
M=
+ 1 [ cm ] image at normal near point (s ' = −25cm)
f
M≡
Magnifying Glass/Virtual Image The Telescope •  A simple telescope contains a large-diameter objective lens which
collects parallel rays from a distant object and forms a real, inverted
image at distance s' = fobj.
•  The focal length of a telescope objective is very nearly the length of
the telescope tube.
•  The eyepiece functions as a simple magnifier.
•  The viewer observes an inverted image.
•  The angular magnification of a telescope is
Telescope α'
f
=− o
α
fe
L = fo + fe
D
Dex = obj the diameter of the exit pupil
M
M=
The Microscope The Microscope •  A specimen to be observed is placed on the stage of
a microscope, directly beneath the objective, a
converging lens with a relatively short focal length.
•  The objective creates a magnified real image that is
further enlarged by the eyepiece.
•  The lateral magnification of the objective is
•  Together, the objective and eyepiece produce a total
angular magnification
Numerical Aperture Paraxial approxima?on sin (θ ) ≈ tan (θ ) ≈ θ
→ NA =
D/2
1
=
f
2f /#
The spa?al resolu?on limit due to diffrac?on ≈ 1.22×f λ /D=0.61×λ/NA [Rayleigh Criterion]. The Microscope: Objec?ve and Numerical Aperture (NA) NA = n sin α
Oil-­‐immersion microscope f-­‐number and irradiance/intensity [W/m2] The light intensity on the detector is
related to the lens’s f-number by
2 ≈ 1.4
A lens both focuses and diffracts the light The ResoluYon of OpYcal Instruments The minimum spot size to which a lens can focus light of wavelength
λ is
where D is the diameter of the circular aperture of the lens, and f is the
focal length.
In order to resolve two points, their angular separation must be greater
than θmin, where
is called the angular resolution of the lens.
Important Concepts Four diffrac?on-­‐limited lenses focus plane waves of light with the same wavelength λ. Rank order, from largest to smallest, the spot sizes w1 to w4. A.
B.
C.
D.
E.
w2 = w3 > w4 > w1
w1 = w2 > w3 > w4
w4 > w3 > w1 = w2
w1 > w4 > w2 = w3
w2 > w1 = w3 > w4
Eye Topics/Keywords: Eye model, Visual Acuity, Cones/Rods accommoda?on, eyeglasses, nearsightedness/myopia, farsightedness/
hyperopia Human Eye – Gullstrand Model Re?na – 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. Par?cle Nature of Light = 4.1357 × 10 −15 eV s
N [#/ s] =
Power
[Watt = J / s]
=
Energy / photon
[J ]
N is the number of photons per second.
See Example 1-­‐2, page 10 in Pedrov3. Wave-­‐Par?cle Duality of Light Maxwell’s theory
•  Maxwell showed that E and
B fields could sustain
themselves (free from
charges or currents) if they
took the form of an
electromagnetic (EM) wave.
•  Maxwell’s theory predicted
that an EM wave would
travel with speed:
v em =
1
ε 0 µ0
v em = c = speed of light
Light is an
electromagnetic
wave!
Anima?on hMp://www.walter-­‐fendt.de/ph14e/emwave.htm Electromagne?c (EM) Waves •  EM waves can travel through
empty space (vacuum); no
medium is necessary!
•  The speed of light c in empty
space is
1
c=
ε0µ0
= 299,792,458 m/s
c = 3x108 m/s
€
•  EM waves carry energy and momentum •  The speed is constant so the frequency f is determined by the wavelength λ and speed of light c: f = c /λ
ProperYes of ElectromagneYc Waves Any electromagne?c wave must sa?sfy four basic condi?ons: 1.  The fields E and B and are perpendicular to the
direction of propagation vem.Thus an
electromagnetic wave is a transverse wave.
2.  E and B are perpendicular to each other in a manner
such that E × B is in the direction of vem.
3.  The wave travels in vacuum at speed vem = c
4.  E = cB at any point on the wave.
ProperYes of ElectromagneYc Waves The energy flow of an electromagne?c wave is described by the PoynYng vector defined as The magnitude of the Poyn?ng vector is The intensity of an electromagne?c wave whose electric field amplitude is E0 is Energy and Intensity S = E×H
•  PoynYng vector describes flows of E-­‐M power •  Power flow is directed along this vector (usually parallel to k) •  Intensity is average energy transfer (i.e. the ?me averaged Poyning vector: I=<S>=P/A, where P is the power (energy transferred per second) of a wave that impinges on area A. sin 2 ( kx − ω t )
= cos 2 ( kx − ω t ) =
1239.85
cε 0 2 cε 0
2
2
S = I ≡| E ( t ) × H ( t ) |=
E =
Ex + E y ) hω[eV ] =
(
λ[nm]
2
2
cε 0 ≈ 2.654 ×10 −3W /V
h = 1.05457266 × 10 Js
example E = 1V / m
−34
I = ? W / m2
1
2
EXAMPLE: The electric field of a laser beam Polariza?on Polarization
is defined
with respect
to the E-field.
Polariza?on & Plane of Polariza?on A Polarizing Filter Malus’s Law Suppose a polarized light wave of intensity I0 approaches a polarizing filter. θ is the angle between the incident plane of polariza?on and the polarizer axis. The transmiMed intensity is given by Malus’s Law: If the light incident on a polarizing filter is unpolarized, the transmiMed intensity is In other words, a polarizing filter passes 50% of unpolarized light and blocks 50%. Polariza?on: Summary ŷ
!
ŷ E

iδ
E = E x eiδ x x̂ + E y e y ŷ
x̂
x̂
linear polariza?on y-­‐direc?on right circular polariza?on lez circular polariza?on lez ellip?cal polariza?on Phase difference Phase difference = 00 r
Ex
r
Ex
ẑ
!
Ey
Phase difference è 90 0 (π/2, λ/4) ẑ
!
Ey
ẑ
ẑ
Phase difference è 180 0 (π, λ/2) r
Ex
ẑ
!
Ey
ẑ
Wave Number, Wave Vector, and Momentum Chapter 4, Pedrov^3) Wave equa?ons in a medium The induced polariza?on in Maxwell’s Equa?ons yields another term in the wave equa?on: 2
2
2
2
∂
E
1
∂
E
∂
E
∂
E
− 2 2 =0
− µε 2 = 0
2
2
∂z
v ∂t
∂t
∂z
This is the Inhomogeneous Wave EquaYon. The polariza?on is the driving term for a new solu?on to this equa?on. ∂2 E
∂2 E
− µ0ε 0 2 = 0
2
∂z
∂t
∂2 E 1 ∂2 E
− 2 2 =0
2
∂z
c ∂t
Homogeneous (Vacuum) Wave EquaYon E ( z, t ) = Re{E0 ei( kz −ωt ) }
= {E0 e
1
2
i ( kz −ωt )
*
0
+E e
Phase velocity }
*Phase velocity can exceed the speed of light in a dispersive medium where the refrac?ve index n is not necessarily >1. − i ( kz −ωt )
=| E0 | cos ( kz − ωt )
c
=n
v
Monochroma?c plane waves Plane waves have straight wave fronts –  As opposed to spherical waves, etc. –  Suppose E ( r ) = E0 eik gr
è − iωt
E
r
,
t
=
Re{
E
(
r
)
e
}
(
)
= Re{E0 eik gr e − iωt }
= Re{E0 ei ( k gr −ωt ) }
–  E0 s?ll contains: amplitude, polariza?on, phase –  Direc?on of propaga?on given by wavevector: λ
By
k = (k x , k y , k z ) where |k|=2π /λ =ω /c
–  Can also define E = ( Ex , E y , Ez )
–  Plane wave propaga?ng in z-­‐direc?on E ( z, t ) = Re{E0 ei( kz −ωt ) } = 12 {E0 ei( kz −ωt ) + E*0 e − i( kz −ωt ) }
Key words: energy, momentum, wavelength, frequency, phase, amplitude… Ex
vz
Spherical waves A spherical wave is also a solu?on to Maxwell's equa?ons and is a good model for the light scaMered by a molecule. Note that k and r are not vectors here! r
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. Young’s double slit interference experiment order m maxima occur at: ym
mλ ≈ a sin θ m ≈ a
s
Interference E ( r ) = E0 e
Michelson Interferometer ik gr
E ( r, t ) = Re{E(r )e − iωt }
= Re{E0 eik gr e − iωt }
= Re{E0 ei ( k gr −ωt ) }
Consider the Op?cal Path Difference (OPD) Or simply the superposi?on 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, AnY-­‐reflecYon coaYng, Fringes of equal thickness, Newton rings. The Michelson Interferometer and Spa?al Fringes x
z
•
•
•
•
Input"
beam"
Suppose we misalign the mirrors Mirror"
so the beams cross at an angle Beam-"
when they recombine at the beam splitter"
spliMer. And we won't scan the delay. Fringes Mirror"
•  If the input beam is a plane wave, the cross term becomes: *
Re• {E 0 exp
i
(
ω
t
−
kz
cos
θ
−
kx
sin
θ
E
[
]
0 exp [ −i (ω t − kz cos θ + kx sin θ ]}
∝ Re {exp [ −2ikx sin θ ]}
∝ cos(2kx sin θ )
Crossing beams maps delay onto posi?on. Fringes (in posiYon) I
x
x
z
Input"
beam"
Mirror"
• Suppose we change one arm’s path length. Beam-"
splitter"
Fringes Mirror"
Re {E0 exp [i (ω t − kz cos θ − kx sin θ + 2kd ] E0* exp [ −i (ω t − kz cos θ + kx sin θ ]}
∝ Re {exp [ −2ikx sin θ + 2kd ]}
∝ cos(2kx sin θ + 2kd )
Fringes (in posiYon) I
The fringes will shiz in phase by 2kd. x
Michelson interferometers: the compensator plate Input"
beam"
Beam-"
splitter"
Output"
beam"
Mirror"
So a compensator plate (iden?cal to the beam spliMer) is usually added to equalize the path length through glass. If reflec?on occurs off the front surface of beam spliMer, the transmiMed beam passes through beam spliMer three ?mes; the reflected beam passes through only once. Mirror"
Interference Fringes and Newton Rings Phase shiz on reflec?on at an interface Near-­‐normal incidence π phase shiz if ni < nt 0 (or 2π phase shiz) if ni > nt DiffracYon Geometry We wish to find the light electric field azer a screen with a hole in it. This is a very general problem with far-­‐reaching applica?ons. 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? Fraunhofer DiffracYon: 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 quadra?c terms << 1, so we can neglect them: 2
2
2
2
r01 = z 2 + ( x0 − x1 ) + ( y0 − y1 ) ≈ z ⎡1 + ( x0 − x1 ) / 2 z 2 + ( y0 − y1 ) / 2 z 2 ⎤
⎣
⎦
kr01 ≈ kz + k ( x02 − 2x0 x1 + x12 ) / 2z + k ( y02 − 2 y0 y1 + y12 ) / 2z
Small, so neglect these terms. ⎡ x12 + y12 ⎤
exp(ikz )
E ( x1 , y1 ) =
exp ⎢ik
⎥
iλ z
2
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 condi?on means going a distance away: z
If D = 1 mm and λ = 1 µ m
, then z >> 3 m. >> kD 2 / 2 = π D 2 / λ
DiffracYon SoluYon The field in the observa?on 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 :
and :
1 exp(ikr01 )
h( x1 − x0 , y1 − y0 , z ) =
iλ
r01
2
2
r01 = z + ( x0 − x1 ) + ( y0 − y1 )
2
Spherical wave A very complicated result! And we cannot approximate r01 in the exp by z because it gets mul?plied by k, which is big, so rela?vely small changes in r01 can make a big difference! Fraunhofer DiffracYon We’ll neglect the phase factors, and we’ll explicitly write the aperture func?on in the integral: ∞ ∞
E ( x1 , y1 ) ∝
∫∫
⎧ ik
⎫
exp ⎨− ( x0 x1 + y0 y1 )⎬ A( x0 , y0 ) E ( x0 , y0 ) dx0 dy0
⎩ z
⎭
−∞ −∞
E(x ,y ) = constant if a plane wave 0 0
This is just a Fourier Transform! Interes?ngly, it’s a Fourier Transform from posi?on, x0, to another posi?on 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! Diffrac?on: Summary Diffrac?on: single, double, mul?ple slits Study Guide: Hecht Ch. 10.2.1-­‐10.2.6 (detailed lengthy discussions), Fowles Ch. 5 (short but clear presenta?on), or Lecture Notes 2
⎛ sin β ⎞
I ( β ) = I ( 0 ) ⎜
⎟
⎝ β ⎠
kb
b
β = sin θ = π sin θ
2
λ
Java applet – Single Slit Diffrac?on hMp://www.walter-­‐fendt.de/ph14e/
singleslit.htm Diffrac?on: Double and Mul?ple Slits 2
2
1
1
⎛ sin β ⎞ ⎛ sin N γ ⎞
I (θ ) = I ( 0 ) ⎜
⎟ ⎜
⎟ β = kb sin θ ; γ = ka sin θ
2
2
⎝ β ⎠ ⎝ N sin γ ⎠
See also hMp://demonstra?ons.wolfram.com/Mul?pleSlitDiffrac?onPaMern/ and hMp://wyant.op?cs.arizona.edu/mul?pleSlits/mul?pleSlits.htm Fraunhofer diffracYon 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
DiffracYon from one-­‐ and two-­‐slit screens One slit Two slits Fraunhofer diffrac?on paMerns Diffrac?on orders Because the diffrac?on angle depends on λ, different wavelengths are separated in the nonzero orders. Diffrac?on 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 deflec?on in each nonzero order. Diffrac?on Gra?ngs • ScaMering ideas explain what happens when light impinges on a periodic array of grooves. Construc?ve interference occurs if the delay between adjacent beamlets is an integral number, m, of wavelengths. ScaQerer a
D
θm
C
θm
Path difference: AB – CD = mλ
a [sin(θ m ) − sin(θi )] = mλ
θi
Incident wave-­‐ a
front A
θi
ScaQerer where m is any integer. A gra?ng has solu?ons of zero, one, or many values of m, or orders. Remember that m and θm can be nega?ve, too. B
PotenYal diffracted wave-­‐
front AB = a sin(θm)
CD = a sin(θi)
The Diffrac?on Gra?ng GraYng EquaYon (OpYcal Path Difference OPD= m λ) a (sin θm − sin θi ) = mλ
a sin θ m = mλ
Normal incidence θi =0 The chromaYc/spectral resolving power of a graYng λ
R≡
= mN
Δλ
m is the order number, and N is the total number of gra?ngs. Uniform Rectangular Aperture a b 2
2
1
⎛ sin α ⎞ ⎛ sin β ⎞
I (θ ) = I ( 0 ) ⎜
⎟ α = ka sin θ ;
⎟ ⎜
2
⎝ α ⎠ ⎝ β ⎠
1
β = kb sin θ
2
Uniform Circular Aperture R R ⎛ 2 J1 ( ρ ) ⎞
I (θ ) = I ( 0 ) ⎜
⎟
ρ
⎝
⎠
ρ = kR sin θ ; k =
2π
λ
A circular aperture yields a diffracted "Airy PaMern," which involves a Bessel func?on. 2
Change in curvature of wavefronts by a thin lens Wave op?cs of a lens The spot diameter is (for an aperture with a radius w,
diameter D).
λf
λ
λf
d = 1.22
= 1.22 = 2.44
w
θ
D
The resolution of the lens as defined by the “Rayleigh” criterion is
d / 2 = 0.61λ / θ
For a small angle θ,
d / 2 = 0.61λ / sin θ = 0.61
λ
NA
Gaussian Beam Op?cs Basic Fourier Op?cs Fourier Transform NotaYon There are several ways to denote the Fourier transform of a func?on. If the func?on is labeled by a lower-­‐case leMer, such as f, we can write: f(t) ==> F(ω)
If the func?on is already labeled by an upper-­‐case leMer, such as E, we can write: or: E (t ) → F {E (t )}
 ω)
E(t) → E(
Example: the Fourier Transform of a rectangle funcYon: rect(t) 1/ 2
F (ω ) =
∫
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 a δελτα φυνχτιον is 1. ∞
∫ δ (t ) exp(−iωt ) dt = exp(−iω [0]) = 1
−∞
δ (t )
1
0 t
ω
∞
And the Fourier Transform of 1 is 2πδ(ω):
∫ 1 exp(−iωt ) dt = 2π δ (ω )
−∞
2πδ (ω )
1
t
0 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
Re
F {exp(iω0t)}
t
0
t
0
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)
t
0
-ω0 0 +ω0 ω Scale Theorem The Fourier transform of a scaled func?on, f(at): F { f (at )} = F (ω /a) / a
∞
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 acYon 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 funcYons f(t)
F(ω)
t
g(t)
F {a f (t ) + b g (t )} =
aF { f (t )} + bF {g (t )}
G(ω)
t
f(t)+g(t)
Also, constants factor out.
ω
t
ω
F(ω) +
G(ω)
ω
Shift Theorem
The Fourier transform of a shifted function, f (t − a) :
F
{ f (t − a)} = exp(−iωa)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) 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)}
y