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Physics 408
Optics
J Kiefer
© 2011
Table of Contents
TABLE OF CONTENTS ............................................................................................................. 2
I.
A.
GEOMETRICAL OPTICS .................................................................................................. 4
Huygens.................................................................................................................................. 4
1. Huygens’ Principle.............................................................................................................. 4
2. “Laws” of Reflection and Refraction .................................................................................. 4
B.
1.
2.
Fermat .................................................................................................................................... 7
Path Length ......................................................................................................................... 7
Fermat’s Principle ............................................................................................................... 8
1.
2.
Refracting Surfaces ............................................................................................................... 9
Spherical surfaces ............................................................................................................... 9
Imagery ............................................................................................................................. 12
C.
D.
Optical Systems ................................................................................................................... 14
1. Ray Tracing ....................................................................................................................... 14
2. Matrices............................................................................................................................. 15
E.
1.
2.
Mirrors ................................................................................................................................. 16
Planar Mirrors ................................................................................................................... 16
Curved Mirrors.................................................................................................................. 17
1.
2.
Prisms ................................................................................................................................... 18
Dispersing Prisms ............................................................................................................. 18
Reflecting Prisms .............................................................................................................. 20
F.
G. Fiber Optics ......................................................................................................................... 21
1. Large Diameter Fiber Optics ............................................................................................ 21
2. Sending Signals on Glass Optical Fiber............................................................................ 23
II. WAVE MECHANICS ........................................................................................................ 27
A.
1.
2.
3.
B.
Wave Equation .................................................................................................................... 27
Wave Profile ..................................................................................................................... 27
Harmonic Waves ............................................................................................................... 27
Three Dimensional Wave Forms ...................................................................................... 28
Fourier Analysis .................................................................................................................. 30
1. Superposition of Waves .................................................................................................... 30
2. Waves Having Different Frequencies ............................................................................... 32
3. Fourier Sums ..................................................................................................................... 33
2
C.
Fraunhoffer Diffraction...................................................................................................... 37
1. Far Field (Fraunhoffer) Diffraction .................................................................................. 37
2. Appertures ......................................................................................................................... 38
D.
1.
2.
Fresnel Diffraction .............................................................................................................. 40
Fresnel Zones .................................................................................................................... 40
Fresnel Lens ...................................................................................................................... 42
1.
2.
3.
Interference ......................................................................................................................... 43
Coherent Interference........................................................................................................ 43
Wave Front Splitting ......................................................................................................... 43
Amplitude Splitting........................................................................................................... 44
E.
F.
Vector Wave Theory ........................................................................................................... 44
G. Polarization .......................................................................................................................... 44
III.
A.
1.
2.
3.
EXTRA............................................................................................................................. 46
Absorption and Emission of Radiation ............................................................................. 46
Transition Coefficients...................................................................................................... 46
Transition Rates ................................................................................................................ 46
Population Inversion ......................................................................................................... 47
B.
Fourier Optics ..................................................................................................................... 47
C.
Lenses, Thick and Thin ...................................................................................................... 47
D.
Aberration ........................................................................................................................... 48
3
I.
Geometrical Optics
A. Huygens
1.
Huygens’ Principle
“Every point on a primary wave front serves as the source of spherical secondary wavelets. The
wavelets advance with speed and frequency equal to those of the primary wave. The primary
front at some later time is the envelope of the wavelets.”
a. Wave front
A wave is a disturbance that propagates from one place to another. The disturbance is described

by a wave function, r, t  . The wave front is the surface of uniform phase, meaning that all
points on the surface share the same disturbance. In the case of light waves, the “disturbance” is
the electromagnetic field, so the electric field has the same magnitude and direction at every
point on the wave front. Say that at t = 0 the wave front has a shape, or profile,  . Then at a
later time, t, the wave front shape is   , possibly different from  .
b. Wavelets
Spreading circles or spheres are easier to deal with than advancing arbitrary shapes. So, we
imagine that a real wave results from the superposition of spherical waves emanating from points
forming the wave front.
The radii of the secondary wavelets are vt , where v is the speed of light in the medium. If the
medium is inhomogeneous, then some wavelets expand faster than others, and the wave profile
changes.
c. Huygens’ construction
The new wave front is constructed by drawing circles of radius vt , each centered on a point on
the primary wave front. The speed, v, need not be the same for all the wavelets.
2.
“Laws” of Reflection and Refraction
We know that, in general, when light impinges on an interface between two media, a portion of
the energy is reflected and a portion is transmitted across the boundary.
4
a. Reflection & transmission at normal incidence
Assume: i) both media are homogeneous and isotropic and ii) both media respond linearly.
Thus,  i   t   , but because vi > vt,  i   t . On the other hand,  r   i and vr = vi, so
r  i .
b.
Snell’s “Law”
 1i strikes point A at t = 0. Secondary wavelets propagate from point A to radii vi t and vt t ,
at which moment.. .
 i2 strikes point A and  1i arrives at point B.
By t  2t , secondary wavelets have spread from point A to 2tvi and 2tvt , and from point B
to tvi and tvt .  i2 has reached point B and  3i hits at point A.
Clean up the picture by removing the wavelets and look at the angles.
5
Measured from the interface,
 i = angle of incidence
 r = angle of reflection
 t = angle of refraction
t = time required for the whole of wave front,  i , to “impact” on the interface.
BD
AC
and sin  r 
. But BD  vi t and AC  vi t , so
AD
AD
sin  i  sin  r . This is the “Law” of Reflection.
We see by inspection that sin  i 
By a similar token, sin  i 
BD
AE
and sin  t 
. But BD  vi t and AE  vt t , so
AD
AD
sin  i BD vi
or vt sin i  vi sin  t . This is the “Law” of Refraction, or Snell’s “Law”.


sin  t AC vt
c.
Index of refraction
c
, where c is the speed of light in vacuum.
v
ni sin i  nt sin t
Define a light ray as a line or arrow indicating the direction of energy flow. If the medium is
isotropic, the light ray will be orthogonal (or normal) to the wave front.
Define the index of refraction, n 
Define corresponding points as the intersections of a specified light ray with successive wave
fronts.
In terms of light rays rather than wave fronts the picture of reflection and refraction looks like
this:
6
In summation, i) all the rays lie in the same plane (by symmetry), ii) i  r , and iii)
ni sin i  nt sin t .
B. Fermat
1.
Path Length
a. Spatial path length
The spatial path length is the total distance traversed by a light wave in propagating from point S
to point P.
b.
Optical path length
The elapsed time in traversing from S to P is t 
s1 s2 s3 1
   s1n1  s2 n2  s3n3  . The optical
v1 v2 v3 c
n
path length (OPL) is the “effective” distance OPL   si ni . If the medium is inhomogeneous,
i 1
P
then OPL   n( s)ds ; sort of like work with a changing force.
S
7
This could be, say, the atmosphere, whose density varies with altitude.
2.
Fermat’s Principle
a. Least action
In going from point S to point P, a light wave traverses a path for which the optical path length is
a minimum.
b.
Snell’s “Law” via Fermat’s Principle
Between S and P OPL  ni h 2  x 2  nt b 2  (a  x) 2 . We wish to minimize this with respect
to x. [Notice that within each medium, if homogeneous, the paths are straight lines—the shortest
distance between two points.]
OPL
0
x
a  x  1  0
x
ni
 nt
h2  x2
b 2  (a  x) 2
ni sin i  nt sin t  0
ni sin i  nt sin t
c.
Principle of Reversibility
8
A light ray going from P to S follows the same path as a ray going from S to P. This follows
from symmetry arguments.
C. Refracting Surfaces
The purpose of optical systems is to manipulate wave fronts, or to take rays emanating from one
location and cause them to converge at another.
1.
Spherical surfaces
a.
Huygens’ picture
The edges of the plane wave front overtake the center, which has already entered the second
medium and slowed.
b.
Fermat’s picture
9
According to Fermat’s Principle, the path SAP of the light ray is a minimum.
OPL  n1 o  n2  i
We want to rewrite this in terms of the polar coordinates of the point A, using C as the origin and
measuring the polar angle clockwise from the optical axis, SP.
By the law of cosines,
 2o  R 2  ( s o  R ) 2  2 Rs o  R  cos 
 2i  R 2  ( si  R ) 2  2 R( si  R) cos 

OPL  n1 R 2  ( s o  R) 2  2 Rs o  R  cos 

1
2

 n2 R 2  ( si  R) 2  2 R( si  R) cos 

1
2
OPL
0

1
n1 2
R  ( s o  R ) 2  2 Rs o  R  cos  2 2 R s o  R sin  
2
1
n2 2
R  ( si  R) 2  2 R( si  R) cos  2  2 R si  R sin   0
2
n1
n2
R( s o  R)  R( si  R)  0
o
i




n1
n
n
n
Rs o  1 R 2  2 Rs i  2 R 2  0
o
o
i
i
n1 n2 n2 si n1 s o



o i
iR oR
This specifies the relationship among the incident ray, R, n1, n2, and the refracted (transmitted)
ray.
10
c. Gaussian optics
If  is small, then the incident ray is called a paraxial ray. In that case, we can make
simplifying approximations by expanding the cosine and sine in power series.
2 4 6
cos   1 



2! 4! 6!
3 5 7
sin   1 



3! 5! 7!
If  is very small, the first order approximation gives cos   1 . Then  o  so and  i  si and
n1 n2 n2 si n1 so 1



 n2  n1 .
s o si
Rs i
Rs o R
That treatment of lenses and light rays that utilizes the first order approximation is called
Gaussian Optics. It is valid for rays arriving at shallow angles to the optical axis.
d. Focal points
There is a focal length on either side of the refracting surface.
n1
R.
n2  n1
n2
The image or second focal length is obtained by setting so   ; then f i  si 
R.
n2  n1
By convention, R > 0 if C is to the right of V.
The object or first focal length is obtained by setting si   . Then f o  so 
e.
Some terminology
i) A light ray originates at the point S, traverses an optical system, arrives at the point P.
S & P are called conjugate points, since their roles could be reversed.
ii) In a perfect optical system, every point in object space is perfectly imaged in
image space.
iii) A real optical system is diffraction limited, since a real system collects only a segment of
a wave front.
iv) In geometrical optics, we ignore diffraction, as if   0 .
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2.
Imagery
a.
focal planes
Off-axis light rays incident normally to the interface will be undeviated and pass through C.
Light rays close to such a ray of either side will be deviated to converge with the undeviated ray.
The ray bundle converges at a focal point lying on a curved focal surface,  .
For paraxial rays, only a small section of the focal surface is involved, centered on the optical
axis. We pretend the surface is flat and call it the focal plane.
It is an approximation of Gaussian Optics that parallel rays entering an optical system are
focused on a plane surface.
b. Imaging extended objects
Each point on an extended object emits light rays in all directions. Those that enter the optical
system are imaged in image space. Tracing only two rays from each point is enough to locate
the image, though often three are used.
Defn. Optical center of a refracting surface or optical system.
Consider parallel surfaces of opposite sides of a lens. A ray passing through those parallel
surfaces will be undeviated in direction, though off set. Where such a ray crosses the optical axis
is called the optical center.
12
Ray tracing rules of thumb
i) paraxial rays passing through the optical center are undeviated.
ii) Incident rays passing through fo emerge parallel to the optical axis.
iii) incident rays parallel to the optical axis emerges to pass through fi.
By convention, the extent of the object (image) is (+) above the optical axis, (-) below.
c.
Magnification
Transverse or lateral magnification M T 
By similar triangles,
yi si  f
s
s

 i . Now, because of the sign conventions, M T   i .
yo
yo
f
so
Longitudinal magnification M L 
We have
yi
.
yo
dxi
.
dxo
y
f
s f
. Cross multiply.
 i  i
so  f
yo
f
f 2  si  f so  f   xi xo
xi 
f2
xo
dxi
f2
  2   M T2
dxo
xo
M L   M T2
13
d. Stops and pupils
Nice sharp images are formed on a flat focal plane only for paraxial rays. Rays entering the
optical system far off-axis will degrade the desired flat image, as they will not focus on the focal
plane, but on the curved focal surface.
Thus, while in general the size of a lens determines the brightness of an image, it may be
desirable to prevent some of the light rays from entering the system for the sake of image quality.
See cheap astronomical telescopes. This is a relative question depending on the size of the
object and its distance from the lens.
Aperture stop: an opening or diaphragm that restricts the amount of light reaching the image. It
may be the rim of the lens itself.
Field stop: an element limiting the angular breadth of the image. It determines the field of view.
Pupils are images of the aperture stop, as seen from different points.
Entrance pupil: the image of the aperture stop as seen from an object point on the optical axis,
looking through any elements preceding the stop. If there are none, then the entrance pupil
coincides with the aperture stop.
Exit pupil: the image of the aperture stop as seen from an axial image point, looking through any
elements following the stop. If there are none, then the exit pupil coincides with the aperture
stop.
Chief ray: a ray from an off-axis point on the object that passes through the midpoint of the
aperture stop.
Marginal ray: a ray directed from an axial object point to the rim of the entrance pupil.
Relative aperture: the ratio of the aperture diameter and the focal length of the optical system.
f-number: reciprocal of the relative aperture. f /# 
D. Optical Systems
1.
Ray Tracing
14
f
.
D
a. Refraction equation
Rewrite Snell’s “Law” in terms of vectors defining the directions of the wave propagation and of
the surface normal.
ni kˆi  uˆ  nt kˆt  uˆ
Next, write the angles of incidence and of refraction in terms of the optical axis rather than the
interface normal. Further, we’ll consider paraxial rays.

 

ni1 i1  nt1 t1
ni1  i1  1   nt1  t1  1 


y 
y 
ni1   i1  1   nt1   t1  1 
R1 
R1 


 n  ni1 

nt1 t1  ni1 i1  y1  t1
R
1


This is the refraction equation for the first surface. The transfer equation takes the ray through
the lens to the second surface.
y 2  y1
 tan  t1   t1
d 21
y 2  y1  d 21 t1  y1  d 21 i 2
b.
Process
i) given  i 1 , y1  at the first interface,
ii) compute  t1 ;
iii) given d 21  V1V2 , solve for y2.
iv) repeat surface to surface through the system.
2.
Matrices
a.
Propagation
15

Each light ray is represented by a column matrix   .
 y
The refraction equation can be rewritten as a matrix multiplication.
 nt1 t1   1  D1  ni1 i1 

  


y
y
0
1
 i1 
 t1  
rt1  R1 ri1
R1 is the refraction matrix for the first surface. Similarly, the transfer equation becomes
1 0 n 
n   


ri 2   i 2 i 2    d 21 1  t1 t1   T21rt1
 y t1 
 y i 2   nt1

where d21 is the approximate distance between the two surface vertices.
b. System matrix
The chain of events is strung together:
1 0 1  D n 
 ni 2  i 2   1  D21  d

1  t1 t1 

  

 21 1 



1  n
1  yt1 
 0
 yi 2   0
 t1

for two surfaces. Any number of such matrices can be multiplied together, in the correct
(reverse) order, so that a many-surface optical system can be traced quickly on a computer.
The system matrix, A, is just the product of all the individual matrices. For the two-surface
system, A = R2T21R1 so that rt2 = Ari1.
E. Mirrors
1.
Planar Mirrors
a.
Ray tracing
By similar triangles, si  so . Also, the image is virtual, so we use a slightly different sign
convention: (+) to the left of the mirror(that is, the side the object is on), (-) to the right(the side
opposite the object).
16
b. Extended objects
A planar mirror produces a virtual image which is perverted, that is, switched left-for-right..
More technically stated, the right-handed coordinate system in object space is transformed into a
left-handed system in image space. This is not simply a rotation. Contrast this result with the
refracting surface, which produces a real image that is inverted both left-to-right and up-to-down.
This called reversion, and is equivalent to a rotation by 180o about the optical axis. The virtual
image produced by a refracting surface is erect—it is not inverted.
c. Magnification
For a planar mirror, of course M T  1 .
2.
Curved Mirrors
a.
Parabolic mirrors
We’ll use Fermat’s Principle to figure out what shape of mirror will focus parallel incident light
rays to a point. The condition is that the optical path lengths of parallel rays striking arbitrary
points on the mirror (A1 and A2) and passing through the focal point F must be the same.
OPL  W1 A1  A1F  W2 A2  A2 F
Establish a base line D1D2 , parallel to the wave fronts (perpendicular to the incident rays) such
that
W1 A1  A1D1  W2 A2  A2 D2
Subtract to eliminate the WAs.
A1F  A1D1  A2 F  A2 D2
We find that the curvature should be such that the quantity AF  AD  constant . This defines a
paraboloid, with focus F and directrix D1D2 .
b. Spherical mirrors
Sadly, a parabolic surface is difficult to make. Spherical mirrors are easier to create. In the
paraxial region, a sphere of radius R = 2f will be very close to the paraboloid surface of focal
17
length f. That is, the shape of a small region of the parabola near the axis is almost the same as
spherical. This can be demonstrated geometrically.
The radius CA bisects the angle SAP, so it divides SP into pieces proportional to SA and to
SC CP
AP . I.e.,
. Notice, too, that SC  so  R and CP  R  si .

SA AP
Now, so and si are both positive, but R < 0, so SC  so  R and CP  R  si  . Near the axis,
SA  soand PA  si , whence
so  R  R  si 

so
si
1 1
2 1
  
so si
R f
This is the Mirror Formula—be careful of the +/- signs. The ray tracing is just like a refracting
surface, although of course the rays do not go through the mirror.
F. Prisms
1.
Dispersing Prisms
a.
Angle of deviation
18
  i1  t1   t 2  i 2 
  i1  t 2  
From Snell’s “Law”
t 2  sin 1 n sin i 2 
t 2  sin 1 n sin   t1 
t 2  sin 1nsin  cos t1  sin t1 cos  
1


t 2  sin 1 n(sin  cos t1  sin i1 cos )
n


 

n 2  sin 2 i1 1

t 2  sin 1 n sin 

sin

cos

i1
2


n
n
 


t 2  sin 1 sin  n 2 sin 2 i1



1
2
 sin i1 cos  

  i1  sin 1 sin  n 2  sin 2 i1

So, if  and n are fixed, then   f (i1 ) .

1
2
 sin i1 cos   

b.
Angle of minimum deviation

 0 , but that’s too messy. Instead we’ll go back to the beginning.
We could carry out
i1
  i1  t 2  
d
d
 1  t2  0
di1
di1
d t 2
 1
di1
To obtain
dt 2
, differentiate Snell’s “Law” at each interface.
d  i1
d sin i1  n sin t1   cos i1di1  n cos t1dt1
d n sin i 2  sin t 2   n cos i 2 di 2  cos t 2 dt 2
From d   t1  i 2   0 , dt1  di 2 . So we can divide the two equations to obtain
19
2
2
 cos i1   cos t1 

  

cos

cos

t2 
i2 


2
2
1  sin i1 n  sin 2 i1

1  sin 2 t 2 n 2  sin 2 t 2
Now, n  1, so this can be true only if i1  t 2 . Therefore, the minimum deviation occurs when
i1  t 2 and t1  i 2 . These mean that the light ray is traversing the prism parallel to its base.
   
sin  m
 2  .
Setting   m and solving for n, we obtain an expression: n 

sin
2
This provides a way of measuring the index of refraction, by observing  m with a spectrometer.
2.
Reflecting Prisms
a. Non-dispersing prism
Let a light ray pass through a prism in such a way as to experience one internal rerflection. The
angle of deviation is independent of wavelength and index of refraction, so there is no
dispersion.
Proof: Evidently,   180o   . On the other hand,
    2  360o
  180o  360o  2  
  180o  360o  2(90o  i1 )  
  2i1  
20
If the triangle isn’t isosceles,   i1  t 2   .
b. Other
Another example is an erecting prism, used in binoculars to un-invert the image. No disoerson,
again.
G. Fiber Optics
1.
Large Diameter Fiber Optics
Compared to the wavelengths of IR, visible and UV light, the diameters of optical fibers are
large. Light pipes are used to reflect and reorient light rays. The communication fibers are used
to transmit signals in the form of light pulses.
a. Total internal reflection
Consider the situation of a light ray approaching an interface between two media. Snell’s “Law”
relates the incident angle to the transmitted angle.
n
sin i  t sin t
ni
There is a critical angle, c , of incidence at which the transmitted angle is 90o.
n
n
sin c  t sin 90o  t
ni
ni
If nt < ni, there exists a critical incident angle for which the wave is entirely reflected; no energy
is transmitted past the interface. This phenomenon is called total internal reflection.
A light ray traveling within a glass cylinder, and striking the side of the cylinder will undergo
total internal reflection the incident angle is greater than or equal to the critical angle. The
critical angle is determined by the indices of refraction of the cylinder, nf, and the medium
21
surrounding the cylinder, na. The light ray will bounce from one side of the cylinder to the other
again and again as it propagates along the cylinder.
The path length traversed by the light ray between successive reflections is h 
D
. The
sin t
D
. If L is the total length of the
cos t
L cos t
cylinder, then the total number of reflections undergone by the light ray is N 
. Finally,
D
the total path length for the light ray is
distance along the cylinder between reflections is g 
 sin 2 i
L 1 

n 2f
L cos t D
cos t

  Nh 
L

sin i
D sin t
sin t
nf




1
2


L n 2f  sin 2 i
sin i

1
2
.
The elapsed time for the light ray to travel the length of the cylinder is
2
2
 n f n f Ln f  sin i  2
.
t 

vt
c
c
sin i
1
b. Acceptance angle
Now the light ray was injected into the fiber through one end. We might ask, at what angle must
the light ray enter the end of the cylinder in order to be totally internally reflected along the
length of the cylinder?
22

sin c 
na
 sin 90o  t
nf
cos t 
na
nf
1  sin  
1
2
t
sin max 
2



na
nf
1 2
n f  na2
na

1
2

The numerical aperture is defined to be NA  na sin max  n 2f  na2

1
2
. If the light ray enters the
end of the cylinder at an angle max , then it will hit the inside surface of the cylinder at c and be
totally internally reflected. Commonly, optical fibers are clad with a material other than air so as
to decrease the NA.
c.
Dispersion
Rays entering the end of the cylinder at different angles travel different total distances along the
cylinder, according to how often the ray bounces. Over a long cylinder, dispersion occurs. The
Ln f
minimum transit time is tmin 
for a light ray traveling along the axis of the cylinder. The
c
longest time occurs for a ray incident at the critical angle, c .
n
L f
2

na Ln f
tmax 


vf
vf
cna
Ln f  n f

  1 .
c  na

A pulse of light injected into the cylinder would be comprised of several light rays, at a range of
angles, so it would spread in space as it propagated along the cylinder. Input pulses must be far
enough apart that they don’t begin to overlap and become indistinguishable by the time they
reach the far end of the cylinder. In optical fibers, a narrow NA is desirable to reduce t .
The intermodal dispersion over the length of the cylinder is t  tmax  tmin 
2.
Sending Signals on Glass Optical Fiber
Optical pulses are generated by a laser and injected into the end of a glass fiber. The light waves
travel through a material medium and therefore the pulses change as they travel along the fiber.
a.
Attenuation
23
Let z be distance along the optical fiber. The power of the light wave will be attenuated along
dP
 P . At the end of a fiber of length L, Pout (t )  Pin (t )e  L . There are several
the fiber:
dz
loss mechanisms. Here are two of them.
i) the molecules (silica) of the glass as well as impurity atoms/molecules absorb EM waves at
different frequencies. There is a spectral loss profile.
ii) Rayliegh scattering occurs due to microscopic fluctuations in the index of refraction. The
scattering depends on the wavelength and the size of the fluctuation regions. In other words, the
glass is not perfectly uniform.
b. Chromatic dispersion
The index of refraction is a function of wavelength. An optical pulse starts out as a (near) square
wave, which is comprised of a sum of harmonic components over a range of wavelengths. Each
component travels at a slightly different speed in the fiber—hence the pulse spreads out as it
propagates. Also, the laser bandwidth is not infinitesimally narrow, itself. At some point
separate pulses may begin to overlap.
c. Bit rates
The optical signals are composed of a continuous sequence or string of on/off pulses. The length
of each bit is the bit period, TB. The reciprocal of the bit period is the bit rate, B. The shorter TB,
the greater is B. However, as TB decreases, the frequency distribution in the pulses widens,
which leads to greater dispersion as the pulse travels down the fiber.
24
Multiple channels of bit strings can be sent into the fiber, at slightly differing carrier
wavelengths, e.g., 1.5, 1.55, 1.45 m . But, if the channel spacing is too close, the dispersion
will cause pulses from different channels to overlap, so the channels are not completely separate
anymore. Information on one channel is mixed with information on another.
d. Engineering the fiber
In designing and manufacturing the optical fiber, the goal is to minimize these effects, and
others.
Minimize attenuation: i) select carrier wavelengths to which the silica is most transparent—IR,
ii) remove impurities, iii) re-amplify periodically along the fiber.
Minimize dispersion: i) single mode fibers—if the fiber core radius is less than about 3 times the
wavelength, then only one mode can propagate along the fiber, rather than many, ii) regenerate
pulses every few kilometers to correct chromatic dispersion (expensive), iii) fiddle with the
composition of the fiber to reduce the dependence of n on wavelength, iv) give initial pulses a
shape that does not disperse—solitons, v) graded index fibers.
The engineering aims are to maximize distance between amplifications and to maximize bit rate.
e. Fiber amplifiers
Here’s something interesting. Ordinary amplifiers require the reading of the optical signal,
conversion to electrical, amplification, re-conversion to optical pulses. This means a relative
bulking and power-hungry gadget, with several possible points of malfunction.
There is a fiber amplifier that does not convert the optical pulses to/from electric signals, and can
be placed in-line with the fiber. A section of fiber itself is the amplifier.
25
The incoming signal stimulates coherent emission in the erbium-doped fiber, thereby exactly
duplicating the incoming signal with increased amplitude.
26
II. Wave Mechanics
Light is an E-M wave, so we’ll review the mathematics of wave motion.
A. Wave Equation
1.
Wave Profile
Define a function, ( x, t )  ( x  vt) which describes the shape and motion of a wave pulse
traveling along the x-axis.
a. General description
We will need the slope and curvature of the function,  ( x, t ) . Let x  x  vt .
 x   


x x x  x 
 2     2 


x 2 x x  x  2
on the other hand, x  x  vt , therefore,
 t 
1 


x  x  t
v t
2
2
2
   
 
 
 
 
2  
2  
  v
 v
v
  v
 v
v
x 
x 
t x
x  
x  
t 2
x  2
x 2
b. Solution
The general solution to the wave equation is   c1 f x  vt  c2 g x  vt , where the f or g is the
wave profile. The solution is a superposition of wave pulses traveling in the +x and –x
directions.
2.
Harmonic Waves
a.
Definitions
x, t   A cos( x  vt  ) or A sin( x  vt  )
Wavelength,  , such that x, t   x  , t  .
Amplitude, A = maximum disturbance.
Propagation number, k, such that k  2 .
Period, , such that x, t   x, t   .
1
Frequency,   . Note that the wave speed, c   .

Angular frequency,   2 .
1
Wave number,   . Notice the distinction we are making now between k and  .

b. Phase velocity
The argument of the cosine is called the phase. The phase is a function of x and of t.
( x, t )  kx   t  
27
If we select a fixed value of phase,  o , then as the wave travels, this point propagates along the
 x 
x-axis with a velocity  
.
 t  o
  o     t 

   c


t 
k
k

c. Complex representation
Another solution to the wave equation is x, t   Aei  t kx    Aei . This is more convenient to
use than the sine or cosine, but it’s complex. The actual, physical wave is Re Aei .

3.
Three Dimensional Wave Forms
a.
Wave equation in three dimensions
2
2
2
1 2
2










x 2
y 2
z 2
c 2 t 2

x  r   x, y , z 

k  k  k x , k y , k z 

Specific forms of  are labeled according to the shapes of the constant-phase surfaces.
b.
Plane Waves

The surface of constant phase is a plane, perpendicular to the wave vector, k . The equation for
such a plane is

r  ro   k  0

where ro is a point in the plane. Write it out
28

r  ro   k  x  xo k x   y  yo k y  z  z o k z
0
 
k  r  xkx  yk y  zk z  xo k x  y o k y  z o k z  a constant
 
The phase of a plane wave, then, is   r  k   t
c. Spherical Waves
The surfaces of constant phase are spherical shells, propagating outward from or inward toward a
central point. How do we write the profile of such a wave?

We require that r   r . We need to write the wave equation in terms of r.
  r

x r x
 2     r  r  2   r 
   r  r
 
  2   


2
r  r x  x r  x 
r  r x  x
x
2
 2   2   r 
  2 r
 2   
r x 2
x 2
r  x 
2
Since r  x 2  y 2  z 2 ,
r x
 2 r 1 x
 1
 and

x  
2
x r
r x
x  r 
x
1  x2 
1  .
r  r 2 
 2   x   2  1  x 2  
 2
 2




1

,
and
similarly
for
and
.
 
x 2  r  r 2 r  r 2  r
z 2
y 2
Add ‘em up
2
2
2
2
2
2
 2   2   2   x   y   z    2  1   x   y   z   


         
 3          
x 2 y 2 z 2  r   r   r   r 2 r   r   r   r   r
2
Therefore,
 2  2  1  2 r 


r 2 r r r r 2
1  2 r  1  2 
 2 
 2 2
r r 2
c t
2
2
 r  1  r 
 2
r 2
c
t 2
We see that the wave function (the solution to the wave equation) has the form f r, t   rr, t  .
For outgoing waves, f r, t   f r  ct  , while for converging waves, g r, t   g r  ct  . For
harmonic waves, the profile is
A
A
r , t   cos k r  ct  or r , t   e ik r  ct  .
r
r
 
A surface of constant phase is defined by k  r  kr  a constant . However, the amplitude of
A
 is , so the profile is not constant.
r

29
d.
Cylindrical waves
x   cos 
y   sin 
zz

1     1  2 


We require that r    . The wave equation is
with solutions
     c 2 t 2
A
A ik r  ct 
.
r , t  
cos k r  ct  or r , t  
e


B. Fourier Analysis
1.
Superposition of Waves
As the wave equation is linear, the sum of two solutions of the wave equation is also a solution.
Query: what is the net disturbance of the medium.
a. General case
Consider two waves, with the same v and  .
E1  Eo1 sin  t  1 
E2  Eo 2 sin  t   2 
where 1  k1x and 2  k2 x   . The net disturbance is E = E1 + E2.
E  Eo1 sin  t  1   Eo 2 sin  t   2 
E  Eo1 cos 1  Eo 2 cos  2 sin  t  Eo1 sin 1  Eo 2 sin  2 cos  t
In general this is a funny, varying wave profile.
b. Flux density
The energy transported by the wave is related to the amplitude squared.
2
2
2
2
E  E1  E2  Eo1  Eo 2  2 Eo1Eo 2 cos 2  1 
The interference term, 2Eo1Eo 2 cos2  1  , depends on the phase difference   2  1 .
c. Optical path difference
The optical path difference is a phase difference arising from difference in path length traverses
by two waves.
30
  kx1  kx2   
2
x1  x2   

2

nx1  x2     ko   
o
If   constant , then the waves are said to be coherent.

d.
Phasor addition
E1  Eo1 sin  t  1 
Graphically, we can represent this expression as a vector of length Eo1 rotating about the origin,
thusly:
Two such vectors are added on the phasor diagram by placing the tail of the second vector on the
head of the first. We can imagine that both are rotating with the same angular frequency, or with
different frequencies.
e. Standing waves
In this case, we have two waaves of the same frequency, but travelling in opposite directions.
E1  Eo1 sin kx   t  and E2  Eo 2 sin kx   t  with Eo1  Eo 2
The net disturbance is E1 + E2.
31
E  E1  E 2  Eo1 sin kx  t   sin kx  t 

 2kx   2 t 
 Eo1 2 sin 
 cos

 2   2 

 2 Eo1 sin kx cos t 
This expression is not of the form of a travelling wave, so it is a stationary of standing wave, not
moving through space. If we fix x, then E  A cos t  .

3
At x  0, , , , sin kx  0 and E  0 for all t. These are the nodes.
2
2
 3
At x  , , sin kx  1 and E  2 E o1 for all t. These are the antinodes.
4 4
Every point (x) in the medium executes simple harmonic motion, of differing but constant
amplitude.
2.
Waves Having Different Frequencies
a. Beats
In this case, we superimpose two waves having different frequencies, but travelling in the same
direction.
E1  Eo1 cosk1 x  1t  and E2  Eo1 cosk 2 x  2 t 
The sum is
E  E1  E 2  E o1 cosk1 x  1t   cosk 2 x   2 t 
1
 1

 2 E o1 cos  k1  k 2 x  1   2 t  cos  k1  k 2 x  1   2 t 
2
 2

 2 E o1 cosk m x   m t  cos k x   t 
The energy transports is proportional to
32
Eo2  4 Eo21 cos 2 k m x   m t   2 Eo21 1  cos2k m x  2 m t  .
The beat frequency is the frequency of the modulation E o2 , namely
1

2 m  2 1   2   1   2 ..
2

b. Group velocity
Look at the modulation envelope Eo x ,t   2Eo1 cosk m x  m t  . This is a wave, with a speed
  
d
 t  x  m 1   2 
vg  




km
k1  k 2
k
dk
  
 x  t
dv
dv
 0 , then v g  v . But in dispersive media,
Since   kv, v g  v  k
. If indeed
dk
dk
c kc dn
c
 k dn 
 v 1 
, so that v g   2
  kv  k
 . [ nk  is the index of refraction.]
n n dk
nk 
 n dk 
3.
Fourier Sums
a. Anharmonic waves
If E1 and E2 have different frequencies and different amplitudes, in general the resultant wave
form won’t be a harmonic wave, but will still be periodic.
Fourier’s Theorem: A periodic function, f(x), having a spatial period  , can be expanded as a
series of harmonic functions whose wavelengths are integral submultiples if  . I.e.,


A
2
f  x   o   Am cos mkx   Bm sin mkx, where k 
, and Am and Bm are expansion
2 m 1

m 1
coefficients.
Now the sine and cosine functions have the orthogonality property.


 cos akx cos bkxdx  2 
ab
0


 sin akx sin bkxdx  2 
ab
0

 cos kx sin kxdx  0
0
We use this property to evaluate the expansion coefficients for a known f(x).

 


0
m 1 0
m 1
m 1
 cos nkxf ( x )dx  0    cos nkxAm cos mkxdx  Bm  0   Am nm  An
33

Similarly, Bn   sin nkxf ( x )dx . For n = 0,
0
For a travelling wave, f kx  t  



f ( x )dx  
0
0

Ao
2
dx  0  0  Ao   f ( x )dx .
2
0


Ao
  Am coskx  t    Bm sin kx  t  .
2 m 1
m 1
b. Non-periodic waves
What can we do if the waveform is aperiodic? We like the Fourier expansion so much that we
still want to use it.
34
c.
Band widths
35
36
C. Fraunhoffer Diffraction
Diffraction occurs whenever a portion of a wave front is obstructed and a change in amplitude or
phase occurs in that portion.
1.
Far Field (Fraunhoffer) Diffraction
a. Far field condition
Fraunhoffer diffraction obtains when the incident and outgoing waves are (nearly) plane waves,
as opposed to spherical.
a2

The pattern observed at the screen,  , changes only in size as the screen is moved back and
forth.
R
b.
Line source
Consider a linear array of point sources of spherical wavelets. If R>>D, then
the short source elements, yi , where
EL EL

for all
ri
R
EL
is the amplitude at P of the spherical wavelet
ri
emanating from yi . The phases of all the spherical wavelets are not the same, though. The
contribution of each wavelet to the total electric field at P is approximately
E
dE  L sin(  t  kr)dy .
R
The radius, r, is a function of y. We’ll have to obtain it from a Maclauren series.
y2
r  R  y sin  
cos 2   
2R
Now, if R>>D, then the third term is negligible for all  . In which case r is a linear function of
y.
Integration over the line source yields the total field at the point P.
37
D
E
E L
R
 sin  t  krdy
D
D
E
EL
R
L
L
L
 sin  t  k R  y sin dy
D
L
Integration over y (t and  are parameters, since P is fixed) yields
 kD

D sin  sin 
E
 2
 sin  t  kR
E L
R   kD 

 
 sin  
 2 

2
The irradiance is <E > (average over time), so

2  kD
sin 
2
2
2 2 sin 
1 EL D
1  EL D  sin 2  1  EL D 
2
2
2


I () 
 
 

 sinc   I (0)sinc 
2
2
2 R 2   kD 
2
R

2
R





 
 sin  
 2 

Keep in mind that  is measured in the plane containing the line source. When D>>  , the
D
sin  becomes very large, namely
irradiance decreases rapidly with increasing  , since  

D
. This means that side fringes are very small, only the central maximum of I(0) remains. The

result is, for a coherent line source with length D>>  , the irradiance is approximately that of a
point source radiating a circular wave in the xz-plane at   0 .
2.
Appertures
a. Single slit
Essentially, we extend the development from the linear source to the second dimension. The slit
is composed of many line sources of length D. Each of these acts like a point source lined up on
the z-axis, radiating outward in the xz-plane.
The calculation is the same as before, except now we integrate over z from 
2
b
b
to , and  is
2
2
 sin  
kb
measured in the xz-plane. I ()  I (o)
 where now   sin  . However, we are
2
  
assuming that b is small, so  is not large. Therefore the side fringes cannot be neglected.
We find that the I(  ) maxima by setting
dI
 0.
d
38
dI
2 sin ( cos   sin )
 I (0)
d
3
This equals zero when sin   0 , or when    ,2,3,
What does this look like? For monochromatic light:
The minima occur at   m ; the maxima m  0 occur when  cos   sin   0 or when
tan    . Notice that the side fringes depend on wavelength. Thus a multi-chromatic light
source gives side fringes at different locations for each colour.
b. Rectangular
Now, do not assume b is small. We have a rectangular opening a x b divided into square area
elements, ds. Wavelets propagate outward from each area element. (X,Y,Z) are the coordinates
of the field point, P.
Let EA be the source strength per unit area, constant over the aperture. The contribution to the
amplitude at P from ds is
E 
dE   A ei  t  kr ds ,
 r 
where r  X 2  Y  y   Z  z 


In the Fraunhoffer regime, all the
EA
E
 A . In the phase,
r
R
2
2
1
2
.
 y 2  z 2 2Yy  Zz  2
r  X  Y  2Yy  y  Z  2Zz  z
 R 1 

 .
R2
R2


2
2
y z
 0 , leaving
The limits on the y and z are the edges of the aperture, so
R2

2
2
2
2
2

1
1
2
 2Yy  Zz  2
r  R 1 
 . This goes in for r in the phase. Now we have “only” to integrate over
R2

the aperture.
1
E
E A
R
E
b a
2 2
 e

  Yy  Zz   
i   t  k  R  1
 
R 2  
 

dzdy
b a
 
2 2
a
2
E A i  t  kR 
e
ae
R

2
39
Zz
ik
R
b
2
dz  e
b

2
ik
Yy
R
dy
bY 
 ik bY
 ik
 e 2R  e 2R 
kbY
sin 
Now,  e dy  b 
, where  
. Similarly,
b

2ikbY

2
R

b



2
2R


a
aZ 
 ik aZ
 ik
Zz
2
2R
2R 
ik

kaZ
e
e
sin 
R
, where  
. Therefore, we obtain
e
dy

a

a


a
2ikaZ
2R




2
2R


E
sin  sin 
E  A ei  t  kR  ab 
R


2
The irradiance at the point P is E .
b
2
ik
Yy
R
2
 sin    sin  
I (Y , Z )  I (0)
 

     
As before, the I(o) is the irradiance at the center of the diffraction pattern, (Y.Z) = (0,0).
2
D. Fresnel Diffraction
Diffraction of spherical wavefronts.
1.
Fresnel Zones
a. Obliquity
In Huygens’ construction, every point a primary wavefront acts as a point source of secondary
wavelets. In order to provide the forward bias to these wavelets (alternatively, eliminate the
unphysical backward wavelet propagation), we introduce a quantity called the obliquity or the
1
inclination factor, K ()  1  cos  ; K (0)  1; K ()  0 .
2
b. Half-period zones
Spherical wave fronts are emitted from the point source, S. In cross section the picture is:
40
The particular wave front is emitted at t = 0. The radius of the wave front at t = t’ the radius is
E
 . At  , the disturbance is E  o cost  k , where Eo is the source strength. Secondary

wavelets from the primary wave at  travel toward the point P.
We define a differential area element, ds, such that all wavelets from ds travel the same distance,
r, to reach P. These wavelets also are all coherent, in phase with each other. Thus, at time t,
wavelets from ds arrive at P with phase  t  k   r .
Eo
. Taking into account the obliquity, we

have the contribution to the amplitude at P from ds as
E
dE  K A cos t  k   r ds
r
We need ds in terms of r. ds  d2sin 
The source strength per unit area on ds is E A  Q
Law of cosines r 2  2    ro   2  ro cos 
2
Take
d
and solve for d .
d
dr
 2  ro sin 
d
rdr
d 
  ro sin 
2r
ds  22 sin 
rdr
2rdr

  ro sin 
  ro
So, from one zone, we integrate over ds.
E 
r
 dE  K 
r 1
r
E A 2 
cos t  k   r dr ,
  ro  r1

 1
.
where, r  ro   and r 1  ro 
2
2
EA  1 
r
E  2K
   sin  t  k   r r1
  ro   k 
2
Now, k 
, so we get

41
E  
K  E A  
 
  1 

) 
sin  t  k (  ro  )  sin  t  k (  ro 
  ro  
2 
2


E   1
 1
2 K  E A
sin  t  k (  ro ).
  ro
k
 1 .]
2
Notice that E has opposite sign from E 1 , but K  K 1 . Therefore wavelets from successive
zones don’t quite cancel each other.
[Used an identity and the fact that sin
Next, we add up the contributions from the half-period zones, up to the mth zone at    , at the
backside of the wave front.
E  E1  E2  E3    Em
The terms alternate signs, so make that explicit (let m be odd)
E  E1  E2  E3  E4    Em
E
E
E  E
E 
E  E
E  E1   2  E2  3    3  E4  5      m  2  Em 1  m   m
2   2
2 
2 
2
 2
 2
E  E 1
Since the zones are narrow, and K  is slowly varying, E   1
. Therefore the
2
E
E
quantities in the parentheses are small, whence, E  1  m . But, for Em, K ()  0 , so Em=0.
2
2
E
E 1
!!
2
2.
Fresnel Lens
a.
Zone plates
E1
, where E1 that due
2
to the first half-period zone. This occurs because the odd and even indexed zones almost (not
quite) cancelled each other. Suppose we made a device that blocked alternate zones—we would
avoid the cancellations and obtain a greater optical disturbance. The problem is to create such an
item. Let’s block the odd zones.
Am  outer edge of the mth zone.
An unobstructed spherical wave gives a total disturbance at the point P of

The zones are defined such that m  rm   o  ro   m . On the other hand,
2
42
m  R  
2
m
2
o

1
2
Rm2
 o 
2o
rm  Rm2  ro2  2  ro 
1
o 
Rm2
2ro
Rm2
R2

 ro  m  o  ro  m
2o
2ro
2
1 1 m
 
o ro Rm2
Rm is the radius of the mth zone on the zone plate. If the odd zones are blocked, the total
disturbance at P is (let’s say including 40 zones)
E
E  E1  E3  E5    E39  20 E1  1
2

Notice that the zones are about
wide, so 40 zones is not very far from the OP line.
2
b. Focal length
With a point source a distance  o from a zone plate, waves converge at the point P, a distance ro
from the zone plate.
1 1 m
 
o ro Rm2
Define a primary focal length: f1 
Rm2
.
m
E. Interference
1.
Coherent Interference
a.
Coherent sources
b.
Phase difference between sources
2.
Wave Front Splitting
a.
Young’s experiment
43
b.
Fresnel double mirror
c.
Fresnel biprism
d.
Lloyd’s prism
e.
Multiple slits
3.
Amplitude Splitting
a.
Double beam interference
b.
Near-normal interference—Newton’s Rings
c.
Michelson interferometer
d.
Fabry-Perot interferometer
F. Vector Wave Theory
G. Polarization
44
45
III. Extra
A. Absorption and Emission of Radiation
1.
Transition Coefficients
a. Absorption
b.
Spontaneous emission
c.
Induced or stimulated emission
2.
Transition Rates
a.
Level populations
b.
Transitions
c.
Dynamic equilibrium
d.
Induced transition rate
46
3.
Population Inversion
a.
Equilibrium level population
b.
Inverting the population
c.
Metastable states
B. Fourier Optics
C. Lenses, Thick and Thin
c. Effective focal length
We would like to use the same Gaussian equation involving so , si , and f to express the net effect
of a system of refracting surfaces. This can be done if the distances are redefined. Consider for
instance a thick lens—two spherical refracting surfaces in sequence.
We define principle planes and principle points H1 and H2, where the optical axis passes through
the principle planes. The distance from the object to H1 is so. The distance from the image to H2
is si. The distance between the principle points is d. With these definitions,
and
where we call f the effective focal length of the combination of two surfaces.
The principle planes are located relative to the vertices of the refracting surfaces.
47
d. Application to thin lenses
For a thin lens,. So for one thin lens, A=R1R2.
For two thin lenses, separated by a gap, d,
.
By inspection, the effective focal length is. We notice, too, that the locations of the principle
points, H1 & H2, relative to the vertices of the two lenses, are
D. Aberration
48