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Optics
Section of PS1220
6 Lectures
2 E-grade problemsets
Lecturer: John Holdsworth
Room P104 Physics Building
School of Mathematical and Physical Sciences
Ph: 4921-5436
E-mail: [email protected]
School of Mathematical and Physical Sciences PHYS1220
Light

Light travels as an electromagnetic wave
E  E y  E 0 sin kx  t 
B  B z  B0 sin kx  t 
cvacuo
2

k
  2f
f   c 

k
n
n  index of refraction
E  hf
Semester 2, 2002
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Spectrum



Optics concerns itself primarily with the visible and near IR region of the
spectrum.
X-ray optics is a new and exciting field.
IR optics is a mature field due to military use.
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Reflections


Reflection from a surface may be diffuse or specular
Specular reflected light has the incident ray, the normal to the surface
and the reflected ray in the same plane. The plane of incidence
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Basic geometry
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Terminology




Object distance d0
Image distance di
Real image: the image is on
the same side of the optical
surface as the outgoing light.
Virtual image: the image is on
the opposite side of the optical
surface as the outgoing light.
Semester 2, 2002
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Spherical aberration and paraxial approximation




Semester 2, 2002
A spherical surface is the
easiest to manufacture
Parabolic surface is required to
focus to a single point
Close to the axis (= paraxial)
the spherical and parabolic
surfaces do not differ much
Quite a useful approximation
in geometric optics
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Spherical mirrors


Within the paraxial approximation, rays parallel to the principal axis will
focus at the focal point
r
Important points: Centre of curvature
f 
Focal point
2
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Image Location Through Ray Diagrams I
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Image location through ray diagrams II

Similar triangles OO’A and II’A
ho d o

hi d i
Similar triangles OO’F and ABF
(within the paraxial approximation) leads to the mirror equation

ho d o  f

hi
f

do do  f

di
f
Define the Lateral Magnification m
Semester 2, 2002
1
1
1
 
do di
f

hi
di
m

ho
do
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Image location through ray diagrams III

Virtual images





di negative (behind the
reflecting surface)
do positive
f and r positive
m positive and >1
Convex Optics




di negative (behind the
reflecting surface)
do positive
f and r negative (behind
the reflecting surface)
m positive and < 1
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Refraction

Regularly observed phenomena




Mirages
“Bent” teaspoon in water
Snell’s Law
n1 sin1 = n2 sin2
“n” index of refraction is defined
by the electro-magnetic properties
of a particular material. It varies
with light of different wavelengths
and is different for most materials
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The spectrum of visible light and dispersion
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Total internal reflection

For light passing from a medium with high index of refraction
one of lower index of refraction, light is bent away from the
normal
n1 sin 1  n2 sin  2
sin  crit


n2
n2

 sin 90 
n1
n1
For light incident at < c some light will be refracted out of the
medium
For light incident at > c no light will be refracted out of the
medium, it is totally internally reflected
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Fibre optics
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Lenses and optical instruments


Refraction at a surface will deflect light according to Snell’s Law. A lens
refracts light a different amount at different distances from the center
of the lens
An array of trapezoids will cause light from a point to refract in 2-D
towards another point and serves to illustrate what a smoothly varying
lens surface can do in 3-D
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Lens: Converging or Diverging?




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Light enters from the left, by convention
Converging lens are thicker in the middle than
at the edges
Diverging lens are thinner in the centre than at
the edges.
Many shapes are possible, each with their own
focusing and aberration properties
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Paraxial and thin lens approximations

An aspheric surface is required to focus parallel
rays of a single wavelength to the smallest point
however a spherical surface is the easiest to
manufacture.




Semester 2, 2002
Spherical aberration
Chromatic aberration
Close to the axis the spherical and aspherical
surfaces do not differ much: Paraxial
approximation.
“Thin Lens” approximation allows one to draw a
ray through the vertex of the lens and simplify
ray tracing. The diameter of the lens is smaller
than the radii of curvature of the surfaces in this
case.
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Focal plane of a lens



Semester 2, 2002
The focal point is the image point for an object
at infinity on the principal axis.
The distance of the focal point from the centre
of the lens is the focal length
The Power of a lens in Diopters is the inverse of
the focal length in metres.
Power (Diopters) = 1/f(m)
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Image location through ray diagrams I


These three rays allow the
imaging properties to be
determined graphically.
The image is real

The image is inverted

The focal length is positive by
convention
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
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On the same side of the lens as
the outgoing light
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Image location through ray diagrams II


The same approach of the three rays can determine the image of a
diverging lens.
The image is virtual


The image distance is negative



It is on the other side of the lens from the outgoing light
It is on the other side of the lens from the outgoing light
The object distance is positive
The focal length is negative by convention as this is a diverging lens
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The lens equation

Similar triangles FBA and FI’I
hi d i  f

ho
f
Similar triangles OAO’ and IAI’
(within the paraxial
approximation) leads to the
lens equation

hi d i

ho d o

di di  f

do
f
Define the Lateral Magnification: m
Semester 2, 2002

1
1
1
 
do di
f
hi
di
m

ho
do
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Virtual image location through ray diagrams

Positive lens





do positive
di negative (opposite
side of the lens from
outgoing light)
m positive and >1
f positive
Negative lens




do positive
di negative (opposite
side of the lens from
outgoing light)
f negative by
convention
m positive and < 1
Semester 2, 2002
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Combination of lens

Use two or more
lens in combination



Telescope
Microscope
Image from first lens
becomes the object
of the second
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Example

Example: Two lens, both bi-convex converging lens with f1=20.0cm and
f2=25.0cm, are placed 80.0cm apart. An object is placed in front of the first
lens so that do = 60.0cm. What is the position di and magnification mtotal of
the final image?
1
1
1
1
1
1
 



di1 f1 do1 20.0cm 60.0cm 30.0cm
1
1
1
1
1
1
 



di 2 f 2 do 2 25.0cm 50.0cm 50.0cm
di1
30.0cm
m1  

 0.5
do1
60.0cm
di 2
50.0cm
m2  

 1
do 2
50.0cm
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mtotal  m1m2  0.5
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Lensmaker’s equation I

How do you make a lens of a particular focal length?

Refraction at a spherical surface
n1 sin θ1  n2 sin θ 2
n1θ1  n2 θ 2


Paraxial approximation: h is small.
= + 2 and 1=  + 
R positive, doand do positive
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n1
h
h
 ,  , 
do
R
di
n1 n2 n2  n1
 
do di
R
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Lensmaker’s equation II

Given a piece of glass in air with a refractive
index of n and radii as shown we can apply
the spherical refraction equation twice to
arrive at the lensmaker’s equation.


L
I1
I1
R2
di
1
1 1
1  1
  n  1   
d o d i
 R1 R2  f

R1
di
d o  d i  L
Thin lens approximation: L<<di L0
O
do
Refraction at the second spherical surface
n 1 1  n n 1
 

d o d i  R2
R2
I
di
do
Refraction at the first spherical surface
n1 n2 n2  n1
1
n
n 1





do di
R1
do  di
R1

O
I
di
L
do
Already included the sign change for R1 and R2 so both radii are entered as
positive values.
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The human eye, a superb optical instrument.




Light enters through the cornea,
where most refraction occurs, passes
through the lens and forms an image
on the retina.
The eye focuses on items of specific
interest by squeezing the perimeter of
the lens to form a crisp image on the
fovea, the region of central and
detailed colour vision.
This ability to accommodate between
a relaxed state and nearby objects
worsens with age as the lens
becomes more crystalline.
Define the near and far points of
vision. Far point is, ideally, infinity
and near point is 25 cm for adults.
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Common problems of the human eye: short-sightedness.

Nearsightedness has the image of an
object at infinity forming before the
retina. Vision of an object at infinity is
corrected by a negative lens forming
a virtual image at the person’s far
point.
1
1
1
1

 
f  15cm 
15cm
Power  1
 6.7 D
f(m)
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Common problems of the human eye: far-sightedness.

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Farsightedness has the image
of nearby object forming
behind the retina. Vision of an
object at 25cm, the standard
near point, is corrected by a
positive lens forming a virtual
image at the person’s near
point.
1
1
1
1



f 25cm  100cm 33cm
Power  1
   3.0 D
f(m)
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Other common problems of the human eye.



Astigmatism: An element of cylindrical correction. The image is distorted
in a particular axis. Corrected by placing a rod-like lens at an appropriate
angle.
Glaucoma: The pressure in the vitreous humor is too high. This causes
shortsightedness and loss of vision through retinal problems.
Macula degeneration: Blood supply in the region of the macula is affected
and vision is lost due to nerves being damaged.
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Magnifying glass I



Close inspection differs from distant viewing
by the size of the image on the retina.
The larger the angle subtended the greater
the detail observed.
When direct vision is inadequate, we use a
magnifying glass to enhance this further.
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Magnifying glass II

 h f N 25cm
Angular Magnification. M    h N  f  f for eye focussed at 
h do N
1 1


 N    for eye focussed at N
h N do
 f N
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Telescopes
Refracting and reflecting telescopes have been made. The limit to
refracting ones is the ability to make large, well corrected glass lens.


For a relaxed
eye, fe = do and
the distance
between the lens
is fo + fe.

M  

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fo
fe
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Telescopes II



Newtonian and Cassegrainian reflecting telescopes are the styles most of
the astronomical telescopes use for the reason that it is possible to make
very large reflecting mirrors to high surface quality.
Hale 200 inch (5.08 m) at Mt. Palomar.
Keck 10 m effective diameter from 36 segments at Mauna Kea Hawaii.
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Telescopes III

Terrestrial telescopes have upright images.


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Galilean
Spyglass folded into binoculars
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Microscopes


Object is placed in front of the objective lens and the object distance
is just longer than the focal point.
Magnification is a product of the lateral magnification of the objective
and angular magnification of the eyepiece.
mo 
hi di l  f e


ho d o
do
Me 
N
fe
 N  l  f e 

M  M e mo   
 f e  d o 
Nl
M
fe fo
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Common aberrations in lens and mirrors


Spherical aberration
 Due to reflecting or refracting
surfaces being ground to spherical
shapes when the ideal shapes are
not spherical. Parabolic for mirrors.
 Off-axis called coma.
Chromatic aberration
 Affects lens due to dispersion in
refractive index. Corrected by
making compound lens from
materials with slightly different RI
and therefore dispersion.
 Does not affect mirrors.
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Huygen’s principle


Every point on a wavefront
may be considered as a
source of tiny wavelets that
spread out in the forward
direction at the speed of the
wave itself.
The new wavefront is the
summation of all the
wavelets and is tangential
to individual wavelets.
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Diffraction



Huygens principle offers an explanation for
the effects of diffraction around objects.
This is supportive of the wave theory of
light.
In general diffraction is not observed with
large aperture optics like windows but you
certainly may observe diffraction with
pinholes.
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Wavefront refraction

Huygen’s principle is consistent
with Snell’s Law of refraction.
v1t
v2 t
sin1 
and sin2 
AD
AD
v1  c and v2  c
n1
n2
n1 sin1  n2 sin2
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Young’s double slit interference
Young’s observation when two slits were uniformly illuminated.


Consistent with interference between wavefronts emerging from
each slit.
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Young’s double slit interference II

Interference arises due to the path difference between light emerging
from each slit varying as d sin.
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Young’s double slit interference III
Constructive interference
d sin  m ,
m  0,1,2,...
Destructive interference
d sin  m  1 2  ,

m  0,1,2,...
Overall intensity envelope is due to diffraction from each single slit.
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Line spacing for double-slit interference


A screen is 1.2m from two slits
0.100mm apart. 500nm
wavelength light is incident upon
the slit. What is the separation
between bright fringes on the
screen?
L
Given: d= 0.100mm = 1x10-4 m, =500x10-9 m, L= 1.20m
d sin 1  m,
m 1
m 500  109
3
sin 1 


5
.
00

10
 1
4
d 1.00  10
x1  L1  1.20m  5.00  103  6.00mm
x2  L 2  1.20m 
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2
 12.00mm
d
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Wavelengths from double-slit interference

White light passes though two slits 0.5mm apart allows the measurement of
wavelengths on a screen 2.5m away. The estimate of the violet and red
wavelengths may be obtained.
d sin 1  m,
m  1,
sin 1  1
d d x


m mL
 red
 violet
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5.0  10 4 m 3.5  103 m

 700nm
1
2.5m
5.0  10 4 m 2.0  103 m

 400nm
1
2.5m
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Coherence

Interference is only possible if there is a fixed phase relationship between
the radiation emerging from each of the two slits.
As the light source for Young’s fringes was a distant
source, the sun, and as both slits “sampled” the
wavefront at the same time, there was interference
between the emerging wavefronts and these slits
are then said to be coherent sources.

Most light sources are incoherent. An incandescent
light bulb will emit light along the length of the
filament. Light emitted at each end of the
filament bears no phase relationship to the light
emitted at the other.
Most lasers are sources of very coherent light. They may have a phase
relationship which extends both across the beam and along the beam. This
high degree of coherence is required to improve the contrast of interference
fringes and in holography.


Semester 2, 2002
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Intensity in the double-slit interference pattern

The electric field is given by the sum of the fields from each slit. These vary with
angle in terms of a phase difference .
E   E1  E2
 E10 sin t  E20 sin(t  )

The intensity in the double slit interference pattern has periodic maxima
corresponding to d sin =m or, in terms of the phase difference :


2
d sin 

The intensity in between these maxima may be determined as a function of angle .
Semester 2, 2002
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Intensity in the double-slit interference pattern II

A phasor diagram illustrates the phase summation of the two equal fields.
E10  E20  E0 ,   
2
E () 0  2 E0 cos   2 E0 cos 

E ()  E () 0 sin t  

2

2
 2 E0 cos  sin t  
2
2

Observe intensity rather than E field

I ()  I 0 cos 2 
2
 d sin  
 I 0 cos 2 




Semester 2, 2002
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Interference in thin films


An interference effect is seen in the reflection of light from water with a
layer of oil on the surface. Rainbow coloured fringes are visible.
Light passing from material of lower to higher
refractive index undergoes a 180° phase change
upon reflection. (Higher to lower has no associated
change.)

The additional optical
path from A to B to C
allows constructive
interference when equal
to a integer multiple of
/noil. (It is assumed that
nair > noil > nwater.)

The film thickness for
normal incidence
corresponding to a
bright fringe is (/noil)/2.
Semester 2, 2002
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Newton’s rings





Semester 2, 2002
Robert Hooke, a contemporary and
antagonist of Newton, observed and
recorded the fringes which now bear
Newton’s name.
The ray, reflected in glass at point B, has
no phase change.
The ray reflected with a phase change at
point C may interfere the light reflected
at B. The optical path BCD is (2 x
physical distance +½ )
When viewed in reflection Newton’s rings
are dark in the centre due to the ½
phase change.
When viewed in transmission Newton’s
rings are bright in the centre due to two
½ phase changes.
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Wedges and optical testing



Fringes may easily be generated by placing a thin
wire or sheet of paper along one edge of a piece
of glass resting on an optically flat surface and
illuminating with monochromatic light.
Bright bands occur at when 2 t =(m+ ½) . First
LH fringe is
Extremely useful in determining the flatness of
optical surfaces.
Semester 2, 2002
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Anti-reflection coating


The deposition of a “quarter-wave” thickness of a mechanically hard
material possessing an appropriate refractive index allows the surface
reflection from glasses, camera lens and optical surfaces to be minimised.
Usually MgF2 is used as it’s refractive index is almost halfway between air
and glass.
Semester 2, 2002
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Michelson interferometer


Extremely precise measurements of
distance are possible using an optical
interferometer. Varying the position
of M1 by as little as 100nm would
alter the fringes of 400nm light from
dark to bright.
By sweeping M1 over a large known
distance from the beam splitter, it is
possible to perform a Fourier analysis
on the allowed frequencies and obtain
very high resolution spectra.
Semester 2, 2002
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Diffraction and polarization


Fresnel proposed a wave theory that predicted and explained diffraction and
interference with non-plane wavefronts. Fraunhofer explained them with
planar wavefronts and this is the more illustrative method to follow.
A counter intuitive aspect of this is that a solid disc will have a bright spot at
the centre as a result of diffraction around the edges.
Semester 2, 2002
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Single slit diffraction


Diffraction is analysed by
looking at the path difference in
light sourced from each
element of the slit.
Minima occur at:
a sin   m, m  1,2,3
Semester 2, 2002
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Intensity in the single slit diffraction pattern I



Semester 2, 2002
Minima are when sin  = /a
The intensity between minima is
given by the vector sum of the
wavefronts emerging from each
segment of the slit Dy.
There is a phase difference in the
waves so the approach may be by
phasor diagrams as with the
interference example.
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Intensity in the single slit diffraction pattern II

There is a phase difference
in the waves from each
segment of the slit given
by:
2
D 


Dy sin
The total amplitude arriving in the centre of the screen, where  is zero, and
all elements are in phase is :
E0  NDE0

The amplitude, when  is small,
between the central maximum and the
first minima is given by a similar sum
but the phase components of each
segment, given below, mean that
E < SDE0 in magnitude.
2
2
  ND  N Dy sin  a sin


Semester 2, 2002
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Intensity in the single slit diffraction pattern III

The phase difference eventually curls around
on itself when  = 2.
2
  2  N Dy sin 


sin  
a

The total amplitude reaches a secondary
maximum when  = 3
Semester 2, 2002
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Intensity in the single slit diffraction pattern IV

As Dy goes to dy, E0 and E
are related by:
E
E0


 r sin and
r
2
2
2
2
sin 2
2
E  E0
where  
a sin 
 2


The intensity is the square of the electric field so the analytical expression
for the intensity is:
  a sin   
2
 sin   
 sin 2 
I  I0 
  I0 


a
sin

 
 2 
 

    
Semester 2, 2002
2
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Diffraction and double-slit interference combined.

The observed two slit
interference pattern
incorporates diffraction
from the slit as well. The
diffraction from slit width
a is convolved with the
interference from the slits
spaced distance d apart.
2
 sin 2  

I  I0 
  cos 
2
  2  

Semester 2, 2002
2
2
2
a sin  and  
d sin 


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Airy disc and Rayleigh criterion


Circular apertures exhibit diffraction too.
This is the limit to performance of optical
systems even when all aberrations have
been removed.
In the limit where the angles are small,
the minima of the diffraction from a
circular aperture are found at:
  sin 
1.22

D

The minimum angular separation that
may be resolved corresponds to when the
maxima from one point source overlaps
with the first minima in the diffraction
pattern. This is the Rayleigh criterion.
Semester 2, 2002
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Telescope resolution


Atmospheric turbulence limits the effective
resolution of Earth based telescopes to about 0.5 arc
seconds.
(1 degree is equivalent to 60 minutes. Each minute
is divided into 60 seconds of arc, so 1 degree =
3,600 seconds.)
Hubble is limited by diffraction as it is above the
Earth's atmosphere. It has a primary mirror
aperture of 2.4m so the resolution limit is:
1.22

D
1.22  550  109 m

2 .4 m
 2.8  107 rad
Semester 2, 2002
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Resolution limit of microscopes


Microscopes are diffraction limited in their
performance if they are well designed.
It is more useful to show the limiting
distance s between two points than the
angular resolution. This is the resolving
power
 s
f
RP  s  f 
RP 

1.22f
D

2
It is not possible to resolve detail of objects smaller than the wavelength of the
radiation being used.
Semester 2, 2002
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Resolution limit of the eye

The eye is principally limited by the spacing of the
cones in the fovea. These are about 3mm apart and
there needs to be one unexcited cone in between two
excited cones for us to distinguish them as being
excited. The limiting angular resolution is therefore:
 s




f
6
6

10
m

2 102 m
 3 104 rad
The diffraction limit ranges from 8-60 x 10-5 rad depending on pupil size and this
corresponds to a resolving power between 2 and 15mm.
Spherical and chromatic aberration contribute a limit about 10mm.
The limit is really the fovea and this translates to us resolving something about 3mm
tall at 10m distance.
When looking at a microscope, the limit is about 500 X magnification. Above this is
not really useful due to diffraction.
Semester 2, 2002
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Diffraction gratings


A diffraction grating has multiple slits
from which interference between
components of an incident wavefront
and diffraction of each component at
each “slit” take place.
The multiple interference narrows the
spectrum considerably but the maxima
still occur at:
sin  

m
d
m  0,1,2,3...
Gratings may be either reflection or
transmission with the number of
lines/mm going as high as 4800. Large
sizes 400mm X 200mm give extremely
narrow lines due to the very large
number of “slits” interfering.
Semester 2, 2002
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Spectra


Spectra of line source and a continuum source are compared above.
There is usually an overlap and interference between 2nd order UV and
first order deep red / NIR wavelengths.
Third and higher order diffraction also occurs at lower efficiencies.
Semester 2, 2002
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X-ray diffraction I


Crystals were shown to diffract light by Max
von Laue. The theory of this was developed by
the Bragg’s. (The father W.H. Bragg and son
W.L. Bragg lived in Adelaide for a time.)
The x-rays reflecting from lower atomic planes
travel an increased distance:
m  2d sin  m  1,2,3...

Things can get complicated as different planes
reflect radiation. Either pure crystals or
polycrystalline powders are required.
Semester 2, 2002
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X-ray diffraction II



Polycrystalline materials or powdered pure crystals produce circular patterns
on film.
Sodium acetoacetate in (b)
X-ray diffraction is a standard analytical chemical technique for structure
determination in new molecules.
Semester 2, 2002
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Polarisation



The polarisation of a light wave refers to the
plane of oscillation of the E field in the E-M
radiation.
Different materials may allow the E field to
propagate or to be blocked depending on their
molecular structure.
Light from incandescent sources is usually
unpolarised while laser light is usually polarised
in either a known or a random direction.
Semester 2, 2002
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Polaroid polarisers


Edwin Land invented polaroid,
which preferentially transmits the
E field perpendicular to the long
molecules comprising the material.
If light is passed through a
polariser at an angle to the optical
axis, the transmitted fraction is:
E  E0 cos 


The transmitted intensity is (Malus Law):
I  I 0 cos 2 
In general an unpolarised source will be, on
average, 50% horizontally polarised and
50% vertically polarised. A polariser will
reduce the intensity to 50% of the initial
value where the remaining intensity is
linearly polarised in the axis of the polariser.
Semester 2, 2002
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Crossed and angled polarisers



If two polarisers have
orthogonal axes then
incident light is totally
absorbed (in the ideal
case).
If there are multiple
polarisers then Malus
Law is applied
successively to the
sequence of polarisers.
For example two
crossed polarisers have
a third polariser
inserted between them
with the axis oriented
at 45° to both.
Semester 2, 2002
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Polarisation by reflection



Light reflected from any non-metallic
surface has a different reflectivity for
the horizontal and vertical
polarisations.
Glare is predominantly horizontally
polarised due to this effect.
Brewster’s Angle
n2
tan  p 
n1
Semester 2, 2002
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