Chromatic Aberration
• The thin lens equation was derived using
Snell’s Law, which was derived using the
wave theory’s explanation of refraction.
• The wave theory’s explanation of color is
based on frequency and wavelength. We
can explain dispersion (rainbow and prism)
by using the wave theory’s explanation of
different colors having different speeds and
hence different indices of refraction for the
same material.
Chromatic Aberration
• Putting these two ideas together, we come up with
the prediction that the focal length of lenses will
be slightly different for different colors!
• This means that a single lens will not be
able to focus white light perfectly. We
actually observe this, and it is called
Chromatic Aberration.
• To reduce this, we can employ two lenses
with different materials to replace a single
lens. This adds to the expense and is one
reason why good cameras are so expensive.
Waves
So far, the wave theory has explained things
very well:
• speed of light
• different colors
• reflection
• refraction (dispersion, thin lenses)
To look at the next property, shadows, we
need to review the wave idea a little bit.
Waves (in general)
Sine waves are “nice”.
Other types of waves (such as square waves,
sawtooth waves, etc.) can be formed by a
superposition of sine waves - this is called
Fourier Series . This means that sine
waves can be considered as fundamental.
Waves (in general)
E = Eo sin() where  is a phase angle
which describes the location along the wave
 = 90 degrees is the crest
 = 270 degrees is the trough
Waves (in general)
E = Eo sin() where  is a phase angle
in a moving wave,  changes with both
– time (goes 2 radians in time T) and
– distance (goes 2 radians in distance )
so = (2/)*x +/- (2/T)*t
– where 2/T = 
and
– where 2/ = k and so
phase speed: v = distance/time = /T = f = /k
Waves (in general)
For nice sine waves:
E = Eo sin(kx +/- t)
For waves in general, we can break the
wave into component sine waves; this
is called spectral analysis.
Property #5: Light and Shadows
Consider what we would expect from
particle theory: sharp shadows
dark
light
dark
Light and Shadows
Consider what we would expect from
wave theory: shadows NOT sharp
crest
crest
crest
dark
dim light dim
dark
Light and Shadows
What DOES happen?
Look at a very bright laser beam
going through a vertical slit.
(A laser has one frequency
unlike white light.)
Double Slit Experiment
We will consider this situation
but only after we consider another:
the DOUBLE SLIT experiment:
note: the different colors are used
only to distinguish between the
light that goes through the
different slits. It should not be
interpreted to mean that different
colors of light go through the
different slits.
Double Slit Experiment
Note that along the solid green lines
are places where crests meets crests
and troughs meet troughs.
crest on crest followed
by trough on trough
Double Slit Experiment
Note that along the dotted lines
are places where crests meets troughs
and troughs meet crests.
crest on trough followed
by trough on crest
crest on crest followed
By trough on trough
Double Slit Experiment
Further explanations are in the
Introduction to the Computer Homework
Assignment Vol 5 #3 on Young’s Double Slit.
crest on trough followed
by trough on crest
crest on crest followed
By trough on trough
Double Slit Experiment
Our question now is: How is the pattern
of bright and dark areas related to the
parameters of the situation: , d, x and L?
bright
x
dim

d
bright
dim
L
SCREEN
bright
Double Slit Experiment
Magnified view of the slit and the direction of
the light to the first (not central) maximum:
sin()= / d
purple angles  90o
note: requires  < d
d

Double Slit Experiment
Magnified view of the slit and the direction of
the light to the first (not central) maximum:
sin()= / d
d

purple angles  90o
note: requires  < d
Double slit: an example
n = d sin(n)  d tan(n) = d (xn / L)
d = 0.15 mm = 1.5 x 10-4 m
x = ??? measured in class
L = ??? measured in class
n = 1 (if x measured between adjacent bright spots)
 = d x / L = (you do the calculation)
Interference: Diffraction Grating
The same Young’s formula works for multiple
slits as it did for 2 slits.
lens
bright
s1
d
s2
s3
s4
s5
4
s2 = s1 + 
s3 = s2 +  = s1 + 2
s4 = s3 +  = s1 + 3
s5 = s4 +  = s1 + 4
bright
Interference: Diffraction Grating
With multiple slits, get MORE LIGHT and
get sharper bright spots.
lens
bright
s1
d
s2
s3
s4
s5
4
s2 = s1 + 
s3 = s2 +  = s1 + 2
s4 = s3 +  = s1 + 3
s5 = s4 +  = s1 + 4
bright
Interference: Diffraction Grating
With 5 slits, get cancellation when s = 0.8;
with two slits, only get complete
cancellation when s = 0.5 .
lens
s1
d
s2
s3
s4
s5
3.2
bright
dark
s2 = s1 + .8
bright
s3 = s2 + .8 = s1 + 1.6
s4 = s3 + .8 = s1 + 2.4
s5 = s4 + .8 = s1 + 3.2
Colors and Wavelengths
for visible light, using the diffraction grating,
we come up with the following:
–
–
–
–
–
–
violet
blue
green
yellow
orange
red
400 - 450 nm
450 - 500 nm
500 - 550 nm
550 - 600 nm
600 - 650 nm
650 - 700 nm
Colors and wavelengths
The wavelengths specified for the colors on the
previous slide are only approximate, and they do
vary slightly with each person.
On any test, I will give you one color leeway, e.g., if the
wavelength falls in the green range (previous slide), but
you answer either blue or yellow, I will count your answer
correct.
Is the way we see different wavelengths
(frequencies) of light similar to the way we
hear different frequencies of sound?
Sound and frequencies
For sound, our ears function like a Fourier
analyzer – we can actually hear each
frequency. If we mix frequencies, we hear
a chord. We can distinguish the base from
the treble; we can distinguish the guitar
from the piano.
Colors and frequencies
For light, our eyes do NOT act this way. If we mix
frequencies, we see only one color which is
different from either of the “pure” colors
(frequencies). We have two different types of
retinal cells: rods and cones.
The rod cells are not that closely packed, and only
send a black and white signal.
There are three different types of cone cells which
act as color light receptors.
The next slide shows only a rough picture. You
should look at more advanced texts to get a more
accurate picture.
Cone cells
Cone cell sensitivity to different wavelengths.
Blue peaks at 437 nm; green at 533 nm; red at 564 nm
(see: https://www.unm.edu/~toolson/human_cone_response.htm)
400 450 500 550 600 650 700 (in nm)
If only the “blue” cone is activated, the color is violet.
If both the “blue” and “green” cones are activated, and the “blue” gives a
stronger signal, the color is blue.
If both the “blue” and “green” cones are activated, and the “green” gives a
stronger signal, the color is green.
Cone cells
Cone cell sensitivity to different
wavelengths
400 450 500 550 600 650 700 (in nm)
If both the “green” and “red” cones are activated, and the “green” gives a
stronger signal, the color is yellow.
If both the “green” and “red” cones are activated, and the “red” gives a
stronger signal, the color is orange.
If only the “red” cone is activated, the color is red.
Colors and Wavelengths
For other types of light (based on the technology
used to make the light):
– radio: f is  1MHz for AM,  100 MHz for FM
so  is
1 km to 10 cm
– microwaves/radar: 10 cm to 1 mm
– infrared:
1 mm to 700 nm
– visible
700 nm to 400 nm
– ultraviolet
400 nm to 10 nm
– x-ray &  ray
10 nm on down
Diffraction: single slit
How can we explain the pattern from light
going through a single slit?
screen
w
x
L
Diffraction: single slit
If we break up the single slit into a top half
and a bottom half, then we can consider the
interference between the two halves.
screen
w
x
L
Diffraction: single slit
The path difference between the top half and
the bottom half must be /2 to get a
minimum.
screen
x
w
L
Diffraction: single slit
This is just like the double slit case, except the
distance between the “slits” is w/2, and this
is the case for minimum: (w/2) sin() = /2
screen
x
w
L
Diffraction: single slit
In fact, we can break the beam up into 2n
pieces since pieces will cancel in pairs.
This leads to: (w/2n) sin(n) = /2 ,
or w sin(n) = n for MINIMUM.
screen
x
w
L
Diffraction: single slit
-2
-1
0
-2
-1
1
2
REVIEW:
• For double (and multiple) slits:
n = d sin(n) for MAXIMUM
(for ALL n)
0
1
2
• For single slit:
n = w sin(n) for MINIMUM
(for all n EXCEPT 0)
Diffraction: single slit
NOTES:
• For double slit, bright spots are equally
separated.
• For single slit, central bright spot is larger
because n=0 is NOT a dark spot.
• To have an appreciable , d and w must be
about the same size as & a little larger than 
• Recall that for small angles, sin)  tan() = x/L
Diffraction: circular opening
If instead of a single SLIT, we have a
CIRCULAR opening, the change in
geometry makes:
the single slit pattern into a series of
rings; and
the formula to be: 1.22 n = D sin(n) .
The computer homework program on
Resolution, CH Vol 5 #4 shows the rings in
several diagrams and the use of this
equation.
Diffraction: circular opening
Since the light seems to act like a wave and
spreads out behind a circular opening, and
since the eye (and a camera and a telescope
and a microscope, etc.) has a circular
opening, the light from two closely spaced
objects will tend to overlap. This will
hamper our ability to resolve the light (that
is, it will hamper our ability to see clearly).
Diffraction: circular opening
How close can two points of light be to still be
resolved as two distinct light points instead
of one? One standard, called the Rayleigh
Criterion, is that the lights can just be
resolved when the angle of separation is the
same as the angle of the first dark ring of
the diffraction pattern of one of the points:
limit = 1 from 1.22 *  = D sin(1) .
Rayleigh Criterion: a picture
The lens will focus the light to a fuzzy DOT
rather than a true point.
D lens
Rayleigh Criterion: a picture
The Rayleigh minimum angle,
limit = sin-1(1.22 /D) = tan-1(x’/s’).
D lens
x’
s’
Rayleigh Criterion: a picture
If a second point of light makes an angle of
limit with the first point, (blue angle = green
angle) then it can just be resolved.
lens
D
x
x’
s
s’
Rayleigh Criterion: a picture
In this case: limit = sin-1(1.22 /D)
= tan-1(x’/s’) = tan-1(x/s) .
D lens
x
x’
s
s’
Rayleigh Criterion: an example
• Consider the (ideal) resolving ability of the
eye
• Estimate D, the diameter of the pupil
• Use  = 550 nm (middle of visible spectrum)
• Now calculate the minimum angle the eye
can resolve.
• Now calculate how far apart two points of
light can be if they are 5 meters away.
Rayleigh Criterion: an example
with D = 5 mm and  = 550 nm,
limit = sin-1 (1.22 x 5.5 x 10-7 m/.005 m)
= 7.7 x 10-3 degrees
= .46 arc minutes
so x/L = tan(limit), and
x = 5m * tan(7.7 x 10-3 degrees) = .67 mm
Rayleigh Criterion: an example
• Estimate how far it is from the lens of the
eye to the retinal cells on the back of the
eye.
• With your same D and and so same limit),
now calculate how far the centers of the two
dots of light on the retina are.
• How does this distance compare to the
distance between retinal cells (approx.
diameter of the cells)?
Rayleigh Criterion: an example
• L = 2 cm (estimation of distance from lens to
retinal cells)
• from previous part, limit = 7.7 x 10-3 degrees
• so x = 2 cm * tan(7.7 x 10-3 degrees)
= 2.7 m .
Limits on Resolution:
• Imperfections in the eye (correctable with
glasses)
• Rayleigh Criterion due to wavelength of
visible light
• Graininess of retinal cells (Note that in low light
where only the rods are activated, we cannot resolve very
well because the rod cells are not packed as closely as the
cone cells are. Also in low light we only see in black and
white – not in color.)
Limits on Resolution:
further examples
• hawk eyes and owl eyes
• cameras:
– lenses (focal lengths, diameters)
– films (speed and graininess)
– shutter speeds and f-stops
• Amt of light  D2 t
• f-stop = f/D
– f-stops & resolution: resolution depends on D
Limits on Resolution:
further examples: microscope
1.22 n  = D sin(n) where 1 = limit ,
so 1.22  = D sin(limit) ; also sin(limit) ≈ tan(limit) = h/s.
Therefore, 1.22  ≈ D*h/s, or h ≈ 1.22 *s/D
where h is the smallest size that is resolvable.
This means that h ≈  
microscopes: smallest size = = 500 nm = .5 m
– can easily see .5 mm, so M-max useful = 1,000
– can reduce  (and hence h) by
– a) immersion in oil (since v and  are smaller in oil) ,
b) use blue light (since blue has smaller  than red).
Limits on Resolution:
further examples
• other types of light
– x-ray diffraction (use atoms as slits)
– IR
– radio & microwave
• surface must be smooth on order of 
Polarization
• Experiment with polarizers
• Particle Prediction?
• Wave Prediction?
– Electric Field is a vector: 3 directions
• Parallel to ray (longitudinal)
– Maxwell’s Equations forbid longitudinal
• Two Perpendicular (transverse)
Polarization: Wave Theory
#1 Polarization by absorption
(Light is coming out toward you)
unpolarized
polarized
light
polarizer
only lets
vertical componet
through
no light
gets
through
polarizer
only lets
horizontal component
through
Polarization: Wave Theory
Three polarizers in series:
Sailboat analogy:
North
wind
sail
force on
sail
boat goes along
direction of keel
Polarization: Wave Theory
#2 Polarization by reflection
– Brewster Angle: when refracted + reflected = 90o
– Sunglasses and reflected glare
incident ray
vertical
horizontal
reflected ray
no problem with horizontal
almost no vertical since vertical
is essentially longitudinal now
surface
vertical can be transmitted
refracted ray
Polarization: Wave Theory
#3 Polarization by double refraction
– different indices of refraction (n’s) in different
directions due to different bonding
#4 Polarization by scattering