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

$Page 1: Chromatic Aberration The thin lens equation was derived using Snell’s Law, which was derived using the wave theory’s explanation of refraction. The wave$
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

lightdark dark

Light and Shadows

Consider what we would expect from

wave theory: shadows NOT sharp

lightdark darkdimdim

crest

crest

crest

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 situationbut 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.


Note that along the solid green lines

are places where crests meets crests

and troughs meet troughs.

crest on crest followedby trough on trough


Note that along the dotted lines

are places where crests meets troughs

and troughs meet crests.

crest on crest followedBy trough on trough

crest on trough followedby trough on crest


Further explanations are in the

Introduction to the Computer Homework

Assignment Vol 5 #3 on Young’s Double Slit.

crest on crest followedBy trough on trough

crest on trough followedby trough on crest


Our question now is: How is the pattern

of bright and dark areas related to the

parameters of the situation: , d, x and L?

d

SCREEN

L

xbright

bright

dim

dim

bright


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


Magnified view of the slit and the direction of the light to the first (not central) maximum:

sin( / d

purple angles 90o

note: requires < dd

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.

d

lens bright

bright

s1

s2

s3

s4

s5

s2 = s1 + s3 = s2 + = s1 + 2s4 = s3 + = s1 + 3s5 = s4 + = s1 + 4

4


With multiple slits, get MORE LIGHT and get sharper bright spots.

d

lens bright

bright

s1

s2

s3

s4

s5

s2 = s1 + s3 = s2 + = s1 + 2s4 = s3 + = s1 + 3s5 = s4 + = s1 + 4

4


With 5 slits, get cancellation when s = 0.8; with two slits, only get complete cancellation when s = 0.5 .

d

lens bright

bright

s1

s2

s3

s4

s5

s2 = s1 + .8s3 = s2 + .8 = s1 + 1.6s4 = s3 + .8 = s1 + 2.4s5 = s4 + .8 = s1 + 3.2

dark

3.2

Colors and Wavelengths

for visible light, using the diffraction grating, we come up with the following:– violet 400 - 450 nm– blue 450 - 500 nm– green 500 - 550 nm– yellow 550 - 600 nm– orange 600 - 650 nm– red 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 frequenciesFor 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

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 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)

Cone cells

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.

Cone cell sensitivity to different wavelengths

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?

w

screen

L

x


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.

w

screen

L

x


The path difference between the top half and the bottom half must be /2 to get a minimum.

w

screen

L

x


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

w

screen

L

x


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.

w

screen

L

x


REVIEW:• For double (and multiple) slits:

n = d sin(n) for MAXIMUM

(for ALL n)

• For single slit:

n = w sin(n) for MINIMUM

(for all n EXCEPT 0)

-2 -1 0 1 2

-2 -1 0 1 2


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 openingIf 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.

lensD


The Rayleigh minimum angle,

limit = sin-1(1.22 /D) = tan-1(x’/s’).

lensD

s’

x’


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

s’

x’

s

x


In this case: limit = sin-1(1.22 /D)

= tan-1(x’/s’) = tan-1(x/s) . lensD

s’

x’

s

x

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.


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


• 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)?


• 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 polarizeronly letsvertical componetthrough

polarizedlight polarizer

only letshorizontal componentthrough

no lightgetsthrough


Three polarizers in series:

Sailboat analogy:

North wind

sail

force onsail boat goes along

direction of keel


#2 Polarization by reflection– Brewster Angle: when refracted + reflected = 90o

– Sunglasses and reflected glare

surface

incident rayvertical

horizontal

refracted ray

reflected rayno problem with horizontal

almost no vertical since vertical is essentially longitudinal now

vertical can be transmitted


#3 Polarization by double refraction– different indices of refraction (n’s) in different

directions due to different bonding

#4 Polarization by scattering

Documents

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