Download Chapter 24 Lecture Notes

Chapter 24 Lecture Notes
Physics 2424 - Strauss
Formulas:
Double slit diffraction:
Bright fringes when
Dark fringes when
Single slit diffraction
Bright spots in diff. grating
Thin Films:
2t = mλfilm
2t = (m +1/2)λfilm
S = S0 cos2θ
tan θB = n2/n1
m = 0,1,2,...
sinθ = mλ/d
sinθ = (m+1/2)λ/d
m = 0,1,2,...
sinθ = λ/W
for first dark fringe
sin θ = mλ/d
m = 0,1,2,...
(NPC ≡ Net Phase Change)
Constructive interference when NPC = 0
Destructive interference when NPC = 1/2λfilm
Destructive interference when NPC = 0
Constructive interference when NPC =1/2λfilm
1. DISPERSION: RAINBOWS AND PRISMS
Dispersion occurs because the index of refraction depends on the wavelength of
the light. White light is made up of colors from the entire spectrum. The
spectrum is ROYGBIV (red, orange, yellow, green, blue, indigo, violet). In a
prism, violet light has a larger index of refraction than red light, so it is
diffracted more. This is called dispersion. Up until now, we have been treating
the index of refraction as being independent of wavelength, but it is really not.
Dispersion is responsible for splitting white light into different colors in a
rainbow, or from a prism.
Red
Violet (n for violet is greater than n for red).
We see a rainbow due to dispersion. Light from the sun (behind you) reflects off
of water drops like in the figure. Violet is diffracted more than red, so you see
the violet part of the rainbow lower than the red part. We see red when we look
up high, and violet when we look down low. (This is figure 24-13 in the book)
1
Incoming Sunlight
Violet
Red
2. THE WAVE NATURE OF LIGHT
2.1 Introduction: Waves and Particles
Waves and particles behave differently. Geometric optics (Chapter 23) primarily
looks at light as a particle. In this chapter we will talk about the wave nature of
light. Waves that are directed exactly perpendicular to a slit will diffract, or
change direction at the slit. However, particles entering perpendicular will either
go straight through or they bounce back. We find that light will diffract and
behave like a wave. What about geometric optics? It only works when the
wavelength of light is small compared to the geometry involved. When the size
of the wavelength is comparable to the size of the physical objects involved, then
light will exhibit wave-like properties.
What happens when a wave or a particle goes through two slits? A particle must
go through either slit but a wave goes through both and produces an interference
pattern.
However, there are cases when I can make light behave like a particle even when
the geometry is small. In fact, I can show light to be both a wave and a particle.
We will talk more about that in chapter 27. However, now we will focus only on
the wave nature of light. (electromagnetic radiation).
2.2 Huygen’s Principle
Christiaan Huygen’s principle predicts where the wavefront will be at a later
time. It says that every point on a wavefront can be considered as a source for
more wavefronts. The new wave will be tangent to all the little wave fronts.
(See figure 24-1, and 24-2). This can be used to explain what happens when light
encounters a slit.
2
2.3 The Principle of Superposition
You must understand this principle to understand chapter 24. When two waves
are in the same region of space, the amplitude of the produced wave is just the
sum of the amplitudes of the two waves. Therefore, waves can exhibit
constructive or destructive interference. We talked about this in chapter 11.
2.3.1 CONSTRUCTIVE INTERFERENCE
Light is a wave and waves have peaks and valleys. If peaks from two waves
align, the E and B fields are added and the intensity increases. This is
constructive interference. The waves are in phase. The light is brighter.
2.3.2 DESTRUCTIVE INTERFERENCE
If a peak lines up with a valley, then the E and B field cancel and we have
destructive interference. The intensity is 0. The waves are 90° out of phase.
There will be no light.
2.3.3 COHERENT SOURCES
Light will only demonstrate noticeable interference effects if it is coherent light.
Two waves of light are from a coherent source if the phase relationship between
the waves does not change. That is the two light rays always have their peaks and
valleys in the same phase relative to each other. If the relative phases stays the
same as time passes, then the constructive or destructive interference remains. In
order to do this, the different waves must have the same velocity, wavelength,
and frequency. Lasers are coherent sources of light, while light bulbs and
fluorescent lamps are incoherent sources. That is why we don’t normally see
interference effects from normal light sources.
2.4 Applications of the Principle of Superposition
Remember that for small angles sinθ ≈ θ when θ is expressed in radians. C = πd
= 2πr. So we see that an arc length a = θr in radians. Therefore, θ = a/r ≈ d/r =
sinθ )
2.4.1 YOUNG’S DOUBLE SLIT EXPERIMENT
If light from a coherent source is incident on two
slits, we may get constructive interference or destructive
interference depending on the difference in the distance
that the light has to travel. Thomas Young used this idea
to demonstrate that light was actually a wave, since
particles will not demonstrate constructive and destructive
interference. Suppose the light from one slit travels a
distance l1 and the light from the other slit travels a
3
l1 = l2
distance l2.
There will be constructive interference, (or a bright spot) where l2 = l1 + λ . For
instance when l1 = l2 (at the center) there will be a bright spot. If the difference
between the two distance is one half a wavelength, there will be destructive
interference, (a dark spot) when l2 = l1 + λ/2.
Now we draw the situation in more detail and do caculations. When the screen is
far compared to the slit distance, the waves are approximately parallel and all
angles θ are the same.
θ
l1
θ
θ
S1
l2
d
θ
S2
∆l
When ∆ l is equal to one wave length, the light is constuctive and bright. When it
is one half wave length, the band is destructive and dark.
So Bright fringes when sinθ = m λ /d for m = 0,1,2,...
Dark fringes when
sinθ = (m+1/2)λ /d for m = 0,1,2,...
The fringes are called higher order maximums (1st order, 2nd order, etc.)
Problem/Demonstration: Determine the wave length of laser from double slit
diffraction pattern using the distance between the zeroth order bright spot and the
first order bright spot.
Suppose we are a distance of 20.0 m from the screen and the distance between the
slits is .250 mm. We measure the distance between the zeroth order maximum
and the first order maximum to be 0.0500 meters. What is the wavelength of the
light?
2.4.2 SINGLE SLIT DIFFRACTION
We talked about Huygen’s Principle before. This can lead to diffraction patterns
from a single slit if the wavelength of light is approximately the same size (or
larger) than the size of the slit. The size is important for diffraction because if
the wavelength is much smaller than the geometry we will not see wave-type
4
phenomena. That is why we don’t see diffraction patterns around doors. Also,
coherence is important for creating visible diffraction patterns. (Most natural
light is incoherent). Diffraction occurs even if the light is incoherent, but
diffraction patterns aren’t seen.
θ
W λ /2
Bottom cancels out the top and so the first
dark fringes are at
sinθ = m λ/W
λ
Problem/Demonstration: We have determined the wavelength of the laser. If we
now know the distance to screen, L, and slit width, W , what will the distance
between the first two dark fringes be?
Illustration: If you don’t have your glasses on and you make a small hole with
your fingers, you can actually focus on distant objects. Why does this work?
(Diffraction can act like a diverging lens).
Because the angle of diffraction depends on the wavelength, we can use a slit to
separate the colors in white light.
2.4.3 DIFFRACTION GRATING
Diffraction gratings put many slits in a row. The bright spots occur at the same
location as for a double slit experiment.
First one travels an extra λ, the second one
and extra 2λ, etc.
They are at sinθ = mλ /d for m = 0,1,2,...
However, the peaks are much sharper and narrower on a diffraction grating.
For two slits when you move slightly off the maximum, you are just slightly out
of phase. When there are multiple slits, and you move off slightly. The light
from a distant slit is already out of phase and you get a dark spot.
Consider the case where I have 100 slits and I move a distance that is 1/100 λ
away from the peak. If I had only two slits the phase difference is only 1/100 λ ,
5
so things don’t change much because the change is only 1(λ + λ/100) = 1.01λ.
However, if I have 100 slits the extra distance from the 50th slit away is 50(λ +
λ/100) = 50.5λ which is one half wave length. So slit 1 cancels slit 51, etc.
Problem: A diffraction grating contains 4820 lines/cm and is used with blue light
(λ = 470 nm). What is the highest order bright fringe that can be seen with this
grating?
2.4.4 THIN FILM INTERFERENCE
air n = 1.0
Gas n = 1.40
Water n = 1.33
Thin film interference will occur because the distance traveled by the light
reflected off the gas is different than the distance traveled by the light reflected
off the water.
The wavelength that matters is the wavelength of light in the gas
since that is where the difference occurs.
n = c/v = (c/f)/(v/f) = λ vacuum /λ film
λfilm = λvacuum /n
So you would think that we get constructive interference when 2t = mλfilm =
m λvacuum /n.
By this is not quite right because when light is reflected from an object
with a larger index of refraction, the wavelength changes by λ/2.
Remember the rope analogy from chapter 11, figure 11-25. We get a net phase
change of 1/2λ when going from a medium with a lower index of refraction to a
medium with a higher index of refraction. We do not get this phase change when
going from an index with a higher index of refraction to one with a lower index
of refraction.
6
First do this attached to wall and loop, then attached to heavy and light strings.
Some goes forward and some is reflected.
So we get constructive interference when
2t = (m+1/2)λfilm =(m+1/2)λvacuum /n.
Destructive interference would be when 2t = mλfilm = mλvacuum /n.
So we see that if the net phase change (NPC) is 1/2λfilm, then we get
2t = mλfilm
Destructive interference
2t = (m +1/2)λfilm Constructive interference
What if the media bottom was not water, but glass with n=1.52. Then there is
also a phase change at the gas, glass boundary, and no net phase change.
We find that for no NPC, we get
Constructive interference
2t = mλfilm
2t = (m +1/2)λfilm Destructive interference
Destructive interference will look dark. Just before the bubble breaks, the
thickness will be 0, and the 1/2 phase difference will cause destructive
interference, and the bubble will look dark at its weakest point.
7
Cancellation of a color occurs when there is destructive interference. Something
looks a certain color when there is constructive interference.
On many cameras and fine quality lenses, the manufacturer will try to eliminate
reflection of the light. To do this, they make compound lense which are lenses
made from a number of different types of glass. For instance, if I coat glass with
magnesium flouride the reflection from the surface can destructively interfere,
and I will get no reflected light of a certain wavelength. MgF2 has n ≈ 1.38 and
glass has n ≈ 1.5 .
Problem: A nonreflective coating of magnesium flouride covers the glass of a
camera lens. Assuming the coating prevents light of 565 nm from reflecting,
what is the minimum thickness of the coating?
2.4.5 AN AIR WEDGE
An air wedge also uses this same principle as thin film interference to see if a
surface is completely flat.
Glass n = 1.52
Air, n = 1.00
Glass n = 1.52
If it is flat, then the wavelength difference will be because of the geometry and
diffraction patterns will be regular. If it is not flat, then the diffraction patterns
will not be regular (see figure 24-27)
Where will the phase of the reflected ray change by 1/2? (At the first glass to air
boundary). What will the fringe look like where the plates touch? (It will be
dark because of the 1/2 λ change.)
3. POLARIZED EM WAVES
Because light is a transverse wave, it can be polarized. Normal light has the
direction of its electric field randomized in space. A polarizer only allows the
electric field going in one direction to pass through.
Put it in a polarizer
Produces
8
The size of slit depends on wavelength of object. For instance, a microwave grid
stops microwaves but not visisble light. A cooling tray will actually polarize
microwaves.
What happens to the intensity of the beam? If I allow unpolarized light to go
through a polarizer, I find that the intensity of the polarized light is cut by 1/2.
When I add another polarizer, the change in intensity depends on the angle of the
second polarizer and is given by
S = S0 cos2θ (Malus’ Law) When θ is 90°, no light passes through.
What happens if I put a third polarizer in.
S1 = (1/2) S0
S2 = S1 cos2θ1
S3 = S2 cos2θ2
S3 = S1 cos2θ1 cos2θ2 = (1/2) S0 cos2θ1 cos2θ2
The polarizer is actually picking out the component of the electric field which is
parallel to the axis of polarization.
Demonstration/Problem: Polarized light.
If I have two polarizers at 90° with respect to each other, then no light will get
through because S2 = S1 cos2θ1 = S1 cos2(90) = 0.
However, if I put a polarizer between them at an angle of 45° with respect to the
two polarizers, I will find,
S3 = (1/2) S0 cos245 cos245 = .125S0 so some of the light gets through.
Light will get through as long as one of the polarizers is not at 90° to the one
directly in front of it.
3.1.1 POLARIZATION BY REFLECTION
When light is reflected off of a surface it is also polarized. Any light reflected
off of a surface will be partially polarized along the horizontal direction.
Polaroid sunglasses have their axis of polarization along the vertical direction to
9
stop glare from reflected light which is always partially polarized along the
horizontal direction. At a certain angle the reflected light will actually be totally
polarized. This angle is called Brewster’s angle and it is given when
tan θB = n2 /n1
The book shows that at this angle, the angle between the reflected and the
refracted rays is 90°.
10