Physics 9
Fall 2009
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DIFFRACTION
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Introduction
In these experiments we will review and apply the main ideas of the interference and diffraction of light. After reviewing the basics, we will use the principles of diffraction to determine
the wavelength and frequency of a laser source. Using this result, we will then use the same
laser source to determine the width of a human hair.
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The Interference and Diffraction of Light
In this section we review the main ideas of the interference of light. Although the discussion
will be given in terms of light, the principles given apply to any system of linear waves such
as sound waves, or waves on a string (water waves, or gravitational waves, turn out to be
more complicated, in general, and their interference properties aren’t necessarily as simple).
2.1
Light as an Electromagnetic Wave
In lecture, we have seen that what we call light is nothing more than wigglings of the electromagnetic field. From Faraday’s law, a changing electric field produces a changing magnetic
field. In turn, by Maxwell’s correction to Ampere’s law, a changing magnetic field produces
a changing electric field. This changing electric field again makes a changing magnetic field,
and so on. These changing fields travel out from a source at the speed of light. As Maxwell
himself said, “We can scarcely avoid the inference that light consists in the transverse undulation of the same medium which is the cause of electric and magnetic phenomena.” Let’s
begin by reviewing some basic properties of a wave.
Consider a wave like that seen in the
figure to the right. In general the wave
has some height, given by the amplitude
A. The tops of the waves are the crests,
while the bottoms of the waves are the
troughs. The distance from peak to peak
(or from trough to trough) is called the
wavelength, λ.
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The whole wave moves to the right with a speed v. If we sit at one spot on the x−axis
and wait, we’ll see the wave passing by. If we watch one peak go by, and count the seconds
until the next peak goes by, then this time is called the period of the wave, T . The number
of peaks that go by every second is called the frequency, f = 1/T . The speed of the wave
is related to the frequency and wavelength by v = λf . In the case of a light ray, then the
speed of the wave is the speed of light, c, and so λf = c, but this is only true for a wave
traveling at the speed of light.
2.2
Interference of Waves
Recall that whenever two waves interact,
they affect each other, giving a single
resultant effect. We say that the waves
interfere with each other. If the waves
interfere in such a way that the amplitude
of the resulting effect is bigger than either
of the two waves then we say that the
waves interfere constructively. If, on the
other hand, the net effect tends to cancel
out the total amplitude, the we say the
waves interfere destructively.
Suppose we have two identical waves, a
“green” one, traveling to the left, and a
“red” one traveling to the right, as seen in
the figure to the right. These two waves
are going to interfere with each other, in
general. The net effect is to give a new
wave, seen in blue. The height of the wave
changes as the waves pass through each
other. Notice that at times t = 81 T and
t = 58 T , the two waves completely overlap.
This gives the large resultant wave in blue,
with the net amplitude bigger than either
of the two incident waves. So, when the
two waves completely overlap, or when the
peaks or troughs overlap, the wave interfere
constructively. However, when the peak of
one wave overlaps with the trough of the
other, as at times t = 38 T , and t = 78 T ,
then the resultant wave is completely canceled! So, we see that when the peak of one
wave overlaps with the trough of another,
the two waves interfere destructively.
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2.3
Diffraction
Whenever we try to force light to pass
through a small hole, it spreads out; we say
that it diffracts through the hole. Suppose
that, instead of a single hole, we have two
small slits spaced very close together in an
opaque sheet. Now, we send a light source,
such as a laser beam, through these slits.
The light passes through the slits and
makes a pattern on a distant viewing
screen as seen in the figure to the right.
Instead of just two bright spots behind
the slits, which would have been the case
if the light just went straight through, we
see an array of bright and dark lines. We
want to see where this pattern comes from.
We can understand the origin of
the pattern by looking at the figure
to the left. Because it’s only one
color (monochromatic), the laser
beam has a specific wavelength,
λ. When the laser passes through
the two slits, each slit acts like
a new source of light. The light
from these two sources spreads out
and overlap with each other. This
overlap causes the two new light
rays to interfere, as we discussed
in the previous section. The bright
spots on the screen occur when the
peaks or troughs of the two waves
overlap (constructive interference),
while the dark spots occur when
the peaks of one wave and the
troughs of the other overlap (destructive interference). In order to
understand the precise placement
of the bright and dark fringes, we
have to look at the geometry of the
situation. Consider the diagram in
Fig. 1.
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Figure 1: The light waves coming from the two slits interfere with each other.
Suppose we look at the light arriving from both slits at some point P on the viewing
screen. The light from the top slit travels a distance r1 from the slit to the point, while the
light from the bottom slit travels a longer distance, r2 , because it’s further away from the
point. So, the bottom light ray travels an extra distance ∆r = r2 −r1 further than the top ray.
If the distance between the two slits is d, then from the geometry of the picture, ∆r =
d sin θ. Now, as we’ve discussed before, depending on how the peaks and troughs overlap,
the light rays can interfere either constructively (a bright spot), or destructively (a dark
spot). If the peaks or troughs of the two waves exactly overlap, then we get a bright spot;
this happens when the extra distance that the two waves travel is exactly a whole number
of wavelengths. In other words, when ∆r = mλ, where m = 0, 1, 2, · · · is a whole number.
On the other hand, if the extra distance is off by a half-wavelength, i.e., ∆r = m + 12 λ,
then we get a dark spot. Recalling that ∆r = d sin θ, we see that we get a bright spot when
d sin θm = mλ,
(1)
where we now have a range of angles θm , one for each value of m. This leads to the multiple
interference fringes as seen before.
From the geometry in Fig. 1 we see that the tangent of the angle is the height, y, of the
point P , divided by the distance to the screen, L. This means that the position of the mth
bright spot is
ym = L tan θm .
(2)
This means that if we measure the distance from the center of the diffraction pattern (when
θ = 0) to one of the bright fringes, then we can determine the diffraction angle, and then if
we knew the wavelength of the light, we could determine the slit spacing, d.
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3
Some Prelab Questions
Before we begin the lab we’ll take some time to understand a little bit more about the
concepts we’ve discussed above. Please box your numerical or algebraic answers.
1. Suppose we had two speakers, one placed some distance directly behind the other and
both pointing along the same direction, facing us. Now, we connect the speakers to
the same source and turn them on producing a single continuous tone of 440 Hz. If the
speed of sound in air is about 343 m/s, then how far would we have to put the back
speaker behind the front one so that we never hear the tone when standing anywhere
directly in front of the speakers? (Note: you can assume that the speakers are small
enough that the front one does not affect the sound from the back one.)
2. What is the difference in angles, ∆θ, between the mth and (m + 1)th bright fringes?
Show further that if θm and θm+1 1, then ∆θ ≈ λd , independent of m. Hint: for very
small angles, sin θ ≈ θ.
3. What is the distance ∆y between the mth and (m + 1)th bright fringes? Using the
ideas above show that, if tan θ ≈ θ for small angles, then ∆y ≈ λd L.
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4
The Lab
Ultimately, what we want to do is to measure the thickness of a human hair using laser
diffraction. Placing a hair in front of the laser beam acts like a double slit diffraction
grating. Thus, we can use the methods described above to determine the slit spacing, which
is just the width of the hair. To determine this thickness, we first need to determine the
wavelength of the laser, which is where we begin.
4.1
Measuring the Wavelength of a Laser
Materials Needed
• Laser pointer
• Diffraction grating
• Tape
• Ruler
• Paper
• Apparatus track with attachments for the pointer, grating, and paper
For this experiment we will actually use a diffraction grating, rather than a two-slit screen.
Rather than only two slits, the grating has an entire screen of tiny grooves, each of which
behaves like a slit. As we’ve discussed in lecture, the diffraction effect for the grating is
exactly the same as for a double slit screen. In particular, Eq. (1) still holds.
Begin by shining the laser pointer through the grating towards the wall. Warning although these lasers are very low-powered, DO NOT shine them into anyone’s
eyes! Always make sure that no one is in the path of the laser beam before
turning it on. Rather than seeing the interference fringes, as in the case of the double slit,
we see a regular array of dots. This does not affect the diffraction effect, at all.
• Why do we see dots, instead of fringes?
• The grating that we will be using has 530 lines per millimeter, meaning that every
millimeter on the screen has 530 tiny little ridges. What is the spacing, d for this
grating in meters? What is it in nanometers (nm)?
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Now, we need to make very precise measurements of the position of these dots. Arrange
the laser pointer on the track such that you can still turn it on. Now, arrange the diffraction
grating on the track such that the laser pointer will shine through the grating screen. The
distance between the laser and the grating can be any value you like.
• Why does the distance of the laser light source to the diffraction grating not matter?
(Note - the laser pointer has a lens which can focus the laser beam a bit - ignore this
in your answer and assume that the beam is always focused.)
Now, place a sheet of paper on the paper screen some distance L behind the grating
such that you can easily see several distinct dots (several orders). The distance between the
grating and the paper does, of course, matter.
4.1.1
First Run
Using the markings on the track measure the distance, in centimeters, from the grating to
the paper as accurately as you can. Record this value in the table below. On the paper,
carefully mark the location of the dots, specifically noting the center (m = 0) dot. Try to
get as many orders as you can. Once you have the positions marked then move the paper to
the table and use the ruler to measure the distances of the dots from the center dot, again
in centimeters, for a given order. Record your results. Using Eqs. (1) and (2) calculate the
angles θm and wavelengths λ for each order. Record these values in the table.
Order, m
0
1
2
3
Length, L
First Run Results
Distance from center, ym Angle, θm
Wavelength, λ
• What is the average value of your wavelengths for this run? Express your answer in
terms of nanometers.
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4.1.2
Second Run
To get better results we will perform the experiment twice, at two different distances. Set
up the apparatus as before, but with a new distance. Choose as different distance as you
can, and perform the same exercises as before, again recording all of your observations and
calculations.
Order, m
0
1
2
3
Length, L
Second Run Results
Distance from center, ym Angle, θm
Wavelength, λ
• What is the average value of your wavelengths for this run? Express your answer in
terms of nanometers.
• What is the average value of your wavelengths for the two runs? Express your answer
in terms of nanometers. This will be the result that you will use for determining the
width of the hair below.
• The wavelength of visible light is 400 − 700 nm, with violet at the low end and red
at the high. Noting the color of the laser beam, does your experimental results agree
with theory? If they don’t, why not?
• What’s the frequency of the laser light?
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4.2
Determining the Width of a Hair
Now that we know the wavelength of the laser, we want to determine the thickness of a
human hair. Some brave soul from the group will need to sacrifice one of their hairs. For
the best results the hairs should be straight. The experiment is performed exactly as before,
but now the hair will be used in place of the grating. Upon shining the laser across the hair
you will see a central bright spot on the paper. This is not really a diffraction effect, but
comes from the fact that the laser beam is thicker than the hair. However, around this you
will see a series of diffraction fringes.
4.2.1
First Run
You can either hold the hair across the beam, or tape it down to pointer. Again, to get the
best results, we’ll perform the experiment twice. Carry out the experiment as before, and
record your results below.
First Run Results
Order, m Length, L Distance from center, ym
0
1
2
3
Angle, θm
Thickness, d
• What is the average value of your thicknesses for this run? Express your answer in
terms of micrometers.
4.2.2
Second Run
Second Run Results
Order, m Length, L Distance from center, ym Angle, θm
0
1
2
3
Thickness, d
• What is the average value of your thicknesses for this run? Express your answer in
terms of micrometers.
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• What is the average value of your thicknesses for the two runs? Express your answer
in terms of micrometers. This is your experimental result.
• The average width of a human hair ranges between 17 to 181 micrometers. Does your
experimental result agree with these values? If they don’t, why not?
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Some Last Questions
1. One can also buy a green laser pointer these days. What if we used a green laser
instead of a red laser in our experiment - would the dots move closer or further? Why?
2. So far we have looked at the diffraction of monochromatic light. What would happen
if we instead use white light? Hint - look at the fluorescent light through the grating.
Don’t look directly at the light, but slightly to the side or you won’t see the effect. Why
does this happen?
3. When you look at the back of a CD or DVD you see a rainbow pattern. Where does
this pattern come from? When you twist and turn the CD the pattern shifts. Why?
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4. A Blu-Ray DVD stores more information on the disc than a conventional DVD, approximately six to ten times more. The name is because the Blu-Ray player uses a blue
laser with a wavelength about 405 nm (a conventional DVD player uses a red laser).
Why does the Blu-Ray player need the blue laser?
5. What happens if the wavelength λ is greater than the slit spacing, d, in the double-slit
experiment? For your answer, consider Eq. (1). What happens mathematically, and
what does that mean physically?
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