Young’s Experiment
YOUNG’S EXPERIMENT
OVERVIEW AND OBJECTIVES
Is a light beam composed of little particles, as Newton proposed? Or is light a wave,
as Huygens suggested? Or is light perhaps something else entirely, neither a particle nor a
wave? And how can we tell the difference?
If we can identify some basic property of waves that light does not possess, then that
fact can be considered one piece of evidence against a wave model of light. Conversely, if
light can be shown to behave in some way that is typical of waves but not typical of particles,
then that fact would be a piece of evidence in favor of a wave theory of light and against a
particle theory.
One fundamental property of waves --- indeed, the fundamental property of waves --is the principle of superposition: two waves impinging simultaneously on the same point can
either reinforce each other ("constructive interference") or cancel each other ("destructive
interference"), depending on whether they arrive "in phase" or "out of phase". By contrast,
particles do not habitually behave in this way: two baseballs do not combine to make zero
baseballs.
Does light exhibit constructive and destructive interference? Can two beams of light
be combined to get darkness? If this were possible, it would be a powerful piece of evidence
in favor of a wave theory of light.
This raises another question: if light is indeed a wave, how do we measure its
wavelength? After all, light moves at a tremendous speed: 186,000 miles per second, or
3 108 m / s (first estimated by Ole Rømer in 1676 using astronomical methods). How do we
measure the length of something moving that fast?
If we assume light is a wave that travels at constant speed, we can devise an
experiment in which we can verify this postulate. In this exercise you will show that light
does indeed exhibit constructive and destructive interference, as was first demonstrated by
Thomas Young in 1802. You will also calculate several wavelength of light, since the
process of interfering waves is wavelength dependant. In this experiment you will pass light
through a diffraction grating --- a very narrowly spaced sequence of slits --- and project the
result on a screen. You will see bright and dark spots, indicating constructive and destructive
interference, respectively. By carefully measuring the position of these bright and dark spots,
you will obtain enough information to calculate the wavelength of the light.
In performing this experiment you will use two different light sources: a red and a
green laser, which provide very pure light of a single wavelength; and an ordinary
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Young’s Experiment
incandescent bulb, emitting white light, which is a mixture of different colors (i.e. different
wavelengths).
Do not forget to bring a calculator to this lab.
MATERIALS

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White screen for observing diffraction pattern.
Diffraction grating: a piece of plastic with thousands of equally spaced grooves
etched into it.
Laser emitting red light.
Laser emitting green light at T.A.’s bench.
White light source (incandescent bulb in a black housing).
Graph paper under the apparatus: 1 box = 1 cm.
WARNING!!!NEVER LOOK DOWN THE LASER BEAM AT THE LIGHT SOURCE.
DON'T EVEN THINK ABOUT IT!
A laser is an extremely intense source of light; looking directly at it, even momentarily, can
permanently damage your retina. It is OK to look at laser light reflected off the screen.
THE PRINCIPLE BEHIND YOUNG’S EXPERIMENT
Any wave is characterized by its wavelength λ and its amplitude A (see Figure 1).
At different places in space, the wave takes different values: thus, at a crest the wave has
value +A; one-quarter of a cycle (i.e. a distance λ/4) farther ahead, the wave has value 0;
another quarter of a cycle farther ahead, the wave is in a trough and hence has value -A; still
another quarter of a cycle farther ahead, the wave again has value 0; and finally, another
quarter of a cycle farther ahead, we have arrived at another crest and the wave again has the
value +A.1 The point in the cycle where the wave happens to be (at a particular point in
space) --- crest, trough, or somewhere in-between --- is called the phase of the wave.
1
You might wonder, what are these "values"? Are they distances, or speeds, or what? The answer
depends on the nature of the wave. For waves on a string, the "value" of the wave is the displacement of the
string from its resting position; it is therefore a distance (measured in meters). For sound waves, the"value" is
the air pressure at a particular point in space. For light waves, the "value" turns out to be an electric (or
magnetic) field. None of our reasoning in this lab will depend on the precise nature of the "value" of the wave.
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Young’s Experiment
Figure 1: A wave is characterized by its wavelength λ, and its amplitude A.
Now suppose we combine two waves (traveling at a constant speed) of the same
wavelength (λ) and the same amplitude (A) but possibly different phases. If the two waves
happen to be exactly "in phase", we get crests of height 2A and troughs of height -2A, so that
the amplitude of the combined wave is 2A. This is called constructive interference. At the
other extreme, if the two waves happen to be exactly "out of phase", they will exactly cancel
each other and we get a wave of amplitude 0 (darkness). This is called destructive
interference. Of course, intermediate situations are possible as well.
Figure 2: (a) Constructive interference
(b) Destructive interference
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Young’s Experiment
Now suppose that we shine light of wavelength λ onto an opaque screen in which we have
cut two narrow slits (A and B) separated by a distance d, and that we allow the resulting light
to be projected onto a white screen sitting a distance L behind the first screen ( see Figure 3).
This was done by Thomas Young in 1800 and he observed an interference pattern of
alternating bright and dark regions, as shown in Figure 3
light interference pattern
incident light
two slits
two
wavefronts
screen
Figure 3: Thomas Young’s experiment to demonstrate the interference of light. Young observed a
pattern of bright and dark bands on the screen.
The explanation for Young’s observation is given in Figure 4.
Figure 4: the origin of the interference patterns in young’s experiment. (a) Constructive interference
occurs when two waves in phase combine. ( b) Although the path length is now longer, constructive
interference still occurs if the two waves are in phase when they reach the screen. (c) Destructive
interference occurs because the wave from the top slit is a half a wavelength out of phase with the wave
from the lower slit.
If the two wave fronts arrive at a particular point in such a way that their wave motion
is in phase, they will constructively interfere and a bright region will be produced. If, at
another point, the two arrive out of phase, they will destructively interfere and a dark
region will exist. As we saw above a shift of 1/2 –wavelength of light is sufficient to change
from constructive to destructive interference. A pattern of alternating light and dark patches
therefore emerges on the screen, corresponding to regions of constructive and destructive
interference.
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Young’s Experiment
Figure 5. The arrangement of the experiment with quantities you will measure to calculate wavelength
In this lab, you will make measurements on the interference pattern, which will be
used to determine the wavelength of light. To do this, we need to find the relationship
between various quantities measured in the experiment and the wavelength that we deduce
from these measurements. Consider, for starters, the point C at the center of the rear screen,
i.e. directly behind the point halfway between the two slits, as shown in Figure 5 (a). The
waves emanating from slits A and B start out in phase; the distances they travel (AC and BC)
are equal; therefore, they will arrive at point C in phase. Therefore, point C is a point on
constructive interference, and we will see a bright spot there.
Next consider a point D on the rear screen lying a distance y off-center, as shown in
Figure 5(b). The distance AD and BD are now unequal. So the waves will not necessarily
arrive in phase. Indeed, if the distance BD - AD is half a wavelength (/2), the two waves
will arrive exactly out of phase, and there will be destructive interference: we will see a dark
spot there. If the difference BD – AD is one full wavelength (λ), the two waves will arrive
exactly in phase, and there will again be constructive interference: we will see a bright spot.
If the difference BD – AD is one and a-half wavelengths (3/2) we will have destructive
interference; if it is two wavelengths (2), we will have constructive interference. And so on.
Let us use a bit of geometry to calculate the position y = y1 of the first off-center
bright spot, i.e. the one corresponding to BD – AD =. To help with the calculation, an
additional line AE has been drawn in Figure 5(b), from Point A to Line BD, so that it is
perpendicular to BD. To simplify the calculation, we make the approximation that d is much
smaller than y and L. (This is definitely the case for your experiment, where L and y are
somewhere around 10 cm, while d is utterly tiny, about 1.67 x 10-4 cm.) With this
approximation, it turns out that the lengths AD and ED are the same (or more precisely, the
difference in lengths AD – ED is much less than the wavelength ), so constructive
interference will result at point D if the length BE = . (Constructive interference will also
occur if BE = 2, 3, etc.)
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Young’s Experiment
Now it also turns out that the triangles ABE and ODC are similar. (You can show this
is true by first showing that the two angles marked by  are the same. See if you can do this.)
For similar triangles, we know that the ratios of corresponding sides are the same. In
particular
BE DC

BA DO
(1)
Since the distances
BE  
(for the first bright spot)
BA  d
DC  y1
DO  L2  y12
(by the pythagorean theorem)
we find from Eq.(1) that the first bright spot occurs when

d

y1
(2)
L  y12
2
If we solve for , we get

d  y1
(3)
L2  y12
What about the second off-center bright spot? It corresponds to BD – AD = 2 (or
BE = 2), so we have
d  y2

,
(4)
2 L2  y22
where y2 refers the distance from the center of the second bright spot, More, generally, the
nth bright spot satisfies BD – AD = n, so that

d  yn
n L2  yn2
,
where yn refers to the distance from the center of the nth bright spot.
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(5)
Young’s Experiment
The plan is now clear: We perform the two-slit interference experiment and carefully
measure L (the separation between the two screens), y1st bright (the position of the first offcenter bright spot) and d (the spacing between the slits). We then plug into equation (3) to
deduce the wavelength λ. If we measure also the position of the second off-center bright spot
and plug into equation (4), we then get an independent estimate of the wavelength λ. And so
on if we are able to measure the third, fourth bright spots. Hopefully these independent
estimates of λ will agree to reasonable accuracy.
A diffraction grating consists, not of two slits, but of a vast array of equally spaced
slits; this turns out to be more efficient. The theory is a bit more complicated, but the
principle of why there is constructive and destructive interference is exactly the same.
Indeed, the quantitative answer turns out to be exactly the same: the location of the first
bright spot is still given by equation (3), that of the second bright spot by equation (4), and
that of the nth bright spot by equation (5).
PART 1: MEASURING THE WAVELENGTH OF LIGHT
The diffraction grating you are using has 6000 slits per centimeter etched into it.
Therefore, the spacing between adjacent slits is
d
1
cm  1.667 104 cm .
6000
(6)
(Don't forget that 1 centimeter equals 1/100 of a meter!)
The results of the following experiment should be recorded on Data Table 1
PROCEDURE
1). Pass the laser beam through the diffraction grating. Do you see a sharp central bright
spot, surrounded by a series of off-central bright spots? Use the metric ruler mounted
on the screen to carefully measure y1st bright. Then calculate the wavelength λ of the
laser light.
2). Measure carefully the distance L between the diffraction grating and the screen.
3). Carefully measure y2nd bright and again calculate the wavelength λ of the laser light.
How do your estimates of λ compare to each other?
4). The actual wavelength of the laser light is λ= 6.328 ×10 -7 meters. Compare your
answer to this number.
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Young’s Experiment
PART 2: PREDICT POSITION OF AN INTERFERENCE PATTERN
In this exercise you will calculate the position for a second order diffraction pattern (y2nd bright)
produced from green laser light. For this calculation, equation 5 has to bee solved for yn.
Doing some algebra shows that equation 5 can be rewritten as:
yn 
nL
d 2  (n ) 2
(7)
The results of the following experiment should be recorded on Data Table 2
PROCEDURE
1). The green laser emits at a 532nm wavelength. Choose your L to be 5cm and calculate
y2nd bright using the equation above.
2). Show the result to your T.A. Once your T.A. approves, you will be given a green laser
to verify your result. Make sure the distance between the grating and the screen is 5
cm. Shine the laser through the grating and onto the screen.
3.) Record the position of the n = 2 green spot. Compare it with your calculation.
PART 3: WHITE LIGHT
White light (e.g. light from an ordinary incandescent light bulb, or from the Sun) is a
combination of many colors, i.e. light of many different wavelengths. One way to show this
is to pass white light through a prism or a diffraction grating, and notice that it becomes
separated into its component colors. The prism and the diffraction grating work differently
though each separates light into its component colors, and both phenomena are explained by
the wave theory of light.
The results of the following experiment should be recorded on Data Table 3
PROCEDURE
1). Pass the beam of white light through the diffraction grating. Do different colors
appear at different places on the screen? Can you explain why this occurs, using the
wave theory of light?
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Young’s Experiment
2). Carefully measure the value of y1st bright corresponding to each of the colors listed in
Data Table 2. Measure at the center of each band. Then calculate the wavelengths
corresponding to each color.
3). Is it true that different colors correspond to different wavelengths? Order the color
spectrum from longest wavelength to shortest.
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Young’s Experiment
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Young’s Experiment
NAME:_______________________
SECTION:________
DATA TABLE 1
Red Laser
SHOW ALL CALCULATIONS TO GET FULL CREDIT
Distance L:
y1st bright:
(meters)
(meters)
λ (calculated from y1st bright):
y2nd bright:
(meters)
(meters)
λ (calculated from y2nd bright):
(meters)
λavarage (average of λ1st bright and λ2nd bright):
(meters)
λ (|calculated – measured avarage|)
(meters)
CALCULATIONS:
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Young’s Experiment
NAME:_______________________
SECTION:________
DATA TABLE 2
Green Laser
SHOW ALL CALCULATIONS TO GET FULL CREDIT
Distance L:
(meters)
n:
:
(meters)
d:
y2nd bright(calculated from equation 7):
y2nd bright(measured):
(meters)
(meters)
y (y2nd bright(calculated from equation 7) - y2nd bright(measured) =
CALCULATIONS:
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Young’s Experiment
NAME:_______________________
SECTION:________
DATA TABLE 3
White Light
SHOW ALL CALCULATIONS TO GET FULL CREDIT
Red
Range of wavelengths λ
(meters)
-9
620 x10 m – 700 x10-9m
Orange
590 x10-9m – 620 x10-9m
Yellow
570 x10-9m – 590 x10-9m
Green
500 x10-9m – 570 x10-9m
Blue
450 x10-9m – 500 x10-9m
Violet
400 x10-9m – 450 x10-9m
Color
y1st bright
(meters)
Wavelength
(meters) using Eq. 5
CALCULATIONS:
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Young’s Experiment
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