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 (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 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
· 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.
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
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
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.
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 (l/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 (3l/2) we will have destructive interference; if it is two wavelengths (2l), 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 =l. 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 l), so constructive interference will result at point D if the length BE = l. (Constructive interference will also occur if BE = 2l, 3l, etc.)
Now it also turns out that the triangles DABE and DODC are similar. (You can show this is true by first showing that the two angles marked by q 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
(1)
Since the distances
we find from Eq.(1) that the first bright spot occurs when
(2)
If we solve for l, we get
(3)
What about the second off-center bright spot? It corresponds to BD – AD = 2l (or BE = 2l), so we have
, (4)
where y2 refers the distance from the center of the second bright spot, More, generally, the nth bright spot satisfies BD – AD = nl, so that
, (5)
where yn refers to the distance from the center of the nth bright spot.
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 off-center 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
. (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.
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:
(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?
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.