Physics 9 Fall 2009 DIFFRACTION

Transcription

1 Physics 9 Fall 2009 NAME: TA: SECTION NUMBER: LAB PARTNERS: DIFFRACTION 1 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. 2 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. 1

2 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 = 1T and 8 t = 5 T, the two waves completely overlap. 8 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 = 3T, and t = 7T, 8 8 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. 2

3 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. 3

4 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 r 1 from the slit to the point, while the light from the bottom slit travels a longer distance, r 2, because it s further away from the point. So, the bottom light ray travels an extra distance r = r 2 r 1 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 + 1 2) λ, then we get a dark spot. Recalling that r = d sin θ, we see that we get a bright spot when d sin θ m = mλ, (1) where we now have a range of angles θ m, 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 m th bright spot is y m = L tan θ m. (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. 4

5 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 m th and (m + 1) th bright fringes? Show further that if θ m and θ m+1 1, then θ λ, independent of m. Hint: for very d small angles, sin θ θ. 3. What is the distance y between the m th and (m + 1) th bright fringes? Using the ideas above show that, if tan θ θ for small angles, then y λ d L. 5

6 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)? 6

7 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 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 θ m and wavelengths λ for each order. Record these values in the table. First Run Results Order, m Length, L Distance from center, y m Angle, θ m Wavelength, λ What is the average value of your wavelengths for this run? Express your answer in terms of nanometers. 7

8 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. Second Run Results Order, m Length, L Distance from center, y m Angle, θ m 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 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? 8

9 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 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, y m Angle, θ m Thickness, d What is the average value of your thicknesses for this run? Express your answer in terms of micrometers Second Run Second Run Results Order, m Length, L Distance from center, y m Angle, θ m Thickness, d What is the average value of your thicknesses for this run? Express your answer in terms of micrometers. 9

10 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? 5 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? 10

11 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? 11