LIGHT: Two-slit Interference

Transcription

1 LIGHT: Two-slit Interference Objective: To study interference of light waves and verify the wave nature of light. Apparatus: Two red lasers (wavelength, λ = 633 nm); two orange lasers (λ = 612 nm); two yellow lasers (λ = 594 nm); one green laser (λ = 543 nm); slide with double slit; screen, compass, ruler, protractor PRECAUTION: The laser beam is not intense enough to hurt your hand, but it can damage your retina if it enters your eye. Work with the laser on the lab table, and your body in a normal standing position. Introduction and Theory: A wave is not an object, but an influence that exists at points in space and varies with time. At a given space point the influence varies continuously over time, from positive to negative to positive to negative, etc. The quantity that varies may be the level of water (in the case of a water wave), the direction of electric and magnetic fields (in the case of a light wave), or something else. Two waves may exist at the same point in space (unlike the case for physical objects), and then the net influence is the algebraic sum of the two influences. Hence, there may be points where the two influences are always opposite, one positive when the other is negative, and vice versa. [Fig.1b] Fig. 1a Fig. 1b Then the influences cancel, and there is no effect that spot, in the case of light waves, is dark, even though both waves are there. This is called destructive interference. There may be other points where the two influences are always in the same direction, and then that spot is bright, brighter than it would be with just one wave. This is constructive interference. [Fig. 1a] The key concept in interference is path length, the distance from the source of light to the point of observation. In Fig. 2 light comes from points S1 and S2 and is observed at point P.

2 Fig. 2 The path length from S1 to P is L1 and the path length from S2 to P is L2. The path difference is = L2 L1. Interference depends on the relation between and the wavelength of the light, λ. If n stands for a positive or negative integer (including 0), the condition is: CONTRUCTIVE INTERFERENCE: = nλ DESTRUCTIVE INTERFERENCE: = (n + ½)λ (1) Part I is an exercise using compass and ruler to find the locus of points corresponding to constructive interference with n = 0 and 1, and destructive interference with n = 0. Here the wavelengths and distances will be of the order of centimeters, not like the real magnitudes in an experiment with light. The object will be to discover how path difference creates geometrical patterns of constructive and destructive interference in space. In Part II we will use a laser and a slide with two slits separated by a small distance, d (Fig. 3). An alternating pattern of light and dark appears on a screen a distance L from

3 Fig. 3 the slide. The center of the pattern will be a bright spot corresponding to constructive interference with n = 0. This is at a point where the perpendicular bisector of the line between the slits intersects the screen. [We have, in Fig. 3, used the perpendicular line from the bottom slit, rather than from the point half-way between the slits. This is to make the mathematics a little simpler in the Appendix. The difference is negligible, since (1/2)d is about 1/10 of a millimeter, while the whole interference pattern (represented by Y in the figure) will be 10 or 20 centimeters in size.] At point O, and all along the line from the slits to point O, the two paths are equal, and = 0. For other points P above or below O, it can be shown that = d sinθ, (2) where θ is the angle shown. A derivation of this important equation is given in the textbook. An alternate derivation is given in the Appendix to this experiment. If P corresponds to the nth bright spot above the central maximum, we have d sinθ = nλ. Since Y will be a few centimeters, and L will be about 2 meters, the angle is small. Hence sin θ differs negligibly from tan θ; one sees from the diagram that tan θ equals Y/L. Hence d(y/l) = nλ, or Y/n = (L/d)λ (3) In Part II we will measure Y for n around 5 or 6, and for each of the four available wavelengths.

4 The lasers will be set at different stations around the lab room, and each student group will take a turn observing and measuring the interference pattern for each wavelength. After taking data, students should proceed individually to analyze the laser data, and to carry out the paper/pencil construction of Part I Part I: The instructor will give a value for d and a value for λ to each group. Each student will do his or her own construction. But when completed, students in each group should compare their results to see if they have any major difference. If there is, you should try to figure out what the problem is, and redo the work if there is some mistake. Near the bottom center of a piece of paper draw two points separated by distance d. [Fig. 4] Draw the line between them, and the perpendicular bisector of that line (using either the protractor or the compass, if you remember how to do that). Fig. 4 That bisector is the locus of all points with = 0; there will be constructive interference at all points on the line. Choose L1 = 6 or 7 cm, and with the point of the compass on S1 draw an arc with radius L1. Now let L2 = L1 + (1/2)λ. With the compass point on S2, draw an arc of radius L2. Find the point (P) where the two arcs intersect. At this point the path difference is (1/2)λ, and there is destructive interference. Now choose a slightly larger value of L1, and repeat the procedure the new value of L2 will equal L1 + (1/2)λ. This gives you a second point (Q) where = (1/2)λ, further out from the sources. You should get at least 4 points with = (1/2)λ, making sure that

5 they spread out over a reasonable fraction of the paper. Draw a smooth curve through the points. Don t expect it to be a straight line. In fact the locus of all points such that the path difference is a constant is a hyperbola. (You may know that the locus of all points such that the sum of the two distances is constant is an ellipse.) What you ve drawn is part of the hyperbola on one side of the bisector. There is a symmetrical curve on the other side of the bisector that corresponds to = (1/2)λ. Both curves correspond to destructive interference. There is also another half of the hyperbola below the points S1 and S2, but that s not relevant, since the light is going up, not down. The curves of the hyperbola become close to straight lines when you look at points far away from the sources (path lengths much larger than d). This is the situation in Part II, when we do an experiment with a small d. Next find the locus of constructive interference corresponding to n = 1. That is, choose a value of L1 to start with, let L2 = L1 + λ, and use the compass to find a point (P ) where = λ. Find a set of at least 4 such points and connect them with a smooth curve. Now if you can imagine a screen across the top of the paper, you can see where the central maximum will be, where the first dark region on the right of this will be, and where the next bright spot will be. The pattern will be the same on the left side of the central maximum. Part II: Mount the slide and the laser so that the beam hits the double slit. Place the screen, with a piece of paper taped to it, at a distance, L = about 2 meters from the slide. Measure L. The distance d on the slide will be given to you. You should see an interference pattern something like Fig. 5. Fig. 5 Locate the central maximum and observe the bright spots. As the interference pattern gets out to larger values of n the spots become a little fuzzy, so choose the largest possible value of n that gives you clear reasonably symmetrical bright spots on both sides of the central spot. Without placing you eye near the pattern, circle the bright spots from your largest n to the same negative n. Take the paper off the screen, and measure D, the distance from the spot on the left to the spot on the right, making an effort to locate carefully the center of each spot. Record n, and your value of Y, which =

6 D/2. [Note: It may be that you re not sure which spot is the central maximum. This will not matter much. Just choose one of the bright spots near the center of the pattern, count n spots to one side and n spots to the other side, and proceed as above.] Repeat this procedure for each of the 4 lasers. You don t have to use the same value of n for all 4 cases. Make a table of λ, n, Y, and Y/n. Note that Y/n is the distance from one bright spot to the adjacent one. By using Y/n we are able to choose different n s for different wavelengths, if that allows us to get the best data. Plot a graph of Y/n vs. λ. (Here you can choose a scale that covers just the range of wavelengths we are using, or you can choose a scale that goes down to λ = 0. (Consider which you think would be better.) According to Eq. (3) this graph is expected to be a straight line, with slope = L/d. Draw the best straight line close to your points. (Count the origin as one of your points if the origin is included in your graph.) Find the slope of the line, and compare it to L/d. Appendix: To derive Eq. (2) we ll use Fig. 6 below. AB is the line between the slits. BO is perpendicular to AB, and O is a point on the screen. Note that, in order to show the geometry, this figure is not drawn to scale. The slit spacing d in the light experiment is about 0.25 mm, while Y, representing the size of the interference pattern on the screen, is of the order of 10 cm. So d is really about 1/400 of Y. (Fig. 3 is closer to the actual scale, but still not very close.) Hence the opening angle φ at point P is very much smaller than the angle θ at point B: φ << θ and also φ << π. (4) In Fig. 6 point C is chosen so that AP = CP. Therefore the line segment BC is the path difference,. Fig. 6

7 Draw point Q to be the midpoint of AC. Then draw PQ which is the perpendicular-bisector of AC. AC is the base of the isosceles triangle ACP. In one of the questions below you are asked to prove the following two relations: γ = π/2 + φ/2, α = θ φ/2. (5) Now use the Law of Sines in triangle ABC. (If you don t remember it, look it up.) Using Eq. (5) this becomes or Since φ is small (Eq. (4)), this simplifies to QED Questions: = d sin θ. 1. Using your value of L and the largest value of Y in your experiment, calculate θ in degrees and radians. Calculate tan θ and sin θ, and find the percent error made when you replace the sine by the tangent (leading to Eq. (3)). [How many places after the decimal point do you have to keep to get an answer?] 2. Verify Eqs. (5) in the Appendix. [Use triangles PQC and ABC.] 3. It is important in this experiment that the spacing d be small. One reason is that the laser beam is narrow and we have to have the beam fall on both slits. But Thomas Young did the experiment in 1800 with an ordinary light source, not a laser. Give another reason why d has to be small. [What would your interference pattern look like if we were able to use d = 1 cm?]