Experiment 107: Interference of Waves

Transcription

1 PHYSICS 10 Experiment 107: Interference of Waves Experiment 107: Interference of Waves Aims The aims of this experiment are to illustrate efficient methods of adding together waves of the same frequency and to apply these methods to understanding interference and diffraction of light. Adding sinusoidal waveforms Many phenomena in Physics involve the combination of two or more sinusoidal signals (i.e., functions of time) of the same frequency, but with different amplitudes and phases. For example, when receiving a television signal, there may be several paths from the transmitting station to the receiver. Besides the direct path, there can be additional paths due to reflections off features in the landscape. These signals interfere with each other, causing the phenomenon of ghosting on the TV screen. Although television uses signals which are more complicated than sinusoids, we first need to understand how to combine these simpler signals. Since coherent, monochromatic light consists essentially of sinusoidal variations of the electric field, the theory we develop is directly applicable to the phenomena of optical interference and the diffraction of light from objects. Recall that a sinusoidal waveform is one of the form f (t) = Acos(ωt + φ) where A is the (real) amplitude, φ is the phase, and ω is the angular frequency (which is π times the frequency). We have chosen to use a cosine rather than a sine, but this does not lead to any loss of generality since sin θ = cos(θ π/). Consider what happens when we add together two sinusoidal waveforms of the same frequency f 1 (t) = A 1 cos(ωt + φ 1 ), and f (t) = A cos(ωt + φ ). If you have not encountered it before, it is perhaps a little surprising that the sum of these signals turns out to be another sinusoidal waveform of the same frequency. (1) Use Matlab to plot graphs of the functions f 1 (t) = 4 cos(t 1), f (t) = 3 cos(t + 0.5) and of f 1 (t) + f (t) on the same set of axes for t in the range π to +π. Observe that the sum is also a sinusoidal waveform of the same frequency, so that it can be written as Acos(ωt + φ). From your graph, estimate the amplitude A and the phase φ. We seek a quick way of finding the amplitude and phase of the sum without adding the two functions point by point over an interval. In order to do this, we note that we can write Acos(ωt + φ) = Re [Aexp i (ωt + φ)] = Re [ Ae iφ e iωt], where i is the square root of 1, and we have made use of Euler s identity, namely that exp(iθ) = cosθ +isinθ. The complex number A = Ae iφ is called the complex amplitude of the sinusoidal waveform. The advantage of writing sinusoidal waveforms like this becomes apparent when we wish to add two of them. If we write A 1 cos(ωt + φ 1 ) = Re [ A 1 e iφ 1 e iωt ] A cos(ωt + φ 1 ) = Re [ A e iφ e iωt ] then A 1 cos(ωt + φ 1 ) + A cos(ωt + φ 1 ) = Re [( A 1 e iφ 1 + A e iφ ) e iωt ].

2 PHYSICS 10 Experiment 107: Interference of Waves 107- This shows that the sum is also a sinsoidal signal whose complex amplitude is A 1 e iφ 1 + A e iφ. We have reduced the problem of adding two waveforms to the problem of adding together two complex numbers. From a complex amplitude A, we recover the real amplitude A from the modulus A and the phase φ from the argument arg (A). In Matlab, the modulus is found using the function abs, and the argument (in radians) is found using the function angle. () The complex amplitudes of the two sinusoidal waveforms added together in step (1) are 4e i( 1) and 3e i(0.5). Use Matlab to add together these two complex amplitudes to find the complex amplitude A of the sum. Calculate A and arg(a), comparing these with your graphical results. Similarly, we can add more than two sinusoidal waveforms of the same frequency. The result is always a sinusoidal waveform of the same frequency. The complex amplitude of the sum is found by adding together the complex amplitudes of the individual waveforms. Young s double slit experiment In 1801, Thomas Young carried out an experiment demonstrating the wave nature of light. He allowed light to pass through two closely set pinholes onto a screen, and observed that the light beams spread out from each hole and overlapped. In the overlapping region, bands of bright light alternated with bands of darkness. This is called an interference pattern, and Young correctly identified its origin as being due to the addition of waves passing through each of the two holes. Today, the experiment is usually carried out using two narrow, closely spaced slits illuminated by plane waves of light. Each point on the screen is illuminated by light from both slits. Since the path lengths from each of the two slits to a point on the screen are generally different, the waves arriving at the point have slightly different amplitudes and phases. These waves add together, and the intensity of the light at the point is proportional to the square of the real amplitude of the sum. Suppose that the complex amplitude of the light at a slit is A. After it travels through a distance r, the signal is given by [ ] Aexp i (ωt kr) Re r where k = π/λ is the wavenumber of the light, and λ is the wavelength. The factor of 1 r indicates that the amplitude decreases with distance, in accordance with the inverse square law for the light intensity. The complex amplitude of the light at the point r away from the slit is thus given by Aexp( ikr)/r. y r 1 P y P y 1 y θ r x D Figure 1: Young s double slit experiment

3 PHYSICS 10 Experiment 107: Interference of Waves In the figure, we see that the complex amplitude of the light at the point P is A P = Aexp ( ikr 1) r 1 + Aexp ( ikr ) r and the intensity of the light at P is proportional to A P. If the coordinates of the slits are (0, y 1 ) and (0, y ) and the coordinates of the point P are (D, y P ), it is easy to see using Pythagorus theorem that r l = D + (y P y l ) for l = 1,. The following Matlab function thus computes the complex amplitude A P at the point P on the screen which is a distance D away from the plane of the slits at an angle θ from the x axis. The wavelength of the light used is λ, and the amplitude A at each slit is taken to be equal to one. Note that the variable yslits is a vector whose elements specify the positions of the slits in the plane x = 0. function Ap = interfere(yslits,d,theta,lambda) k = *pi/lambda; yp = D*tan(theta); nslits = length(yslits); Ap = 0; for l = 1:nslits rl = sqrt(d.^+(yp-yslits(l)).^); Ap = Ap + exp(-i*k*rl)./rl; end (3) Consider slits at y 1 = m and y = m (so that their separation is 1 mm), illuminated with green light of wavelength 500 nm. A screen is placed 10 m away. Using the function interfere, calculate and plot the intensity of light on the screen at a set of 501 angles over the range 5 mrad θ 5 mrad. You should find that the function interfere will work even when theta is a vector of angles, so you do not need to use a for statement to loop over the points on the screen. Note that the Matlab function linspace is useful for generating a vector of equally spaced points between two specified values. (4) From your graph, note the angles at which there are maxima and minima in the interference pattern. Explain why it is that there is a maximum at the centre of the screen. Is the spacing between successive maxima and between successive minima equal? Interference with more slits It is now straightforward to calculate the interference from more than two slits, since we simply need to add together the complex amplitudes at the screen of the waves which arise from each slit. The different path lengths lead to different phases (and slightly different amplitudes) for the sinusoidal signals which combine to give the total at the screen. The amplitude at the point P then becomes A P = S Aexp( ikr l ) r l, (1) where S is the number of slits. The intensity at P is (proportional to) A P, as before. (5) Use the function interfere to calculate the intensity of light on the screen when the double slit is successively replaced by a collection of S = 3, 5, 7 and 9 slits. The first and last slits should still be at ±5 10 4, and the remaining slits should be divided evenly between these. A convenient way of doing this in Matlab is to set yslits=linspace(-5e-4,5e-4,s).

4 PHYSICS 10 Experiment 107: Interference of Waves (6) Notice that the interference patterns consist of a number of large primary maxima between which are a number of smaller secondary maxima. From your results, what is the relationship between: (a) The angular separation between the primary maxima, and the separation between the adjacent slits (which is 1 mm / (S 1) for the case of S slits)? (b) The number of secondary maxima between a pair of primary maxima and the number of slits? Interference in the far-field In the examples considered so far, the distance D between the slits and the screen is so large that two approximations can be made which simplify the analysis. These are: Although the real amplitude of the waveform from slit l at the point on the screen is actually A /r l, when D is large, these can all be well-approximated by A /D. In the diagram below, consider the circle centred at P which passes through the origin. This circle has radius R = D/ cosθ. If the circle is large enough, it may be approximated by a straight line in the vicinity of the slits. By considering the geometry at the slits, we see that where y l is the y coordinate of the l th slit. r l R y l sinθ y y l sin θ r l P y l y l θ R x Approximate geometry for large R D Figure : Geometry for far-field interference We can use this to get a handy approximate formula for A P. Substituting r l R y l sin θ in the exponential and r l D (!) in the denominator and using exp( ikr) = 1 we get A P = S Aexp( ikr l ) r l S A D exp( ikr)exp( iky l sin θ) A P A D S exp(iky l sinθ). ()

5 PHYSICS 10 Experiment 107: Interference of Waves (7) Using Eq. () derive an approximate formula for the expected value of A P at primary maxima (for example, at θ = 0). How does this change with S? Does this agree with your results for Question 5? (8) Suppose that we have two slits, one at y 1 = d/ and the other at y = +d/. Show that A P 4 ( ) A kdsin θ D cos Plot a graph of this function for A = 1, D = 10 m, d = 1 mm, λ = 500 nm at a set of 501 angles over the range 5 mrad θ 5 mrad. Compare this graph with that obtained in step (3). Analytically find the values of θ at which the minima and maxima of the intensity arise, and compare these with your previous results. What further approximation is needed in order to conclude that the successive maxima (or minima) are equally spaced in θ? Now let us consider a situation where there are an odd number of slits symmetrically placed around the origin. Suppose that we have S = M + 1 slits located at y coordinates Md, ( M + 1)d,..., (M 1)d, and Md, so that the separation between the slits is d. The intensity at a point P on the screen is now A P A D exp [ik ( M)dsin θ] + exp[ik ( M + 1)dsinθ] + + exp[ikmdsin θ] (3) (9) Show that the sum of exponentials in equation (3) is a geometric series, i.e., that it may be written in the form a + ar + ar + + ar N 1 for a suitable choice of a, r and N. Find expressions for a, r and N appropriate to this series. (10) Using the formula for the finite sum of a geometric series, namely ( ) 1 r a + ar + ar + + ar N 1 N = a 1 r show that A P A D 1 exp [ik (M + 1)dsin θ] 1 exp [ikdsin θ] (11) [Optional] By making use of Euler s identity, show that this may be further simplified to A P A sin [( ) ] M + 1 kdsin θ D sin [ (5) 1kdsin θ] (1) Using either (4) or (5), plot a graph of the intensity on the screen for a system of S = 7 slits corresponding to the parameters used in step (5). Note that in this case, d = 1 mm / (S 1) = 1 6 mm. Compare your graph with the one obtained in step (5) for this situation. (4) Diffraction from a single wide slit So far, each slit is assumed to have been very narrow, so that it may be regarded as a source of waves centred at that slit. We can treat a wide slit by supposing it to be made up of many narrow slits spaced so closely together so that they merge into a single slit. In order to model a slit of width w, we can consider M + 1 slits separated by d = w/ (M), letting M, so that the distance between the first and last slit always remains at w. We also need to decrease the amplitude A from each slit as M becomes large, so that the intensity on the screen remains finite. (13) Starting from equation (5), consider what happens as M, d 0 and A 0 in such a way that w = Md and A/d = 1 remain constant. Using the fact that sinx x when x is small (and is measured in radians), conclude that A P sin [ 1 kw sin θ] D [ 1 k sin θ] (6)

6 PHYSICS 10 Experiment 107: Interference of Waves (14) Plot a graph of this function for D = 10 m, w = 1 mm, λ = 500 nm at a set of 501 angles over the range 5 mrad θ 5 mrad. The function interfere can also be used to calculate an approximation to the diffraction pattern from a wide slit by using a large number of closely-spaced narrow slits. For example, if we set yslits=linspace(-5e-4,5e-4,01) this is a good approximation to a single slit of width w = 1 mm. An advantage of using the function interfere is that we can obtain the diffraction pattern when the distance between the slits and the screen is no longer large. In this situation, it becomes much more difficult to obtain analytical expressions for the diffraction pattern, although the principle of its formation (i.e., the addition of sinusoidal waveforms of varying phase) is still the same. (15) Use the function interfere with the 01 closely-spaced slits approximating a single wide slit of width w = 1 mm. Using λ = 500 nm at a set of 501 angles over the range 5 mrad θ 5 mrad, plot graphs of the intensity at the screen for D = 10 m, 3 m, 1 m and 0.3 m. Also obtain a graph for D = 0.1 m, but you will need to change the range of θ in this case to 0 mrad θ 0 mrad. Instead of plotting the pattern as a function of θ for this last case, it is interesting to plot it as a function of y P = D tan θ. Notice how the graph for D = 10 m resembles that which you found in step (14). As D is decreased, the pattern changes gradually. What can you conclude about the pattern for small D?