10.5 General Two Source Wave Interference

Transcription

1 38 0 Stage Physics 0.5 General Two Source Wave Interference 0.5. Conditions for Constructive and Destructive Interference Consider two sources of waves each emitting identical waves into the same medium. The two sources are monochromatic (i.e. emitting waves of one and the same frequency, and hence wavelength), and they are in phase with each other (i.e. when source emits a crest then so does source, when source emits a trough then so does source ). For simplicity s sake, we will assume that the waves emitted by the two sources are of the same amplitude. The two sources are depicted in Figure 0.0 as are the waves that they are emitting. Note that the waves are in phase with each other as they leave the two sources S and S. C 4 C 3 C C S T 3 T T C 4 C 3 C C S T 3 T Fig. 0.0 T Note that the waves are not being emitted in straight lines (i.e. in one dimension) as depicted here but in three dimensions. A two dimensional representation of these waves would be a set of circular wavefronts i.e. a circular set of crests and troughs as depicted in Figure 0.. This is most easily visualised as water waves on the surface of a pond. In three dimensions it would a set of spherical wavefronts emanating from each source. Now consider what will be the combined effect of the two waves at some point, distant from the two sources, as depicted in Figure 0.. Q S M P S Fig. 0. Point P is equidistant from the two sources S and S (i.e. on the perpendicular bisector of the line segment joining S and S ). Waves leave sources S and S in phase with each other and arrive at point P in phase with each other, because they have travelled equal distances. Thus at point P the two waves will reinforce each other and we will have constructive interference. Consider what will happen at some point Q, which is not equidistant from the two sources. The two waves travel different distances in going to Q. Thus there is a path difference between the distances travelled by the two waves on arrival at point Q. 38

2 Interference THE of INTERFERENCE Light OF LIGHT 39 In Figure 0. the path difference to Q = QS QS. If the path difference is equal to one wavelength then, when crest C (see Figure 0.0) from source S arrives at Q, crest C from source S, which was emitted at the same time, is still one wavelength away from Q. When crest C from source S arrives at Q it will arrive simultaneously with crest C from source S. When trough T from source S arrives at Q it will arrive at the same time as trough T from source S. Thus the two waves arrive at point Q in phase with each other and therefore they will reinforce each other. i.e. there will be constructive interference at point Q. If the path difference is equal to two wavelengths then, when crest C from source S arrives at Q, crest C from source S, which was emitted at the same time, is still two wavelengths away from Q. When crest C from source S arrives at Q it will arrive simultaneously with crest C 3 from source S. When trough T from source S arrives at Q it will arrive at the same time as trough T 3 from source S. Thus the two waves arrive at point Q in phase with each other and therefore they will reinforce each other. i.e. there will be constructive interference at point Q. Similarly if the path difference from the two sources to point Q is a whole number of wavelengths then waves from the two sources will arrive at point Q in phase with each other; the two waves will reinforce each other and the result is constructive interference at point Q. If the path difference is equal to one half of a wavelength then, when crest C from source S arrives at Q, crest C from source S, which was emitted at the same time, is still one half a wavelength away from Q. When crest C from source S arrives at Q it will arrive simultaneously with trough T from source S. When trough T from source S arrives at Q it will arrive at the same time as crest C from source S. Thus the two waves arrive at point Q half a wavelength (or 80 ) out of phase with each other and, therefore, they will annul each other. i.e. there will be destructive interference at point Q. If the path difference is equal to one and a half wavelengths then, when crest C from source S arrives at Q, crest C from source S, which was emitted at the same time, is still one and a half wavelengths away from Q. When crest C from source S arrives at Q it will arrive simultaneously with trough T from source S. When trough T from source S arrives at Q it will arrive at the same time as crest C 3 from source S. Thus the two waves arrive at point Q half a wavelength (or 80 ) out of phase with each other and, therefore, they will annul each other. i.e. there will be destructive interference at point Q. Similarly if the path difference from the two sources to point Q is a whole number of wavelengths plus a half a wavelength, then waves from the two sources will arrive at point Q half a wavelength (or 80 ) out of phase with each other; the two waves will annul each other and the result is destructive interference at point Q. Thus if the path difference to point Q is a whole number of wavelengths there will be constructive interference at Q; if the path difference to point Q is a whole number of wavelengths plus a half a wavelength there will be destructive interference at Q; if the path difference to point Q is other than one of the above there will be partial interference at point Q. The above discussion may be summarised as follows; For two monochromatic sources in phase with each other, the waves at a point some distance away in a uniform medium: Interfere constructively when the path difference between the two waves at the point is m where m = 0,,, 3, 4,. and is the wavelength of the wave. Interfere destructively when the path difference is m where m is an integer and is the wavelength of the wave. 39

3 40 0 Stage Physics 0.5. The Two Source Wave Interference Pattern in Two Dimensions We can perform a geometrical construction to determine the effect in a uniform medium of two sources emitting waves of the same frequency (and hence of the same wavelength). i.e. the two sources are emitting identical monochromatic waves and in our construction they are in phase with each other. This situation is depicted in Figure 0.. In this diagram the solid circles represent a line of crests emanating from each source and the dotted circles represent a line of troughs emitted from each source. A 0 P P A A P P A A A 3 A 3 A 4 A 4 A 5 A 5 S S B 5 B 5 Q 5 Q 5 B 4 B 4 Q 4 Q 4 B 3 B 3 Q 3 Q 3 B B Q Q B B Q Q B 0 Fig. 0. Note that along the lines dark solid lines labelled AB crests and troughs arrive simultaneously at any given point. The two waves arrive in phase at all points on these solid lines and will continue to arrive in phase at all times. Thus, at all points on the solid lines the waves reinforce each other and we have constructive interference. These are lines of maximum amplitude of the wave, i.e. of maximum wave energy. They are also sometimes called anti-nodal lines, using the terminology of nodes and antinodes, as used in one-dimensional interference as studied at Stage, when you discussed the production of standing waves in strings and air columns. Along the broken lines PQ, crests from one source arrive simultaneously with troughs from the other. At all points along these broken lines the two waves arrive half a wavelength (or 80 ) out of phase with each other. Thus, at all points on these broken lines the waves annul each other and we have destructive interference. These are lines of zero amplitude of the wave, i.e. of zero wave energy. They are also sometimes called nodal lines. 40

4 Interference THE of INTERFERENCE Light OF LIGHT 4 In between these lines of maximum wave energy (amplitude) and zero wave energy (amplitude) there is partial reinforcement. Thus a stationary pattern is set up where we seemingly have the wave energy propagated in lines (AB) interspersed by lines of zero energy (PQ). Note that along all the lines of maximum wave energy the path difference from the two sources to any point on the lines of maximum wave energy is a integral number of wavelengths. Along A 0 B 0 the path difference is zero; along A B the path difference is one wavelength; along A B the path difference is two wavelengths; along A 3 B 3 the path difference is three wavelengths etc. Along the lines of zero wave energy the path differences from the two sources to any point along these lines is an integral number of wavelengths plus a half. Along P Q the path difference is one half a wavelength; along P Q the path difference is one and a half wavelengths; along Q 3 the path difference is two and a half wavelengths etc. Figure 0. represents what we would see if the waves were waves in ripple tank or on the surface of a pond of water. We would see this (stationary) pattern of nodal and antinodal lines. Now, consider what we would detect if the waves were sound or light waves. For this refer to Figure 0.3. X Y S S Fig. 0.3 Let the two sources be loudspeakers connected to a single audio oscillator by wires of equal length then they are emitting identical waves of the same frequency and amplitude and the two sources are in phase with each other. If we move along the line XY with a sound wave detector (our ear), where the line XY intersects a line of maximum wave energy we will hear a loud sound and where it intersects a line of zero wave energy we will hear no sound (silence). Thus, as we move along the line XY we will hear a waxing and a waning of the sound intensity. i.e. the sound varies from loud to soft to loud to soft etc. Let the two sources be coherent light sources in phase with each other and emitting identical light waves of the same frequency and amplitude. If we move along the line XY with a light wave detector (our eyes), where the line XY intersects a line of maximum wave energy we will see bright light and where it intersects a line of zero wave energy we will see no light (darkness). Thus, as we move along the line XY we will see the light intensity progressively changing from bright to dark to bright to dark etc. 4

5 4 0 Stage Physics 0.6 The Two Slit Interference of Light 0.6. The Apparatus Used The two source interference pattern for light was first observed in 80 by English physicist Thomas Young. There are difficulties involved in observing this pattern for it is not easy to set up two coherent sources of light. Remember that light from an incandescent source is not coherent (Section 0.3.3). Young set up two coherent sources of light by an arrangement of a single slit followed by a double slit just after it, and this is the common way that we do this in school laboratories to this day. A schematic of this arrangement is depicted in Figure 0.4. Light from a monochromatic source falls on the single slit S 0 from the left hand side. This slit is very fine (thin) and as the light passes through it diffracts (spreads out) as shown in Figure 0.4. These diffracted wave fronts impinge on the double slits S and S. Because these two slits are equidistant from slit S 0, crests emitted from slit S 0 arrive at S and S simultaneously as do troughs. Therefore as the light passes through slits S and S and diffracts out into the region to the right of the double slits, these two slits effectively form two coherent sources of light. These two sources are in phase with each other. The resulting pattern can be observed on the screen C. It is necessary to have the single slit S 0 in front of the double slits S and S. We cannot create a two slit interference pattern by Fig. 0.4 shining light from the monochromatic source directly onto the double slits. Light from an incandescent source (or from a vapour discharge tube) does not come from a single source but from multiple sources, which have no constant phase relationship with each other (as discussed in Section 0.3.3). Thus if light falls on the double slits from such a source there would be no constant phase relationship between the light emanating from the double slits on the right. The effect of the single slit is to create a single source of light, so that when this impinges on the double slits we have two coherent sources of light created. A typical Young s double slit interferometer as used in school laboratories is depicted in Figure 0.5. It consists of light tight tube about 30cm long and cm in diameter. At one end is the single slit and inside are the double slits. At the other end is a magnifying eye-piece through which we observe the interference pattern on a screen. Single slit Double slits Cross hairs Micrometer Magnifying eye-piece Before the magnifying eye-piece is a set of cross Fig. 0.5 hairs and a micrometer gauge. As we turn the micrometer, the cross hairs move across the screen and, by positioning the cross hairs on lines of reinforcement and reading the micrometer gauge we can determine the distance between positions of reinforcement. 4

6 Interference THE of INTERFERENCE Light OF LIGHT The Two Slit Interference Pattern for Light As seen from the constructions in Figures 0. and 0.3 and the discussion in Section 0.5. we expect to see a series of alternating bright and dark bands. This is what we do see and this pattern of bands is depicted in Figure 0.6. Fig. 0.6 These bright and dark bands caused by the interference of light from the two sources are usually referred to as interference fringes. The bright fringes are the result of constructive interference and the dark fringes are the result of destructive interference. The relationship of the fringes to the slits and path differences from the two slits can be seen in Figure 0.7. Single slit Double slits Screen Monochromatic light source S 0 S S P 8 P 7 P 6 P P P 0 P P P 6 P 7 P 8 Fig. 0.7 At the point P 0 on the screen, which lies on the perpendicular bisector of the line segment joining the slits S and S, we have a bright fringe. This is because the point P 0 is equidistant from the slits S and S. Thus there is zero path difference between the waves from slits S and S when they arrive at P 0. Therefore they arrive at P 0 in phase with each other, and so they reinforce each other, (i.e. constructive interference results) and there is a bright band at P 0. When we move away from P 0 to some other point P on the screen, either up or down the screen, there is now a path difference from the slits S and S to that point. The further we move from P 0 the greater that path difference. 43

7 44 0 Stage Physics The path difference is given by path difference s where PS and PS PS, if P lies above the point P, if P lies below the point P. 0 0 Therefore as we move up or down the screen away from the point P 0 (where path difference = 0), eventually we come to the point P where the path difference from the two slits S and S is half a wavelength ( ). The waves from slits S and S arriving at this point are half a wavelength out of phase with each other, and so they annul each other (i.e. destructive interference occurs), and so there is no resultant wave energy at this point, and hence a dark fringe (band) results. As we move further away from P 0 the path difference keeps increasing until we get to the point P where the path difference is one wavelength. The waves arriving from S and S are in phase with each other again, and so they reinforce each other (i.e. constructive interference occurs), and so there is resultant maximum wave energy at this point, and so a bright fringe (band) results. As we keep moving further away from P 0 the path difference keeps increasing until we get to the 3 point where the path difference is one and a half wavelengths ( ). The waves arriving from S and S are half a wavelength out of phase with each other again, and so they annul each other (i.e. destructive interference occurs), and so there is no wave energy at this point, and so a dark fringe (band) results. As we keep moving further away from P 0 the path difference keeps increasing until we get to the point where the path difference is two wavelengths ( ). The waves arriving from S and S are in phase with each other again, and so they reinforce each other (i.e. constructive interference occurs), and so there is resultant maximum wave energy at this point, and so a bright fringe (band) results. As we keep moving further away from P 0 the path difference keeps increasing until we get to the 5 point where the path difference is two and a half wavelengths ( ). The waves arriving from S and S are half a wavelength out of phase with each other again, and so they annul each other (i.e. destructive interference occurs), and so there is no wave energy at this point, and so a dark fringe (band) results. And so the progression continues as we move up and down the screen from point P 0. The above can be succinctly summarised as follows: As we move up or down the screen from point P 0, the path difference from the slits S and S to any 3 5 point on the screen progressively increases from 0 3 etc. Hence we pass through alternating regions of reinforcement (constructive interference) and annulment (destructive interference). Thus we get an alternating series of bright and dark interference fringes (bands). Intensity distribution for two slit interference pattern of monochromatic light The two slit interference pattern for monochromatic light is usually represented diagrammatically, not by drawing the bright and dark fringes as in Figure 0.7, but by drawing a graph of the intensity of light at various parts of the screen as in Figure 0.8. In this graph the maxima of intensity correspond to the centres of the bright fringes on the screen and the zeros of intensity correspond to the centres of the dark fringes on the screen. The points P 0, P, P,, etc. are labelled exactly as they are in Figure 0.7 so that you can draw the parallels between the two representations, i.e. between the interference fringes in Figure 0.7 and the intensity graph in Figure

8 Interference THE of INTERFERENCE Light OF LIGHT 45 Single slit Double slits Screen Monochromatic light source S 0 S S P 8 P 7 P 6 P P P 0 P P P 6 P 7 P 8 Intensity Fig. 0.8 Note that there are no defined cut off points between a bright fringe and a dark fringe. The intensity is maximum at the centre of the bright fringes, where the waves from slits S and S are in phase with each other. As we move either side of these points the intensity progressively decreases until the path difference has changed by half a wavelength, when the intensity has decreased to zero. This effect can also be seen in Figure 0.6, where the edges of the bright fringes are definitely not distinct, but just fade away into darkness Derivation of Formulae for Path Difference and Fringe Separation In the following we will derive two formulae the formula for the path difference to any point on the screen and the formula for the distance between adjacent maxima or minima on the screen (i.e. the fringe separation or band width). In these derivations it is important to bear in mind that we are deriving this for the pattern as observed in the Young s double slit interferometer apparatus as depicted in Figure 0.5. The representation of the pattern as depicted in Figures 0.7 and 0.8 is greatly exaggerated in the vertical direction. The dimensions of the apparatus are such that the angular spread of the pattern is nowhere near as great as depicted. The pattern spreads out no more than about 3 from the central line. Because the angular spread is small the distance between the centres of adjacent maxima on the screen is constant, to a very good approximation. Obviously the distance between adjacent minima on the screen is the same as the distance between adjacent maxima. This distance is often termed the fringe separation of the interference fringes, or the band width of the pattern of bright and dark bands on the screen. 45

9 46 0 Stage Physics Formula for Path Difference to Any Point on the Screen Consider any point P on the screen that is not on the perpendicular bisector MP 0 of the line segment joining the two slits S and S. Let the angular displacement of this point up the screen be, as depicted in Figure 0.9. i.e. is the angle between MP 0 and MP. Remember that this diagram is greatly distorted. The angle is very small, no more than several degrees. Also note that the distance d between the slits S and S is also very small (less than mm). Single slit Double slits Screen P S S 0 d M P 0 Monochromatic light source S D The path difference to the point P from the slits S and S is given by path difference s Construct the line S D, such that PD = PS. path difference s SD In triangle DS S, the angle DS S = This is so because the angle between two lines is equal to the angle between their normals (perpendiculars). The original angle lies between the lines MP and MP 0. Angle DS S lies between DS and S S. DS is normal to MP and S S is normal to MP 0. Therefore angle DS S = angle PMP 0. In triangle S PD, angle S PD is very small (close to zero, less than 3 ). This is so because the distance d between the slits is very small and the angle is very small. angles PS D and PDS are both approximately 90 ( 89 +) triangle S S D is right angled at D (to a very good approximation) therefore,in S S D Fig. 0.9 i.e. SD sin S S S D d sin i.e. path difference s dsin 46