Week 7: Interference

Week 7: Interference Superposition: Till now we have mostly discusssed single waves. While discussing group velocity we did talk briefly about superposing more than one wave. We will now focus on superposition in more detail now. Superposition of waves and the resulting consequences are crucial for understanding a large number of phenomena involving interference and diffraction. At the outset, let us recall that the principle of linear superposition holds for the fields associated with the waves under consideration (eg. electric and magnetic fields for an electromagnetic wave. This is because the wave equation is linear and a linear sum of two solutions is also a solution. We must keep this in mind as we go along. Let us first discuss the simple case of adding the electric fields of two plane electromagnetic waves of the same frequency, in vacuum, but moving along different directions k 1 and k. We write the electric fields as E 1 = E 01 cos ( k1 r ωt + ǫ 1 E = E 0 cos ( k r ωt + ǫ (1 ( ǫ 1, are arbitrary constant phases. The resultant electric field is just E = E 1 + E. What is the intensity or irradiance due to the superposed field? We know that I = ǫ 0 c E T (3 with T much greater than τ (the time period of the field oscillations. We shall forget the factor ǫ 0 c henceforth and only look at I = E T. This is because we are not very interested in the absolute values of the intensities. Let us calculate E first. This turns out to be, E = ( E1 + E ( E1 + E = E 01 cos ( k1 r ωt + ǫ 1 + E01 cos ( k r ωt + ǫ + E 01 E 0 cos ( k1 r ωt + ǫ 1 cos ( k r ωt + ǫ (4 In the last term, we isolate the time-dependent parts by breaking the cos ( k1 r ωt + ǫ 1 and cos ( k r ωt + ǫ. Thereafter, taking a time-average, we obtain I = 1 E 01 + 1 E 0 + E 01 E 0 cos (( k1 k r + ǫ1 ǫ (5 1

With redefinitions and assuming E 01 and E 0 are parallel vectors, we can write the above as I = I 1 + I + I 1 I cosδ (6 where δ is defined as δ = ( k1 k r + ǫ1 ǫ (7 and I 1 = 1 E 01, I = 1 E 0. Note that if E 01 and E 0 are perpendicular to each other I = I 1 + I. The term associated with the phase δ is known as the interference term. The interference term is the cause behind the variations in the intensity. When δ = mπ we see that I = I 1 + I + I 1 I whereas when δ = (m + 1π we have I = I 1 + I I 1 I. If we have I 1 = I then the intensities become 4I 1 or zero. For visible wavelengths, we will therefore see regions where there is no light (dark and regions where there is a lot of light (bright. The redistribution of light will be regular and is quantified. Such a regular redistribution of the light intensity is known as a fringe in optics. We shall deal with the occurence and the properties of such fringes which can arise in various diverse situations. Precise measurements of the properties of such fringes can lead to precise measurements of wavelengths, the thickness of films and various other quantities in science and engineering. In the above, we have talked about plane waves. If we have spherical waves how do things change? The electric fields can now be written as E 1 = E 01 (r 1 e i(kr 1 ωt+ǫ 1 ; E = E 0 (r e i(kr ωt+ǫ (8 where r 1 and r are the radii of the overlapping wavefronts at the point of observation P (see Figure 1. The δ in this case is δ = k (r 1 r + ǫ 1 ǫ (9 We now make a crucial assumption the separation between the sources is small and the region of interference is also small. With these assumptions, the E 01 and E 0 can be assumed to be independent of r 1 and r. Therefore, assuming E 01 = E 0, we have I = 4I0 1 cos [k(r 1 r + ǫ 1 ǫ ] (10 as the expression for the intensity. We have maxima and minima for (assuming ǫ 1 = ǫ, r 1 r = π m k = m (maxima ; r 1 r = π m k = m (minima (11

FIG. 1: Superposition of two spherical waves (figure from E. Hecht (Optics with m = 0, ±1, ±... and m = ±1, ±3... What are these surfaces? They are hyperboloids of revolution (see Figure 1. If we assume a screen perpendicular to the m = 0 plane we get somewhat straight fringes as shown in Figure 1. On the other hand, if we place a screen parallel to the m = 0 surface, we will see approximately circular fringes. We now turn to the topic of coherence which is intimately related to the question: when do we really see the fringes? 3

Coherence (qualitative: Coherence is understood w.r.t. the spatial and temporal prop- FIG. : Spatial and temporal coherence erties of emissions from sources and therefore we shall talk about spatial and temporal coherence separately. Loosely speaking, temporal coherence is associated with the finite frequency band-width of a source, i.e. how close a source is to being monochromatic. Spatial coherence is related to the finite spatial dimensions of a source and hence to the shape of the wavefront as it propagates. Let us now try to understand these features pictorially. Figure shows perfect coherence on the L. H. S. (top-temporal, bottom-spatial. On the R. H. S (top we have a superposition of different frequencies which result in the formation of wave groups or wave-trains they are approximately sinusoidal over a limited period of time this time is the coherence time τ c and the corresponding temporal coherence length l c = cτ c. Both these quantities are shown explicitly in Figure 3 once again. In Figures and 3 (bottom figures we illustrate spatial coherence (L. H. S and spatial in-coherence (R. H. S. Spatial in-coherence, as mentioned earlier arises due to the finite dimensions of any source (for example, finite width for a slit source. Thus, a spatially incoherent wavefront is distorted in shape (as shown if Figs. and 3. In Figure 4, we illustrate the four possible cases: (i spatially and temporally 4

coherent, (ii temporal coherence but spatial incoherence (iii spatial coherence but temporal incoherence (iv spatial and temporal incoherence. The figures are self-explanatory. Figures -4 are taken from Saleh and Teich, Fundamentals of Photonics (007. We will discuss spatial and temporal coherence in a more quantitative way later. FIG. 3: Coherence time,length, spatial coherence Huygens principle: Long before we had a proper understanding of wave optics, Huygens, based on pure intuition formulated a principle which, remarkably, holds till today. Here is what Huygens said. Every point on a propagating wavefront serves as a source of spherical secondary wavelets, such that the wavefront at a later time is the envelope of these wavelets. If the propagating wave has a frequency ν and is transmitted through the medium at a speed v t, then the secondary wavelets have the same frequency and speed. Later, Fresnel added the notion of interference to Huygens principle and Kirchoff put this principle on firm mathematical ground using Maxwell s theory of electrodynamics. For a spherical waves, while using Huygens principle, we just draw hemispheres, ignoring the backward wave. This issue was indeed addressed and solved by Kirchoff. A discussion 5

FIG. 4: Different situations on this is however beyond the scope of this course. Young s double slit: Let us now turn to Young s double slit experiment which is the first example of an interferometer or more precisely a wavefront splitting interferometer. Generically, interferometers are devices which produce a regular redistribution of light known as a fringe. Thomas Young made this device his use of two slits produced two coherent beams which could produce observable interference. Figure 5 (top provides a three dimensional view of what is happening in the Young s double slit experiment. Note the incident cylindrical wavefront which is split at the two slits and which then interferes to produce the new intensity distribution. The lower figure in Figure 5 sets up the details in a transverse section. We may work out the details as follows. Assume the coordinates of the various points in the lower figure in Fig. 5 as follows (choosing the plane of the figure as the xy plane. The origin is at the centre between the two slits. The top slit has coordinates (0, d and the bottom one has (0, d. The coordinates of the point of observation in the screen are (D, y 6

FIG. 5: Young s double slit, figure from Hecht, Optics and the distance from the origin to the point P is r, the line joining the origin to P makes an angle θ with the x axis. The unit vectors representing the propagation vectors of the two waves from slit 1 and slit are ˆk 1 = Dî + (y d ĵ D + (y d (1 ˆk = Dî + (y + d ĵ D + (y + d (13 The vector r = Dî + yĵ. Therefore one can evaluate ( k k 1 r = k (ˆk ˆk 1 r. Using the approximation that D is large, i.e. the screen is very far away, we can assume D + (y d r D + (y + d. This finally gives ( k k 1 r dyk r Since y = r sin θ and D = r cosθ and further, assuming θ is small so that sin θ θ, we get the phase δ as δ = πdθ 7 (14 (15

Therefore, we finally obtain the intensity as ( I = I 0 1 + cos πdθ (16 The bright fringes appear for and the fringe-width is y m = mr d y = r d ; θ m = m d (17 (18 Figure 6 shows further details about Young s double slit including a photograph of the actual fringes that occur. There are several other devices which are similar to the Young s double slit set-up these include Fresnel s mirrors, Fresnel biprism and Lloyd s mirror. The analysis of interference in each of them appear as problems in Tutorial 6. FIG. 6: Young s double slit, figure from Hecht, Optics The Young s double slit experiment can be used to measure the spatial coherence of a source. Let us assume that the source from which light is incident on the double slit system is wide. Its angular width is α, i.e. w.r.t the origin it spreads over the angles α to α. Let us consider a point at an angular position β (where β 0. If we consider the intensity distribution due to this point source we will find that the intensity changes to ( πd (θ + β I(θ, β = I 0 1 + cos 8 (19

S1 P α/ α/ S β θ S Young s double slit: wide slit, spatial coherence. Screen FIG. 7: Young s double slit, wide slit, spatial coherence Exercise: Show that the above expression is indeed correct. If we now superpose the intensities due to the point sources across the width α (note we are not considering interference between these sources, all the point sources are mutually incoherent, we will get I(θ = 1 α α α Therefore, we have I(θ, βdβ = I 0 + I 0 α sin πd ( θ α sin πd ( θ + α πd [ = I 0 1 + cos πdθ ] ( I = I 0 1 + (sinc u cos πdθ sin πdα πdα where u = πdα. Note that we once again get a sinc function in the final expression. What does this sinc function do. If u 0 then the sinc function becomes 1 and we get back our earlier expression for the intensity. u 0 means that α 0 and the slit is very close to a point. But if u = mπ (m 0 then we find that the so-called interference contribution vanishes (i.e. there are no fringes at all and I = I 0! Thus, the sinc u piece seems to decide whether we will see clear fringes or not in other words, it tells us about the visibility of fringes. If we rewrite the intensity as ( I = I 0 1 + C 1 cos πdθ 9 (0 (1 ( (3

then, we can show quite easily that In general, the visibility is defined as C 1 = I max I min I max + I min = V (4 V = C 1 = I max I min I max + I min (5 4 3 1 0 0 5 10 15 FIG. 8: Young s double slit, spatial coherence, u = 1 (red, u = (green, u = 10 (blue If V = 1 then the fringes are sharply visible. If V = 0 they are not visible. Why do the fringes loose visibility? The cause seems to be the finite width of the source which results in various different wavefronts which illuminate the double slit thereby destroying spatial coherence. The visibility is thus a measure of spatial coherence. While doing the Young s experiment, one can measure the maximum and minimum intensities and calculate V it will surely never be exactly equal to one and its value will be able to tell us about the extent of the primary source. Division of amplitude interferometry: In Young s experiment we achieved interference effects by dividing the waverfont. Alternatively, one may consider dividing the amplitude. A simple way of dividing the amplitude is by considering the superposition of different reflected waves. For example, to begin, let us consider a parallel plate or a film of uniform thickness d (transparent plate/film and refractive index n f. The refractive index outside the plate is n 1. Light is incident obliquely 10

(at an angle, say θ i on this film. It is reflected and refracted at the point A. The refracted ray travels inside the plate/film and gets reflected again. It is once again refracted into the medium n 1. We would like to see what happens if we superpose these two waves the one directly reflected and the other refracted-reflected-refracted. S θ t A θ i D C B n n n 1 f 1 P d FIG. 9: Division of amplitude, parallel film. It is easy to consider the effect of superposition by looking at the net path/phase difference. Look at Figure 9. What is the optical path difference between the paths SADBP and SAP? We have We know that Therefore Λ = n f (AD + DB n 1 (AC (6 AD = We also have AC = AB sin θ i and AB = d tanθ t. Therefore, d cosθ t = AB (7 Λ = n fd cosθ t n 1 AC (8 Λ = n fd cosθ t n 1 d sin θ i tan θ t (9 11

FIG. 10: Division of amplitude, parallel film, figure from Hecht, Optics. From Snell s law of refraction n 1 sin θ i = n f sin θ t. Using this, we get Λ = n f d cosθ t (30 If the angle of incidence is such that it is less than θ p (recall the polarisation, Brewster angle and n 1 < n f (say, air and glass or air and water then for both components of the incident and reflected fields (the in-plane of incidence and to plane of incidence, there will be a net phase difference of π between the two reflected waves meeting at P. Therefore, we have the net phase difference as: δ = 4πd cosθ t f π (31 Exercise: Work out the path difference if n 1 > n f and θ i < θ p, n 1 < n f and θ i > θ p. Since, this is a superposition of two waves, we will have the maxima and minima governed by δ = mπ and δ = (m + 1π, respectively. Hence, we have d cosθ t = (m + 1 f 4 (Maxima d cosθ t = (m f 4 (Minima (3 The interference fringes will be seen over a small region if the source is point source and the lens used to focus the rays has a small aperture. For an extended source, light will reach the lens from all directions and the maxima-minima will be spread out over a larger region of the plate/film. Figure 10 shows how the waves interfere to form the fringes. 1

What controls δ in this case? It is the value of θ t or θ i. Thus, the fringes appearing at P are for a specific θ i. If we change θ i the maximum or minimum appears at another point. This leads to the nomenclature fringes of equal inclination. For an extended source, each point on it is a source which is incoherent w.r.t. any other point. Therefore, when we see the image of the extended source in reflected light, we find dark and bright bands. Each band is on the arc of a circle centred on the intersection of a perpendicular dropped from the eye to the film. Note that we have considered the interference of only two waves whose amplitudes are obviously different due to reflection and transmission. In reality, of course the process of successive reflections and transmissions continue. We have made an assumption that the subsequent reflected and transmitted waves do not contribute because their amplitudes have become very small. This assumption does seem to work. When we increase d, the quantity AB increases and one of the rays may not reach the eye. In such an event, we will not see any interference fringes. A larger lens has to be used. We can also reduce AB by reducing θ i (nearly normal incidence. The fringes of equal inclination seen for nearly normal incidence and with thick plates are known as Haidinger fringes. Since θ i is close to zero, we have circular symmetry and you see concentric circular fringes in the field of view. Figures 11, 1 show the various cases for an extended source. Figure 13 illustrates Haidinger fringes. Another type of fringes can be observed in reflected light. This involves the case of a wedge shaped film (see Figure 14. The fringes seen here are known as fringes of equal thickness since the dominant parameter here is the optical thickness n f d. The colours of thin films, oil slicks upon illumination by white light are largely understood as fringes of equal thickness. If we assume normal incidence (Fizeau fringes, then θ t = θ i = 0. If the wedge angle is α, then we know that the thickness at the horizontal location x (see Figure 14 is given by d = xα (33 Using the path/phase difference calculated earlier, we find that the condition for a maximum 13

FIG. 11: Division of amplitude, parallel film, figure from Hecht, Optics. at a specific d m (or x m is, ( m + 1 0 = n f d m = n f αx m (34 We can rewrite this as Therefore, as we move along the wedge, after every f 4α the separation between the maxima will be x m = m + 1 α f (35 we will encounter a maximum. Hence x = f α (36 The thickness of the film at the maxima will be d m = ( m + 1 f (37 14

FIG. 1: Division of amplitude, parallel film, figure from Hecht, optics. FIG. 13: Haidinger fringes, figure from Hecht, Optics which is an odd mutiple of the quarter of a wavelength (take m = 0, 1,... If you hold a soap film in a frame vertically and illuminate it with white light you notice that there are various colours but the top portion turns out to look black. This is because the thickness of the top part has become less than f 4 (due to drainage caused by gravity 15

Extended source P 1 n 1 α nf x n FIG. 14: Wedge shaped film and, therefore, there is no scope of having an intensity maximum there. The most well known application of the fringes of equal thickness is the case of Newton s rings that you see in your laboratory class. Figure 15 shows the set-up. If R is the radius of curvature of the lens and d, x as shown in Figure 15, then, x = R (R d Rd (38 since R is much greater than d. If we, as before, look at only the first two reflected rays (see Figure 15, then the mth order interference maxima occur when the thickness follows the relation n f d m = ( m + 1 0 (39 The radius of the mth bright ring is therefore found as x m = [( m + 1 ]1 f R (40 Similarly, for a dark ring, we have x m = (m f R 1 (41 Hence, the central fringe m = 0 will clearly turn out to be dark. It will be a maximum (bright if we observe in transmitted light. Newton s rings are another example of Fizeau 16

FIG. 15: Newton s rings, figure from Hecht, Optics fringes. An application of Newton s rings, which you do in your lab, involves experimentally determining the radius of curvature of the lens from the radii of the rings, provided the wavelength is given. In the same way, if the radius of curvature is known, one can find the wavelength. If the region between the lens and the lower plate is filled with a liquid of a certain refractive index, then one can find the refractive index by determining the radii of the bright and dark rings. 17