Waves and Matter: Polarisation Part II & Interference in Optics PDF

B27FF – Waves and Matter Waves Week 5 Richard Carter page 1 of 23 5. Polarisation Part II & Interference in Optics For the textbook see Sears and Zemansky chapter 33 “The Nature and Propagation of Light”...

B27FF – Waves and Matter Waves Week 5 Richard Carter page 1 of 23 5. Polarisation Part II & Interference in Optics For the textbook see Sears and Zemansky chapter 33 “The Nature and Propagation of Light” plus chapter 35 “Interference”. 5.1 Circular Polarisation, Photoelasticity and Scattering A: Circular and Elliptical Polarisation Last week we looked at linear polarisation and the use of linear polarisers for converting and/or filtering this kind of polarisation. However other polarisation states exist. The most important of these is circular polarisation. Circular polarisation is formed when we have wave components of the same amplitude and frequency but orthogonal linear polarisation and a phase difference of 𝜋/2. To explain this let us first reconsider a case from last week. Two identical waves which have orthogonal polarisation. Figure 5.1 Illustration of linear polarisation. Two waves of equal amplitude, frequency and phase but orthogonal polarisation are shown propagating down the x axis. At any point the combined amplitude and polarisation can be found by adding the two vectors. B27FF – Waves and Matter Waves Week 5 Richard Carter page 2 of 23 Figure 5.1 shows the situation. There are two polarisation states one aligned with the z axis (blue) and the other along the y axis (green). The two waves have the same wavelength, amplitude and start with the same phase. At any point along the z axis the combined 𝑬 ⃗ field vector can be easily shown (in red) as either in the positive or negative direction at 45° to the y,z axis. At some points ⃗ field vector as the amplitude of both components (0, 𝜋. 2𝜋 𝑒𝑡𝑐. ) along the propagation there is no 𝑬 is zero at these points. 𝜋 Now let us see what happens if we add a 2 phase difference to the two waves: Figure 5.2 Illustration of circular polarisation. Two waves of equal amplitude, frequency but with a phase difference of 𝜋/2 and orthogonal polarisation are shown propagating down the x axis. At any point the combined amplitude and polarisation can be found by adding the two vectors. Again it the combined wave vector is easy to plot at any point down the propagation axis but this time we can see the ⃗𝑬 field vector is not confined to a single line but rather goes round a complete circle in one wavelength (2𝜋 phase). The points plotted in Figure 5.2. happen to be those where only one of the two waves has amplitude to add to the wave vector but if we plot more of these we can see the complete circle (Figure 5.3) B27FF – Waves and Matter Waves Week 5 Richard Carter page 3 of 23 Figure 5.3 Illustration of the polarisation rotation in circularly polarised light. Here one wavelength is illustrated at 8 points in one wavelength period, T. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1121, 2015) Unlike the linear polarisation system the ⃗𝑬 field vector has a constant magnitude ⃗ field vector rotates one complete circle in one wavelength. The direction of the 𝑬 The direction of rotation depends on if there is a 𝜋/2 or −𝜋/2 phase difference That last point means there are two forms of circularly polarised light right hand and left hand circular polarisation. The difference being if the polarisation vector rotates clockwise (as in the image, which is right hand) or anti-clockwise (left hand) when looking down the z axis toward the origin. Sometimes the phase difference between the two components is not exactly 𝜋/2 or possibly the amplitude of the two components is not the same. In these cases, the perfect symmetry of the system is broken and instead of describing a circle the ⃗𝑬 field vector will describe an ellipse giving elliptically polarised light. Converting linearly polarised light to circularly polarised light (or visa versa) required introducing a specific phase shift between the two polarisation components. This can be done using a birefringent material. This is one which has two different refractive indices, one for each polarisation axis. This implies that there is a difference in the wavelength/wave speed of the two polarisations in the material. Hence the two axis are often called fast and slow axis. To make a filter we cut the material to have an exact number of wavelengths for one axis and an exact number plus ¼ for the other axis. This sort of filter is referred to as a quarter wave plate as it adds ¼ of a wavelength of difference. With this filter (in the right orientation) you can convert circular polarisation to linear and vice versa. The other common form of waveplate is a half wave plate. This is also birefringent and, as the name implies, introduces ½ of wavelength difference. The effect of this plate is rather different, it rotates the angle of both polarisation states. For linear polarisation this will rotate the angle, for circular it has no effect. B27FF – Waves and Matter Waves Week 5 Richard Carter page 4 of 23 B: Photoelasticity Truly birefringent materials are reasonably uncommon however several materials have effects where the refractive index varies according to some physical stimulus. The two common examples are the thermo-optic effect (temperature), the stress-optic effect (stress). The stress-optic effect is particularly interesting as the change in refractive index is different for polarisations aligned with and against the stress in the material. This effect is referred to as photoelasticity and is particularly common in polymers (plastics). The process can be usefully applied to convert stress into a visually resolvable effect and there are instruments which can use this to quantitatively measure the stress in transparent materials. Generally, this is done by placing the material between two crossed polarisers (Figure 5.4). The first polariser will, of course, polarise the light, if nothing affects the angle of polarisation then the second polariser will filter it out and no light will be seen. However if the material is birefringent then it will act something like the half-wave plate, rotating or otherwise altering the polarisation and some light will pass through the second polariser. The amount of light, which is passed, and the colour (if the material is also dispersive) can be used to gauge the amount and the direction of the stress in the object. Figure 5.4 Example of the photoelastic effect. Here a plastic model (a cross section of a hip replacement) is placed between crossed polarisers and bent. The fringes indicate the direction and severity of the stress in the object. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1122, 2015) C: Scattering Last week we worked out how rainbows are formed. Now we turn to the question - why is the sky blue”. The answer is due to scattering but this requires some explanation as there are two separate effects to consider. B27FF – Waves and Matter Waves Week 5 Richard Carter page 5 of 23 Scattering is the overall effect when light is absorbed by an atom and re-radiated. The key point is that scattering is not entirely random but depends on the frequency and the radiated light may be polarised. The frequency effect is simpler to consider. Experiments demonstrate that1 the intensity of scattering is proportional to 𝑓 4 hence higher frequency light (blue) is much more likely to be scattered than low frequency light (red). Since the light from the sun is approximately white it means that un-scattered light will appear mostly red and scattered light is mostly blue. If you look up at the sky away from the sun (do not look directly at the sun!) the only light you can see is that scattered by the atmosphere, i.e. light which has not taken a direct route from the sun to your eye. Hence it is blue. By contrast if you look at a sunrise then it is light which is mostly un- scattered which will make it through the atmosphere for you to see hence it will appear red. Clouds appear white because they are made of a relatively high concentration of water vapour. As such there is a massive amount of scattering in a cloud and all frequencies will end up randomly scattered to more or less the same degree, hence they are white. The second effect is one of polarisation. Figure 5.5 shows a sketch of this effect whereby the un- scattered light is unaffected as is still unpolarised but the scattered light will be polarised in a specific direction – one which is orthogonal to the direction of the sun. Figure 5.5 Illustration on how scattered light is polarised. Note that light is absorbed equality for all polarisations, so the transmitted light is unpolarised but the scattered light is polarised according to the direction it is scattered to. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1123, 2015) As mentioned scattering is the effect caused by the absorption and re-radiation of light from particles/molecules. Many of the molecules in the air are able to be polarised (i.e. have unbalanced electric charges) and hence act as a dipole with a specific, but random, orientation. In effect these 1 This is really much more complex as it depends on the size of the particles causing the scattering. What is stated here is true for air molecules and hence why the sky is blue on earth. However, for larger particles (dust) the effect is reversed which is why the atmosphere on Mars appears red, this means a Martian sunset is blue not red and NASA have the photos to prove it. B27FF – Waves and Matter Waves Week 5 Richard Carter page 6 of 23 dipoles act as a mini antenna which re-radiates the light out in a direction and with a polarisation which is determined by the orientation of the dipole. Since light is a transverse wave the dipole will be excited transverse to the direction of propagation (this is the direction the ⃗𝑬 field is oscillating in) so none of the dipoles will be excited in the x direction but rather somewhere in the y,z plane. These oscillations can then be resolved into two components, along the y and z axis. Since an oscillating dipole (or antenna) will not re-radiate in the same direction as the oscillation none of the z oscillation components will radiate light toward the ground. Instead only the light radiated by the y components will be seen on the ground and hence the light seen on the ground will be polarised in the z direction, Figure 5.5. Note that the un-scattered light is still randomly polarised, the scattering effect does not act as a filter. 5.2 Interference, Young’s Slits and Thin Film Interference A: Interference in 2 and 3 Dimensions You will recall from earlier in the course that interference refers to any situation where two or more waves overlap and is governed by the principle of superposition. Provided that the polarisation is not orthogonal we can get constructive and destructive interference depending on the relative phase of the two waves at any given point. The principle here is general and will apply to EM waves just as easily as to sound or waves on a string. So far we have looked at two special cases of interference: 1. Two identical waves counter propagating along a line creating a standing wave. 2. Two non-identical waves propagating along a line generating a beat frequency. However both of these forms of interference are essentially one dimensional since they assume the waves are travelling in the same, or opposing, directions. We did briefly look at what happens when waves propagate from two point-sources and determined that we can obtain constructive or destructive interference depending on the path length difference. Let us now revisit and complete this. We can start by plotting the wavefronts of a point source at some “snapshot” in time (Figure 5.6) B27FF – Waves and Matter Waves Week 5 Richard Carter page 7 of 23 Figure 5.6 A snapshot of a sinusoidal wave emitted from a point source. The blue lines are wavefronts representing maximum intensity (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1185, 2015) For this section of the course we will use the wavefronts corresponding to wave crests as in our plots. Hence each wavefront is separated by one wavelength and represent a maximum amplitude of the wave. The waves propagating from the source in Figure 5.6 (source 1, 𝑆1 ) do so in a circular pattern indicating that there is a constant wave speed in all directions. If 𝑆1 was radiating an EM wave this implies that the refractive index is a constant around it (this gives the constant wave speed) and that it is a monochromatic source (since there is one specific frequency and thus wavelength being emitted here). While single frequency sources of sound are relatively common monochromatic light sources are not, this is really limited to lasers and some other very specific sources of light generated by fluorescence. The plot in Figure 5.6 is also clearly in 2D, or it is a 2D slice of a 3D image. There is nothing stopping us from making a 3D plot of the same sort other than the difficulty in visually representing it on a piece of paper. Everything we are about to discuss will work perfectly well in 2 or 3D but we will keep using 2D diagrams for simplicity. B: Constructive and Destructive Interference Now let how the diagram changes when we add a second source. To do this we need to be careful to define what sort of wave this source is generating. To keep things simple this will also be monochromatic, have the same amplitude, frequency and be driven at the same phase as the first source. Hence our two sources are in phase with one another. This plus the same frequency makes them coherent. We will also assume that the two sources have the same polarisation to allow them to interfere. B27FF – Waves and Matter Waves Week 5 Richard Carter page 8 of 23 Figure 5.7 Illustration of two source interference. a) Wavefronts from two sources (green and blue) with three points a, b, c noted and the paths from the two sources. b) Analysis of the path lengths to point b (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1186, 2015) The result, Figure 5.7, shows the expanding wavefronts from both sources. Because there is a constant phase relation between the two sources and they have the same frequency adjacent wavefronts have a constant radial spacing between them. If this was not the case the two waves would be incoherent and the interference pattern would be highly randomised – to the point that it would average out and we would say the two waves cannot interfere. Drawn on Figure 5.7 is a line bisecting the two sources, x. At any point on this line (e.g. a) the distance between the two sources is the same so we would expect to see constructive interference. Notice that this line joins up points where the wavefronts cross. If we choose another point on the plot, we will need to work out the path lengths and hence the phases of the light reaching that point from the two sources. As an example at b we have a point where the wavefronts cross on the diagram. At this point there is an exact number of wavelengths from 𝑆1 → 𝑏 = 𝑟1, and 𝑆2 → 𝑏 = 𝑟2. It does not matter that there are a different number of wavelengths the two waves will arrive in phase and constructively interfere (Figure 5.7 b). Conversely at c we have a position which lies on one of the wavefronts from 𝑆1 but exactly half way between the wavefronts from 𝑆2. This means there is an exact number of wavelengths from 𝑆1 → 𝑏 = 𝑟1 , but an exact number and a half wavelengths from 𝑆2 → 𝑏 = 𝑟2. Again it does not matter what the exact number of wavelengths is what matters is there is a difference of a half and hence the waves are out of phase and will destructively interfere (Figure 5.7 c). This gives us a fundamental principle for constructive and destructive interference. For constructive: 𝑟2 − 𝑟1 = 𝑚𝜆 (𝑚 = 0, ±1, ±2, ±3 … ) 5.1 For destructive: 1 5.2 𝑟2 − 𝑟1 = (𝑚 + ) 𝜆 (𝑚 = 0, ±1, ±2, ±3 … ) 2 B27FF – Waves and Matter Waves Week 5 Richard Carter page 9 of 23 In both cases m is an integer number. For Equations 5.1 and 5.2 to hold the sources must be coherent, i.e. of the same frequency and phase relation but also be emitted at the same phase. Have a think – how would equations 5.1 and 5.2 be altered if the two sources were coherent but the two waves were not emitted in phase? Now recall that x represents the line where the path lengths 𝑟1 = 𝑟2 i.e. when 𝑚 = 0 for equation 5.1. It is now possible to draw on lines representing the positions for 𝑚 = ±1, ±2, ±3 etc. for equations 5.1 and 5.2. If we do this for the constructive interference case these are antinodal curves. For destructive interference these are nodal curves. The terminology here is the same as for standing waves: nodes never see amplitude i.e. they are points of destructive interference and antinodes sit between them. To keep things simple, we will draw only the antinodal curves (constructive interference) if you look at Figure 5.7a you should be able to work out where these go, we already have the 𝑚 = 0 curve and the higher order lines will similarly connect points where the wavefronts cross: Figure 5.8. Figure 5.8 Plot of antinodal curves for two coherent, in phase sources. Note that there are more orders of interference but we may need a larger diagram to plot them. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1187, 2015) C: Two Source Interference of Light Creating two coherent sources of sound or in water etc. is relatively straightforward as we can drive two systems with the same signal. However, for light this is much more challenging. Even if we have two sources with the same signal it is impossible to keep them in phase due to quantum mechanical B27FF – Waves and Matter Waves Week 5 Richard Carter page 10 of 23 effects. Instead, it is usually necessary to take a coherent light source (one which has a constant frequency and phase) and then split it into two. These secondary sources will maintain coherence over a period of time i.e. separation distance (although not forever as there will be slight variations in phase over time). The distance at which the two beams can remain coherent is called the coherence length of the source and we will have more to say about that later. For the moment we will assume the following experiment takes place within this coherence length. Figure 5.9 illustrates one of the first examples of interference using a pair of secondary coherent light sources. A single slit, only a few µm wide is used to create a long thing light source through diffraction2. A second pair of slits repeats this process creating two new secondary line sources which are coherent. Note that we assume the two slits are placed an equal distance from the first slit, so that the light that hits and is emitted from them is in phase. Figure 5.9 Young’s Slit experiment. a) Overall schematic. b) Actual geometry. c) Our simplified geometry. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1188, 2015) 2 This concept will be covered in future optics courses if you are taking Physics. For now we can just accept that light exits the slit in an extended cylinder shape of which we depict a 2D slice. B27FF – Waves and Matter Waves Week 5 Richard Carter page 11 of 23 As the light from these two slit sources propagates forward, they will overlap and hence bands of interference can be seen on a screen. Light and dark bands indicate differing orders of constructive and destructive interference. This experiment was performed in 1800 by Thomas Young and as such is known as the Young’s Slits (Experiment). Clearly we expect to see constructive interference when the path length difference from the two slits (𝑆1 𝑎𝑛𝑑 𝑆2 ) fits equation 5.1. The actual geometry (Figure 5.9 b) requires that for a given position on the screen, 𝑦, the rays coming from 𝑆1 𝑎𝑛𝑑 𝑆2 will be travelling at different angles. However, in this experiment the distance between the two slits, d, is extremely small in comparison to the distance between the slits and the screen, 𝑅. Hence the two angles will be very similar and the difference so small is a reasonable approximation to ignore them. This allows for the approximation in Figure 5.9 c where we assume two parallel rays which will nevertheless “meet” on the screen. Hence the only path length difference is at the slit end formed from a right-angle triangle. Basic geometry shows this path length difference is: 𝑟2 − 𝑟1 = 𝑑 𝑆𝑖𝑛[𝜃] 5.3 Where 𝜃 is the angle of the rays shown in Figure 5.9 b. Combining this with Equations 5.1 and 5.2 allows us to work out the angles associated with constructive and destructive interference: For constructive: 𝑑 𝑆𝑖𝑛[𝜃] = 𝑚𝜆 (𝑚 = 0, ±1, ±2... ) 5.4 For destructive: 1 5.5 𝑑 𝑆𝑖𝑛[𝜃] = (𝑚 + ) 𝜆 (𝑚 = 0, ±1, ±2... ) 2 This will appear on the screen as a series of light and dark bands like those in Figure 5.10 and are often called interference fringes. B27FF – Waves and Matter Waves Week 5 Richard Carter page 12 of 23 Figure 5.10 Example image of interference fringes produced by the Young’s Slit experiment. The orders of interference are indicated on the image, the “0” order will be in line with the centre of the two slits as the point of equal optical path length. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1189, 2015) However it is generally more useful to be able to measure the distances on the screen as opposed to the angles of the rays and this. This requires a reference point for 0 distance which will be the m=0 constructive band. In this case we will run the calculation for the constructive (bright band) interference. Again, simple geometry tells us: 𝑦𝑚 = 𝑅 𝑇𝑎𝑛[𝜃𝑚 ] 5.6 Where 𝑦𝑚 is the distance from the centre of the 0 order bright interference band (which will be in line with the centre of the two slits) and the centre of the 𝑚𝑡ℎ order bright interference band , 𝑅 is the distance between the slits and the screen and 𝜃𝑚 is the angle of the rays corresponding to the 𝑚𝑡ℎ order interference band. B27FF – Waves and Matter Waves Week 5 Richard Carter page 13 of 23 Combined with Equations 5.4 and 5.5 this gives: 𝑚𝜆 5.7 𝑦𝑚 = 𝑅 (𝑚 = 0, ±1, ±2... ) 𝑑 Note that we have applied a second approximation here. We have taken the small angle approximation that 𝑇𝑎𝑛[𝜃] = 𝑆𝑖𝑛[𝜃] = 𝜃. As the name implies this approximation will hold provided that 𝜃 is a small angle3. This provides an equation which is easily used in practice since we can measure 𝑅 𝑎𝑛𝑑 𝑑, count of orders of 𝑚 on the screen (see Figure 5.10). Since the distance 𝑦 is also easily measured this provides a simple way to calculate the wavelength – which is what the experiment was first used to do. Notice that 𝑦𝑚 is inversely proportional to 𝑑 so if the slits are more closely spaced the interference pattern will spread out. Should we need to calculate the positions of destructive interference we can either replace m with (𝑚 + 1/2) in the equation above or simply note they will be exactly halfway between the constructive interference bands. 5.3 The Principle of Interference by Division of Amplitude, Thin Film Interference and Wedges A: Optical Path Length Difference Note that this section deviates from the textbook which has a longer and more complex explanation for optical path length which is not required here. We have previously described optical path length in terms of numbers of wavelengths. In this form is it simple enough to determine the phase difference between two optical paths as any whole numbers of wavelengths will result in a 2𝜋 phase shift – i.e. no overall phase change and only a fractional component will contribute to a phase difference. This is perhaps clearest if we try an example. Consider two optical paths for a wavelength of 1.1 µm: 1) 2 cm in air 2) 3 cm in air The two optical path lengths will simply be the number of wavelengths within the path: 𝐿 5.8 𝑝= 𝜆 Where 𝑝 is the optical path and 𝐿 the physical length. In this case the path length for 2 cm is 1818.182 𝜆 and for 3 cm 2727.273 𝜆. 3 You can try the approximation yourself. It is accurate to 3dp for angles below ~0.4 rad B27FF – Waves and Matter Waves Week 5 Richard Carter page 14 of 23 The optical path length difference is then the difference between these: Δ𝑝 = 𝑝1 − 𝑝2 5.9 Which for this example will be: 2727.273 𝜆 − 1818.182 𝜆 = 909.091𝜆 5.10 The phase difference can then be determined by ignoring the whole numbers of wavelengths and only looking at the fraction 0.091𝜆 since one whole wavelength is 2𝜋 this will correspond to: 0.091 × 2𝜋 = 0.572 𝑟𝑎𝑑 5.11 Since this is neither 0 𝑛𝑜𝑟 𝜋 it will not be full constructive or destructive interference, but it is also closer to 0 than to 𝜋 so we should expect partially constructive interference. Effect of Refractive Index The example above – and all the examples so far – have assumed our two optical paths are in the same material and therefore use the same refractive index. This is not always the case, and the types of interference we are about to look at next require different materials to be involved. There is more than one way we can consider the effect of refractive index but by far the simplest is to recall that this will alter the wavelength from the vacuum value 𝜆0 to a new value in that material 𝜆𝑛. This will clearly have an effect on the number of wavelengths along a path. Recall the equation for wavelength in a material 𝜆0 5.12 𝜆𝑛 = 𝑛 This modifies the optical path length (Equation 5.X) to be: 𝑛𝐿 5.13 𝑝= 𝜆0 In fact nL, is often used and quoted as the optical path length. This would give an optical path length in m as opposed to multiples of wavelength and is usually a convenient substitution for 𝐿 in most equations. Clearly, we would need to use the wavelength of the light in a vacuum to convert the optical path length in this format to a phase. As an example, let us suppose that the 2 cm path is now in glass, which has a refractive index of 1.5 while the 3 cm path is still in air. The 3 cm path will still be 2727.273 𝜆 but the 2 cm path is now also 2727.273 𝜆 hence the path length difference and phase difference is now 0. This is even clearer if B27FF – Waves and Matter Waves Week 5 Richard Carter page 15 of 23 we use nL to calculate the optical path length in terms of m. For 3 cm this is 3 cm (𝑛 = 1) for the 2 cm path this is 1.5 × 2𝑐𝑚 = 3𝑐𝑚 hence the optical path length difference is clearly 0. B: Thin Film Interference If you have ever looked at a soap bubble or oil on water, you will have seen a very characteristic appearance consisting of bright bands of colour (Figure 5.11). These bands are created by a specific form of interference: thin film interference. Figure 5.11 Thin film interference. Left as seen in a soap bubble. Right) schematic drawing for a flat thin film. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1195, 2015) Like all interference effects this requires two (or more) optical paths with different optical path lengths. In this case the two optical paths are created by two reflections from a thin film (or layer). Figure 5.11 shows the general principle. 1. The first optical path is generated by light from a reflecting at an angle from the air-film surface at b directly into the eye i.e path 𝑎 → 𝑏 → 𝑐 → 𝑝 2. The second optical path is generated by the light refracted from the air-film interface. Part of this will be reflected from the second film interface, d, (in the case of oil on water this is the oil-water interface. For the soap bubble it is a film-air interface). This second reflection is refracted a second time on the way out of the film, e, and is directed into the eye. i.e. path 𝑎→𝑏→𝑑→𝑒→𝑓→𝑝 Clearly the physical path length of ray 2 is longer than ray 1, not only this but it is also partially (from b to e) travelling in a material with a refractive index 𝑛 > 𝑎𝑖𝑟. For the oil on water example this is exactly what is happening. The different colour bands come from the fact that the film is dispersive so the optial path length (partially a funciton of the refractive index of the film and partly the angles of refraction) will be different for different colours. Hence different locations on the film will appear as light or dark bands of interference depending on the colour. However, the details of where we expect to see different bands of colour depend on what type of reflections we will get from the thin film. Let us illustrate this by considering another simple case. B27FF – Waves and Matter Waves Week 5 Richard Carter page 16 of 23 Monochromatic light inidcent on a pair of glass microscope slides which are placed at an angle to one another (using a wedge of some sort) and an observer looking almost straight down onto the slides, Figure 5.12. Figure 5.12 Thin film interference generated by an air wedge. Note that the approximations used for the analysis require the rays to be incident on the glass at close to normal and the angle between the two glass plates to be small. Both are exaugurated here for clarity. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1189, 2015) In this case the two reflections are generated from the bottom of the first glass slide and the top of the second glass slide (the other surfaces will generate reflections but we will ignore these for simplicity). Since both optical paths share the same path up to the reflection from the bottom of the first glass slide this will not contribute to an optical path length difference and we can ignore this. In addition if we follow through the laws of reflection and refraction we will discover that the two rays are parallel on their return path through the top glass slide. Here we will make an approximation. As the observer is almost direcly on top of the glass slides all the rays in the diagram will be almost vertical. We can therefore ignore any optical path length difference generated between the glass slides and the eye as the paths will be near identical. So the optical path length difference is entirely due to the extra distance travelled by one of the rays in the air gap. This will be approximately 2𝑡 since the path length in the air gap will be approximatly twice the gap thickness for vertical rays. For a monochromatic light source we would expect to see bands of light and dark depending on if 1 2𝑡 = 𝑚𝜆 or 2𝑡 = (𝑚 + 2) 𝜆 according to Equations 5.1 and 5.2. But if we were to do this experiment, we would see that this is not the case since the pattern we will see has a dark fringe at the point the two glass plates touch. At this point the optical path length has decreased to 0 hence B27FF – Waves and Matter Waves Week 5 Richard Carter page 17 of 23 we expect this to be 2𝑡 = 0𝜆 i.e. 𝑚 = 0 producing constructive interference but exactly the opposite has happened. Why? The answer is in the two reflections. One is from a glass air interface (𝑛𝑎 > 𝑛𝑏 ) and the other from an air-glass interface (𝑛𝑎 < 𝑛𝑏 ). The difference is significant. Maxwell’s equations for EM radiation will give a fully rigorous answer to this but this is outside the scope of the course. Instead we will just look at the relevant equation for the reflected E field amplitude from a surface: 𝑛𝑎 − 𝑛𝑏 5.14 𝐸𝑟 = 𝐸 𝑛𝑎 + 𝑛𝑏 𝑖 Where 𝑛𝑎 and 𝑛𝑏 take on their usual meanings of the refractive index of the two materials and 𝐸𝑟 is the reflected amplitude. From this we can see an interesting observation: Where (𝑛𝑎 > 𝑛𝑏 ), Er is positive Where (𝑛𝑎 < 𝑛𝑏 ), Er is negative This is just like the case we saw with a wave on a string back in week 2 where we had two types of boundry conditions – one where the refelcted wave had the same amplitude sign (phase) as the incident wave and one where the reflected wave has the opposite amplitude sign (phase) as the incident wave. Here we can see that the boundry coniditon depends on if (𝑛𝑎 < 𝑛𝑏 ). In fact the same principle applies for reflections of mechanical waves anywhere there is a change in the wave speed, Figure 5.13. Figure 5.13 Examples of the boundary conditions in reflection for boundary giving a change in wavedspeed. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1196, 2015) For a wave on a rope the wave speed depends on the tension and the mass per unit length. Since there is no real way to abruptly change the tension on a rope this is more likely to occur from a B27FF – Waves and Matter Waves Week 5 Richard Carter page 18 of 23 change in the thickness or density of the rope. For an EM wave the wave speed depends on the refractive index. If we consider this change in amplitude sign in terms of phase it corresponds to a 𝜋 𝑜𝑟 180° phase shift (from positive to negative amplitude). This phase shift is the equivalent of 𝜆/2 in wavelength. So from a reflection we gain a phase shift of: 0 𝑖𝑓 (𝑛𝑎 > 𝑛𝑏 ) 5.15 𝜆/2 𝑖𝑓(𝑛𝑎 < 𝑛𝑏 ) This additional phase shift from the reflection needs to be added to that from the optical path length difference before we can work out interference. Of course if both reflections are of the same type these will cancel and will not contribute to the phase difference but it will if they are of different type they will. So if both reflections are of the same type: For constructive: 2𝑡 = 𝑚𝜆 (𝑚 = 0, ±1, ±2... ) 5.16 For destructive: 1 5.17 2𝑡 = (𝑚 + ) 𝜆 (𝑚 = 0, ±1, ±2... ) 2 As expected from Equations 5.1 and 5.2. However if the two reflections are of different type, and hence have a 𝜆/2 phase shift between them: For constructive: 1 5.18 2𝑡 = 𝑚 (1 + ) 𝜆 (𝑚 = 0, ±1, ±2... ) 2 For destructive: 2𝑡 = 𝑚𝜆 (𝑚 = 0, ±1, ±2... ) 5.19 Which is the reverse. C: Thin and Thick Films In the previous section it was emphasised that interference was to be expected in reflections from thin films. Where does this emphasis on “thin” come from and what would be the difference when using a “thick” film? The answer comes from the previously mentioned concept of coherence length. To allow for interference the two rays involved must be coherent, which is to say they need to have the same frequency and maintain a definite and constant phase relationship. B27FF – Waves and Matter Waves Week 5 Richard Carter page 19 of 23 The details of coherence length are somewhat complex and so we will consider a somewhat simplified version here. Light emitted from any source can be considered to be emitted in short “bursts” of radiation. Light within the burst has a constant frequency and phase relationship and is therefore coherent with itself. Quantum mechanical effects mean that adjacent bursts will not be coherent with one another due to a shift in phase and/or minor fluctuations in frequency. So for interference to be seen the path length difference needs to be less than the length of these bursts – this is the coherence length. The effective length of these bursts, and therefore the coherence length, depends on the bandwidth of the light source, where the bandwidth is a measure of the range (width) of frequencies emitted. A monochromatic light source has a low bandwidth and hence a large coherence length – possibly as long as m or km. A white light source has a high bandwidth and hence a short coherence length, in the order of a few 𝜇𝑚. Therefore when illuminated by a white light (like the sun) our film must be “thin” – of the order of micrometres – to allow for the short coherence length of the source (Figure 5.14). This is why you do not see an interference effect from a glass window, the reflections are present but the path length difference on a mm thick glass plate is too large. Notably if we used a much longer coherence length source (like a laser) we can still see interference with much thicker films even up to cm or m thick “films”. B27FF – Waves and Matter Waves Week 5 Richard Carter page 20 of 23 Figure 5.14 The effect of coherence length on interference from thin and thick films. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1197, 2015) D: Newton’s Rings Interference fringes have several useful applications in metrology since they effectively provide an optical “ruler”. One application is to compare two surfaces by using interference to measure the gap between them. If we use a reference flat surface (Figure 5.15) this would allow us to measure the curvature of a lens by creating a series of interference rings. These are centred on the point the two pieces of glass touch and each one represents an additional 𝜆/2 gap between the two pieces of glass. B27FF – Waves and Matter Waves Week 5 Richard Carter page 21 of 23 Figure 5.15 Illustration of Newton’s rings. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1199, 2015) This sort of interference is referred to as Newtons Rings (named after Isaac Newton) and can be seen whenever two differently curved surfaces4 with reflections are within the coherence length of the light source. A variation on this method is to use a “Master” with a specific shape to compare to a manufactured piece. This is common for the manufacture of optics. A perfectly made piece will perfectly fit the master and no fringes will be seen. If the optic is not perfect then the fringes will show where, and by how much the manufactured piece differs from the master (Figure 5.16). Figure 5.16 Example of interference used to gauge the accuracy of an optic. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1199, 2015) 4 a flat is a curve with infinite radius in this context B27FF – Waves and Matter Waves Week 5 Richard Carter page 22 of 23 E: Coatings For the final section in this part of the course we can consider another useful use for thin film interference. High and low reflectivity coatings. The most common use for this technique is in antireflection or nonreflective coatings on optical surfaces but the opposite can be used to create dielectric mirrors (i.e. a mirror that does not use a metal). These coatings are made by placing a specific thickness of thin, hard5, transparent material onto a piece of glass. In the case of an anti-reflection coating this is chosen to have a refractive index between air and the glass. In this way both reflections from the thin film will have the same phase relation. If the film is made to have an optical thickness (i.e. 𝑛 × 𝑡 in this case) of exactly 𝜆/4 then the second reflection will be 𝜆/2 or exactly half a cycle out of phase and the two reflections will destructively interfere with one another (Figure 5.17). The result is that no amplitude is reflected and hence the reflected intensity is zero – i.e. there is no reflection. Figure 5.17 Illustration of an antireflection coating. (Sears and Zemansky University Physics with Modern Physics, Pearson Education, p1199, 2015) If the light is incident on the glass at higher angles then the optical path length difference will be slightly higher than 𝜆/4 and the antireflection effect will not be perfect. This error is more significant for thicker films and it is for this reason that 𝜆/4 is preferred instead of higher orders like 1 1 (2 + 4) 𝜆, 𝑜𝑟. The same is true if the material is dispersive and the coating is intended to work for more than one wavelength. Clearly the film can be exactly 𝜆/4 for only one wavelength and there will be a similar optical path length errors for other wavelengths. The thinner the film is the wider the range of wavelengths that can have acceptable levels of antireflection. This is the reason that coatings on 5 Hard to stop it scratching or wearing off. B27FF – Waves and Matter Waves Week 5 Richard Carter page 23 of 23 things like binoculars look purple (a combination of red and blue). The visual spectrum is centred around yellow-green (550 nm) and the coating is designed to work best at this wavelength. When considering high-reflective coatings, we need a more complex arrangement. The refractive index contrast between air-film and film-air will determine the amount of light reflected. Typically, this is in the region of 4% for each interface. So for an uncoated piece of glass, we expect 4% reflection, not enough to really use it as a mirror. Adding a coating 𝜆/2 thick (so the reflections constructively interfere) with a refractive index between air and glass might give 6% (not as high as two glass-air interfaces as the refractive index contrast is not as high). We can add more layers each with a different refractive index, but this will not really work as each layer we add will further decrease the refractive index contrast between layers (they all need to be in order from the index of air to the index of glass). Instead we will want to alternate high and low refractive index layers of 𝜆/4 thickness. This will ensure all the reflections are in phase and we can stack as many 4% reflections as we like. By using different types of material with different dispersions it is possible for these stacks of layers to work across reasonably broad ranges of wavelengths or even angles, but this is clearly more challenging, and therefore expensive, to manufacture than the antireflection coatings. Summary We have completed our look at polarisation by considering circular and elliptical polarisation states. The effects of birefringence and considered optical scattering. We have completed our model for two source interference and considered how this can work with an optical system using secondary sources like the Young’s Slits experiment. We have considered different forms of thin film interference and derived equations explaining where we expect to see constructive and destructive interference. We have also looked at the boundary conditions for reflections from an interface and seen that we can obtain either a 0 or 𝜋 phase shift from an interface. Finally we have looked at the concept of coherence length and its effect (in optical systems on interference). This concludes the Waves part of the course. The next topic (after consolidation week) will be on Matter.