# 4 Waves G482 Electricity, Waves & Photons 4 Waves G482 Electricity, Waves & Photons 2.4.1 Wave Motion 2.4.1 Wave Motion Mr Powell 2012 Index 2.4.2. EM.

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4 Waves G482 Electricity, Waves & Photons 4 Waves G482 Electricity, Waves & Photons 2.4.1 Wave Motion 2.4.1 Wave Motion Mr Powell 2012 Index 2.4.2. EM Waves 2.4.2. EM Waves 2.4.3 Interference Part A p150-151 Part B p152-157 2.4.3 Interference Part A p150-151 Part B p152-157 2.4.4 Stationary Waves 2.4.4 Stationary Waves 1

Mr Powell 2012 Index Trig Series Trigonometric functionsTrigonometric functions can be expanded in power series, which facilitates approximations of the functions in extreme cases. The angle x must be in radians.power series radians

Mr Powell 2012 Index Small Angle Approximation... One of the most important applications of trigonometric series is for situations involving very small angles.trigonometric series For such angles, the trigonometric functions can be approximated by the first term in their series. This gives the useful small angle approximations: Examples of the use of the small angle approximation are in the calculation of the period of a simple pendulum, and the calculation of the intensity minima in single slit diffraction.simple pendulumsingle slit diffraction This approximation is used in most of the common expressions of geometrical optics which are built on the concept of surface power for lenses. This approximation sin x = x reaches a 1% error at about 14 degrees

Mr Powell 2012 Index Maths puzzling.... Construct an excel spreadsheet to explore and prove the concept that for the situation where  <10 and measured in radians; DegreesTheta RadSin ThetaDifference 00.00 10.02 0.00 20.03 0.00 30.05 0.00 40.07 0.00 50.09 0.00 60.10 0.00 70.12 0.00 80.14 0.00 90.16 0.00 100.17 0.00 200.350.34-0.01 300.520.50-0.02 400.700.64-0.06 500.870.77-0.11 601.050.87-0.18 701.220.94-0.28 801.400.98-0.41

Mr Powell 2012 Index Practical Skills are assessed using OCR set tasks. The practical work suggested below may be carried out as part of skill development. Centres are not required to carry out all of these experiments.  Students should gain a qualitative understanding of superposition effects together with confidence in handling experimental data.  Students should be able to discuss superposition effects and perform experiments leading to measurements of wavelength and wave velocity.  Use an oscilloscope to determine the frequency of sound.  Observe polarising effects using microwaves and light.  Investigate polarised light when reflected from glass or light from LCD displays.  Study diffraction by a slit using laser light.  Study hearing superposition using a signal generator and two loudspeakers.  Study superposition of microwaves.  Determine the wavelength of laser light with a double-slit.  Determine the wavelength of light from an LED using a diffraction grating.  Demonstrate stationary waves using a slinky spring, tubes and microwaves.  Determine the speed of sound in air by formation of stationary waves in a resonance tube.

Mr Powell 2012 Index 2.4.3 Interference (Part B) Assessable learning outcomes... (h) describe the Young double-slit experiment and explain how it is a classical confirmation of the wave-nature of light (HSW 1); (i) Select and use the equation = ax/D for electromagnetic waves; (j) describe an experiment to determine the wavelength of monochromatic light using a laser and a double slit (HSW 1); (k) describe the use of a diffraction grating to determine the wavelength of light (the structure and use of a spectrometer are not required); (l) select and use the equation dsinθ = nλ; (m) explain the advantages of using multiple slits in an experiment to find the wavelength of light. Assessable learning outcomes... (h) describe the Young double-slit experiment and explain how it is a classical confirmation of the wave-nature of light (HSW 1); (i) Select and use the equation = ax/D for electromagnetic waves; (j) describe an experiment to determine the wavelength of monochromatic light using a laser and a double slit (HSW 1); (k) describe the use of a diffraction grating to determine the wavelength of light (the structure and use of a spectrometer are not required); (l) select and use the equation dsinθ = nλ; (m) explain the advantages of using multiple slits in an experiment to find the wavelength of light.

(h) describe the Young double-slit experiment and explain how it is a classical confirmation of the wave-nature of light (HSW 1);  What is the general condition for the formation of a bright fringe?  Are Young’s fringes equally spaced?  What factors could be… a) increased b)decreased, to increase the fringe spacing? (i) Select and use the equation = ax/D for electromagnetic waves; (h) describe the Young double-slit experiment and explain how it is a classical confirmation of the wave-nature of light (HSW 1);  What is the general condition for the formation of a bright fringe?  Are Young’s fringes equally spaced?  What factors could be… a) increased b)decreased, to increase the fringe spacing? (i) Select and use the equation = ax/D for electromagnetic waves; a = s= slit spacing separation D = distance to screen x or w = fringe width = wavelength NB: all distances in “m” 7

Mr Powell 2012 Index x Maxima to Maxima a = slit spacing separation D = distance to screen x = fringe width = wavelength

Mr Powell 2012 Index Change of colour on fringe separation 9 a = slit spacing separation D = distance to screen x = fringe width = wavelength ROY-G-BIV Blue to shortest Red is longest

Mr Powell 2012 Index Two emissions of coherent light, interfere.... a = s = slit spacing separation D = distance to screen x or w = fringe width = wavelength

13.4 Double slit interference 11

From one source and two gaps 1 st bright fringe 1 st bright fringe central fringe - the closer the slits become the wider the fringe separation. - waves of short wavelength give narrow fringe separation because blue light diffracts less. 12

Path difference Phase difference 13

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W First bright fringe Central fringe screen Double Slit interference - theory 15

W First bright fringe Central fringe screen Double Slit interference - theory Ə D = wavelength = 600nm D = slit to screen distance = 2m a = slit separation (or s) = 0.5 mm x = fringe width (or w) = 1mm 16

For constructive interference at the screen: ( ie a bright fringe ) Wave fronts from S1 and S2 must arrive at the screen in phase with a path difference of a whole number of wavelengths For destructive interference at the screen: ( ie a dark fringe ) Wave fronts from S1 and S2 must arrive out of phase with a path difference of half a wavelength 17

For constructive interference at the screen: ( ie a bright fringe ) Wave fronts from S1 and S2 must arrive at the screen in phase with a path difference of a whole number of wavelengths For destructive interference at the screen: ( ie a dark fringe ) Wave fronts from S1 and S2 must arrive out of phase with a path difference of half a wavelength 18

soh cah toa 19

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Mr Powell 2012 Index Focus on Coherence… Double slits and light are a coherent source as they emit light waves of a constant frequency with a constant phase difference. (same as water) so you get a regular pattern. Two light bulbs don’t produce the same effect as the emission of the light is random and the phase difference changes depending on the times when the waves interfere.

Mr Powell 2012 Index For the double Slit experiment the fringe spacing depends on the wavelength used (see formulae) x or w . Using white light, fringes appear from all the various wavelengths present and do not overlap exactly, hence coloured fringes. (laser produce more collimated dots) Inner fringes are tinged with blue on the inside and red on the outside (red diffracted more as longer wavelength). Where all the colours overlap it produces white light. λ red » λ blue White Light Fringes

Mr Powell 2012 Index Exam Question…. single slit then double slits to the right (1) single slit and double slits labelled (2) Single slit Double slit

Mr Powell 2012 Index Exam Question

Mr Powell 2012 Index = wavelength D = distance slit to screen a = slit separation x = fringe width 5 fringes = D = 3.1m a = 0.5 x 10 -3 m 5x = 20.65 x 10 -3 m slit thickness = 0.15mm The best way to look at this phenomenon is to try some readings. A laser is projected through a double slit onto a screen. Here are the findings… Task: use my findings to check the wavelength of the light...... 20.65 x 10 -3 m (j) describe an experiment to determine the wavelength of monochromatic light using a laser and a double slit (HSW 1);

Mr Powell 2012 Index Single Slit OR Double Slit? (Extension) 1.A diffraction pattern formed by a real double slit. The width of each slit is much bigger than the wavelength of the light. This is a real photo. 2.This idealised pattern is not likely to occur in real life. To get it, you would need each slit to be comparable in size to the wavelength of the light, but that's not usually possible. This is not a real photo. 3.A real photo of a single-slit diffraction pattern caused by a slit whose width is the same as the widths of the slits used to make the top pattern.

Mr Powell 2012 Index Single Slit – Diffraction (Extension) This is an attempt to more clearly visualise the nature of single slit diffraction. The phenomenon of diffraction involves the spreading out of waves past openings which are on the order of the wavelength of the wave.single slit diffraction diffraction

Mr Powell 2012 Index Slit Width (Extension) One of the characteristics of single slit diffraction is that a narrower slit will give a wider diffraction pattern as illustrated in the images.

Mr Powell 2012 Index Single Slit Diffraction…(Extension) Slit of width equal (a) to wavelength of an incident plane wave. Slit of width (a) equal to five times the wavelength of an incident plane wave. = wavelength D = distance slit to screen a = slit separation x = fringe width

Mr Powell 2012 Index a = λ Semi circular wave fronts a = 2 λ First minima & maxima become visible a = 4 λ Diffraction is the spreading of wavefronts around corners and obstacles. If the slit gets narrower diffraction increases. If the wavelength increases diffraction increases. Or can view as a water wave (single slit diffraction)

Mr Powell 2012 Index What happens…..

Mr Powell 2012 Index What happens when changes… Longer so the x is larger Shorter so x is smaller

Mr Powell 2012 Index What happens when slit width changes… Larger a so the x is smaller smaller a so the x is larger

(k) describe the use of a diffraction grating to determine the wavelength of light (the structure and use of a spectrometer are not required); (l) select and use the equation dsinθ = nλ; (m) explain the advantages of using multiple slits in an experiment to find the wavelength of light. (k) describe the use of a diffraction grating to determine the wavelength of light (the structure and use of a spectrometer are not required); (l) select and use the equation dsinθ = nλ; (m) explain the advantages of using multiple slits in an experiment to find the wavelength of light. n = order of diffraction i.e. 0,1,2,3... = wavelength in m d = 1/lines per mm or spacing in m Sin  = angle deviation from 0  35

Mr Powell 2012 Index HUYGEN’s CONSTRUCTION FOR A PLANE WAVEFRONT Original wavefront Secondary sources Subsequent position of wavefront “Every point on a wavefront acts as a source of secondary waves which travel with the speed of the wave. At some subsequent time the envelope of the secondary waves represents the new position of the wavefront.”

Mr Powell 2012 Index Application to Grating…. n=0 Each slit effectively acts as a point source, emitting secondary wavelets, which add according to the principle of superposition n=2 n=1 n=1 corresponds to a path difference of one wavelength n=2 corresponds to a path difference of two wavelengths n=3 corresponds to a path difference of three wavelengths

Mr Powell 2012 Index Path differences…. Because the screen is so far away from the slits which are small. We consider the wave fronts to be parallel. Each slit adds another wavelength to the pattern. For formula creates another maximum each time n = 1,2,3,…..

Mr Powell 2012 Index If we consider the triangle as shown in the zoomed in diagram we can see that an angle  can be worked out from knowing QP and QY and then similar triangles rule used to find out the angle of the rays as they appear on a screen. We can use the idea of constructive interference and take the idea that from trigonometry. However, we also know that the length QY should be the length of a whole wavelength and also that QP is the slit spacing….. Light through a diffraction grating...

Mr Powell 2012 Index We can also now think about the idea that this happens not only once but several times or orders. Each time we have a whole wavelength we get the construction of a new wave front or “n” wave front. Our formulae can be changed to; Light through a diffraction grating... d = slit spacing (m) or 1/lines per m  = angle from normal n = order of diffraction = wavelength of the light (m) NB: max number of orders is given by n = d/ where sin(90) = 1

Mr Powell 2012 Index Practical Setup a laser light source and use the formula below to check the wavelength of the light. Your quoted value is 632.8nm for red (632.8 x 10 -9 m). Your teacher will give you a slit which has 300 lines per mm or a spacing d = 3.33 x 10 -6 m. Use this one at first to check the calculation. Then try and take some other readings for 100, 600 and repeat with the green laser beam. d = slit spacing  = angle from normal n = order of diffraction = wavelength of the light

Mr Powell 2012 Index 42

Mr Powell 2012 Index Example Results These results were obtained using 100 lines/mm grating and small handheld laser levels. The dots were projected onto a screen 1.735m away and the distances measured using a tape measure (+/-0.1cm) and ruler (+/-1mm). HSW Task: Can you work out an overall estimated error for the experiment?

Mr Powell 2012 Index Exam Practice… (2 marks) d

Mr Powell 2012 Index Exam Practice… (2 marks) d d = slit spacing  = angle from normal n = order of diffraction = wavelength of the light 1.Angle will decrease (get smaller) if decreases. sin   2.Path difference must be smaller as is less so angle must also be less for constant d sin  

Mr Powell 2012 Index Laser Diffraction Extension! TASK: Can you work out the spacing of lines on a CD if light is 632.8 nanometres. (or similar) L = 229.5cm Opp = 55.5cm 57.1cm 58.3cm (three repeats 2 nd order) Model = 1 x 10 -6 m 1.6 microns,dvd is.74 microns Experimental 5.4 microns

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