Presentation on theme: "Interference of Light Waves"— Presentation transcript:
1 Interference of Light Waves Chapter 3Interference of Light Waves
2 To observe interference in light wave, conditions are: The sources must be coherent that is, they must maintain a constant phase with respect to each other.The sources must be monochromatic that is of a single wavelength.The superposition principle must apply.
3 Huygen’s Principle, Wavefronts and Coherence Examples of coherence are:Laser lightSmall spot on tungsten filamentWavefrontkMost light is incoherent:Two separate light bulbsTwo headlight beams on a carSun is basically incoherent
4 Interference is the combination of two or more waves to form a composite wave, based on the principle of superposition
5 Out of Phase by 180 degrees or p radians or l/2 In betweenIn Phase
6 Young’s Double Slit Experiment, The narrow slits, S1 and S2 act as sources of wavesThe waves emerging from the slits originate from the same wave front and therefore are always in phase
7 Resulting Interference Pattern The light from the two slits form a visible pattern on a screenThe pattern consists of a series of bright and dark parallel bands called fringesConstructive interference occurs where a bright fringe appearsDestructive interference results in a dark fringe
8 Fringe PatternThe fringe pattern formed from a Young’s Double Slit Experiment would look like thisThe bright areas represent constructive interferenceThe dark areas represent destructive interference
9 Interference Patterns Constructive interference occurs at the center pointThe two waves travel the same distanceTherefore, they arrive in phase
10 The upper wave has to travel farther than the lower wave The upper wave travels one wavelength fartherTherefore, the waves arrive in phaseA bright fringe occurs
11 The upper wave travels one-half of a wavelength farther than the lower wave The trough of the bottom wave overlaps the crest of the upper waveThis is destructive interferenceA dark fringe occurs
12 Interference Equations The path difference, δ, is found from the tan triangleδ = r2 – r1 = d sin θThis assumes the paths are parallelNot exactly parallel, but a very good approximation since L is much greater than d
13 For a bright fringe, produced by constructive interference, the path difference must be either zero or some integral multiple of the wavelengthδ = d sin θbright = m λm = 0, ±1, ±2, …m is called the order numberWhen m = 0, it is the zeroth order maximumWhen m = ±1, it is called the first order maximum
14 The positions of the fringes can be measured vertically from the zeroth order maximum y = L tan θ L sin θAssumptionsL>>dd>>λApproximationθ is small and therefore the approximation tan θ sin θ can be used
15 When destructive interference occurs, a dark fringe is observed This needs a path difference of an odd half wavelengthδ = d sin θdark = (m + ½) λm = 0, ±1, ±2, …
16 Interference Equations, final For bright fringesFor dark fringes
17 Intensity distribution of the double-slit interference pattern The total electric field intensity at point P on the screen is the vector superposition of the two waves.
18 We suppose the two slits represent coherent sources of sinusoidal waves. Hence, the waves have the same angular frequency ω and a constant phase difference φ .
19 Assuming the two waves have the same amplitude Eo , the electric field intensities at P due to each wave separately is :E1 = Eo sin ω t ,andE2 = Eo sin ( ω t + φ ) Although the waves are in phase at the slits, their phase difference φ at P depends on the path differenceδ = r2 - r1 = d sin θ .
20 Because a path difference of λ (constructive interference) corresponds to a phase difference of 2 π rad,a path difference of λ/2 (destructive interference) corresponds to a phase difference of π rad, we obtain the ratio:
21 Ep = 2 Eo cos (φ/2) sin (ω t + φ/2) Using the superposition principle we can obtain the resultant electric field at PEp = E1 + E2 = Eo [sin ω t + sin (ω t + φ)]sin A+sin B = 2 sin[(A + B) / 2] cos[(A – B)/2]Taking A = ω t + φ , and B = ω t,Ep = 2 Eo cos (φ/2) sin (ω t + φ/2)the electric field at P has the same frequency as the light at the slits, but its amplitude is multiplied by the factor:2 cos (φ /2).
22 the intensity of a wave is proportional to the square of the resultant electric field at that point. I α Ep2= 4 Eo2 cos2 (φ /2) sin2 (ω t + φ/2)the time average value of sin2 (ω t + φ/2) over one cycle is 1/2,Iav = Io cos2 ( φ / 2)where lo is the maximum possible time average light intensity
23 Iav = Io cos2 ( π d sin θ /λ ) since sinθ ≈ y/ L for small values of θIav = Io cos2 ( π yd /λ L) Constructive interference, which produces intensity maxima, occurs when the quantity( π yd / λ L) is an integral multiple of π , corresponding to :y = ( λ L / d) m.This is consistent with the equation of ybright .
24 Intensity distribution versus d sin θ or the double-slit pattern when the screen is far from the two slits (L » d).
25 Interference in Thin Films Interference effects are commonly observedin thin filmsExamples are soap bubbles and oil on waterThe interference is due to the interaction of the waves reflected from both surfaces of the film
26 An electromagnetic wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1There is no phase change in the reflected wave if n2 < n1The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum
27 Ray 1 undergoes a phase change of 180° with respect to the incident ray Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave
28 Ray 2 also travels an additional distance of 2t before the waves recombine For constructive interference2nt = (m + ½ ) λ m = 0, 1, 2 …This takes into account both the difference in optical path length for the two rays and the 180° phase changeFor destruction interference2 n t = m λ m = 0, 1, 2 …
29 Two factors influence interference Possible phase reversals on reflectionDifferences in travel distanceThe conditions are valid if the medium above the top surface is the same as the medium below the bottom surfaceIf the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversed
30 Be sure to include two effects when analyzing the interference pattern from a thin film Path lengthPhase change
31 Newton’s RingsAnother method for viewing interference is to place a planoconvex lens on top of a flat glass surfaceThe air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness tA pattern of light and dark rings is observedThis rings are called Newton’s RingsThe particle model of light could not explain the origin of the ringsNewton’s Rings can be used to test optical lenses
32 The interference effect is due to the combination of ray l, reflected from the flat plate, with ray 2, reflected from the lower part of the lens.Imperfection in lens
33 Ray 1 undergoes aphase change of 180oupon reflection,because it is reflected from a medium of higher refractive index, whereas ray 2 undergoes no phase changethe conditions for constructive and destructive interference are given, with n = 1 because the film is air.Point O is dark, because ray 1, reflected from the flat surface, undergoes a 180o phase change with respect to ray 2
34 Expressions for the radii of the bright and dark bands can be obtained in terms of the radius of curvature R and wavelength λ.For example, the dark rings have radii given by: r ≈ √ m λ / n .By measuring the radii of the rings, the wavelength can be obtained, provided R is known.Conversely, if the wavelength is accurately known, this effect can be used to obtain R.
35 The Michelson Interferometer A beam of light provided by a monochromatic source is split into two rays by a partially silvered mirror, M, inclined at 45o to the incident light beam. One ray is reflected vertically upward toward mirror M2, while the second ray is transmitted horizontally through M toward mirror M2.
36 Hence, the two rays travel separate paths L1 and L2. After reflecting from M1 and M2, the two rays eventually recombine to produce an interference pattern, which can be viewed through a telescope.The interference condition for the two rays is determined by their path length differences.When the two mirrors are exactly perpendicular to each other, the interferencepattern is a target pattern of bright and dark circular fringes, similar to Newton’s rings.
37 As M1 is moved, the fringe pattern collapses or expands, depending on the direction in which M1 is moved.For example, if a dark circle appears at the center of the target pattern (corresponding to destructive interference) and M1 is then moved a distance λ /4 toward M0, the path difference changes by λ /2. What was a dark circle at the center now becomes a bright circle.
38 As M1 is moved an additional distance &/4 toward M0, the bright circle becomes a dark circle again. Thus, the fringe patternshifts by one-half fringe each time M1 is moved a distance λ /4. The wavelength of light is then measured by counting the number of fringe shifts for a given displacement of M1.If the wavelength is accurately known, mirror displacements can be measured to within a fraction of the wavelength.
39 Anti-Reflecting Coatings We can apply a thin film of a material with the proper thickness so that it will cause destructive interference of reflected light.Thus all the light must be transmitted through the film into the lens.
40 Problem Solving Strategy with Thin Films Identify the thin film causing the interferenceDetermine the indices of refraction in the film and the media on either side of itDetermine the number of phase reversals: zero, one or two
41 The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λThe conditions are reversed if one of the waves undergoes a phase change on reflection
42 Problem Solving with Thin Films Equation1 phase reversal0 or 2 phase reversals2nt = (m + ½) lconstructivedestructive2nt = m l
43 Example An example of different indices of refraction A coating on a solar cellThere are two phase changes
44 CD’s and DVD’s Data is stored digitally A series of ones and zeros read by laser light reflected from the diskStrong reflections correspond to constructive interferenceThese reflections are chosen to represent zerosWeak reflections correspond to destructive interferenceThese reflections are chosen to represent ones
45 CD’s and Thin Film Interference A CD has multiple tracksThe tracks consist of a sequence of pits of varying length formed in a reflecting information layerThe pits appear as bumps to the laser beamThe laser beam shines on the metallic layer through a clear plastic coating
46 Reading a CDAs the disk rotates, the laser reflects off the sequence of bumps and lower areas into a photodectorThe photodector converts the fluctuating reflected light intensity into an electrical string of zeros and onesThe pit depth is made equal to one-quarter of the wavelength of the light
47 When the laser beam hits a rising or falling bump edge, part of the beam reflects from the top of the bump and part from the lower adjacent areaThis ensures destructive interference and very low intensity when the reflected beams combine at the detectorThe bump edges are read as onesThe flat bump tops and intervening flat plains are read as zeros
48 DVD’s DVD’s use shorter wavelength lasers The track separation, pit depth and minimum pit length are all smallerTherefore, the DVD can store about 30 times more information than a CD