2BE-PHYSICS- INTERFERENCE-2010-11 TopicsTwo source interferenceDouble-slit interferenceCoherenceIntensity in double slit interferenceInterference from thin filmMichelson’s InterferometerText Book:PHYSICS VOL 2 by Halliday, Resnick and Krane (5th Edition)MIT-MANIPALBE-PHYSICS- INTERFERENCE
3What is an Electromagnetic wave (EM)? 900Electric field (E)Magnetic field (B)The electromagnetic waves consist of the electric and magnetic field oscillations.In the electromagnetic waves, electric field is perpendicular to the magnetic field and both are perpendicular to the direction of propagation of the waves.
4Radio and television Waves Properties of electromagnetic waves (EM)Electromagnetic waves are non-mechanical waves i.e they do not require material medium for propagation.They are transverse waves. ie. They travel in the form of ‘crests’ and ‘troughs’.Examples:Light wavesHeat wavesRadio and television WavesUltraviolet wavesGamma rays, X- rays
5They differ from each other in wavelength (λ) and frequency (f). In vacuum, all electromagnetic waves (EM) move at the same speed and differ from one another in their frequency (f).Speed=c=Frequency x wavelengthi.e c= f x λc= 3 x 108 m/s
6Electromagnetic spectrum NameFrequency range (Hz)Wavelength rangeGamma rays(γ-rays)5 x x 1019nmX-rays3 x x 10160.1 nm-30nmUltraviolet light1 x x 101430nm-400nmVisible light8 x x 1014400nm-800nmInfra-red4 x x 1013800nm-30000nmRadio frequencies3 x x 104nmMore frequency (f) more energy (E), and lesser wavelength(λ).
7Dual Nature of LightAlbert Einstein proposed that light not only behaves as a wave, but as a particle too.Light is a particle in addition to a wave-Dual nature of light.Light as a stream of particlesDual nature of light -treated as1) a wave or2) as a particleDual nature of light successfully explains all the phenomena connected with light.
8When light behaves as a Wave? fThe wave nature of light dominates when light interacts with light.The wave nature of light explains the following properties of light:Refraction of lightReflection of lightInterference of lightDiffraction of lightPolarization light
9When light behaves as a stream of particles? The particle nature of light dominates when the light interacts with matter (like solids, liquids and gases).Particle nature - Photoelectric, Compton Effect, Black body radiation..Light as a stream of particlesQuantum (bundles/packets of energy)(Photon/quantum)The light is propagated in bundles of small energy, each bundle being called a quantum.Each quantum is composed of many small particles called quanta or photon.Photon energyE = hfh = Planck’s constant = 6.626x10-34Jsf = frequency of radiation
10Energy of a photon or light wave: Light as a wave:c= f Light as a particle:E = hfphotonEnergy of a photon or light wave:Where h = Planck’s constant = 6.626x10-34Jsf = frequency of a light wave -c = velocity of lightλ= wavelength of a light wave -distance between successive crests
11Visible Light400–435 nm Violet435 nm-440nm Indigo440–480 nm Blue480–530 nm Green530–590 nm Yellow590–630 nm Orange630–700 nm RedThe color of visible light is determined by its wavelength.White light is a mixture of all colors.We can separate out individual colors with a prism.1 nm = 10^-9 m
12Wave Function of Sinusoidal Waves y(x,t) = ym sin(kx-wt)ym: amplitudekx-wt : phasek: wave numberw: angular frequency
13PRINCIPLE OF SUPERPOSITION When two waves traveling almost in the same direction superpose, the resulting displacement at a given point is the algebraic sum of the individual displacements.i.e. when waves, y1=A sin ωt & y2=A sin (ωt + ) superpose, the resultant displacement isy= y1+y2= a sin (ωt) + a sin (ωt+)
15TWO-SSOURCE INTERFERENCE When identical waves from two sources overlap at a point in space, the combined wave intensity at that point can be greater or less than the intensity of either of the two waves. This effect is called interference.ORWhen two waves of same frequency (or wavelength) with zero initial phase difference or constant phase difference superimpose over each other, then the resultant amplitude (or intensity) in the region of superposition is different than the amplitude (or intensity) of individual waves.
16At certain points either two crests or two troughs interact giving rise to maximum amplitude resulting in maximum intensity.(Constructive interference).At certain points a crest and a trough interact giving rise to minimum or zero amplitude resulting in minimum or zero intensity.(Destructive interference).
17TWO-SOURCE INTERFERENCE Constructive InterferenceDestructive interferenceTwo waves (of the same wavelength) are said to be in phase if the crests (and troughs) of one wave coincide with the crests (and troughs) of the other. The net intensity of the resultant wave is greater than the individual waves.(Constructive interference). If the crest of one wave coincides with the trough of the second, they are said to be completely out of phase. The net intensity of the resultant wave is less than the individual waves. (Destructive interference).
18TWO-SOURCE INTERFERENCE INTERFERENCE PATTERN PRODUCED BY WATER WAVES IN A RIPPLE TANKMaxima: where the shadows show the crests and valleys(or troughs).Minima: where the shadows are less clearly visibleMIT-MANIPALBE-PHYSICS- INTERFERENCE
19Path difference corresponds to phase difference of 2. PHASE AND PATH DIFFERENCEPhase: Phase of a vibrating particle at any instant indicates its state of vibration.BFλπ/23π/22ππOAC3λ/4EGt=0λ/4λ/2λλDPhase may be expressed in terms of angle as a fraction of 2π.Path difference corresponds to phase difference of 2.
21Constructive interference Maximal constructive interference of two waves occurs when their:path difference between the two waves is a whole number multiple of wavelength.ORPhase difference is 0, 2, 4 , … (the waves are in-phase).
23Complete destructive interference of two waves occur when the path difference between the two waves is an odd number multiple of half wavelength.Orthe phase difference is , 3, 5, … (the waves are 180o out of phase).
24Coherence – necessary condition for interference to occur. Two waves are called coherent when they are of :same amplitudesame frequency/wavelengthsame phase or are at a constant phase difference
25A SECTION OF INFINITE WAVE COHERENCEA SECTION OF INFINITE WAVEA WAVE TRAIN OF FINITE LENGTH LFor interference pattern to occur, the phase difference at point on the screen must not change with time. This is possible only when the two sources are completely coherent.No two independent sources can act as coherent sources, because the emission of light by the atoms of one source is independent of that the other.If the two sources are completely independent light sources, no fringes appear on the screen (uniform illumination). This is because the two sources are completely incoherent.
26A SECTION OF INFINITE WAVE COHERENCEA SECTION OF INFINITE WAVEA WAVE TRAINOF FINITE LENGTH LCommon sources of visible light emit light wave trains of finite length rather than an infinite wave.The degree of coherence decreases as the length of wave train decreases.MIT-MANIPALBE-PHYSICS- INTERFERENCE
27BE-PHYSICS- INTERFERENCE-2010-11 COHERENCELaser light is highly coherent whereas a laboratory monochromatic light source (sodium vapor lamp) may be partially coherent.Common sources of visible light emit light wave trains of finite length rather than an infinite wave.MIT-MANIPALBE-PHYSICS- INTERFERENCE
28COHERENCE Methods of producing coherent sources: (1). Division of wave front: In this method, the wave front is divided into two or more parts with the help of mirrors, lenses and prisms.The common methods are:Young’s double slit arrangement, b. Lloyd's single mirror method.(2) Division of Amplitude: In this method, the amplitude of the incoming beam is divided into two or more parts by partial reflection with the help of mirrors, lenses and prisms.These divided parts travel different paths and finally brought together to produce interference.The common methods areNewton’s rings, b. Michelson’s interferometer.
29DOUBLE SLIT INTERFERENCE If light waves did not spread out after passing through the slits, no interference would occur..Two narrow slits (can be considered as two sources of coherent light waves).If the widths of the slits are small compared with the wavelength distance a (<<) - the light waves from the two slits spread out (diffract) – overlap- produce interference fringes on a screen placed at a distance ‘D’ from the slits.dScreen
30DOUBLE SLIT INTERFERENCE A monochromatic light source produces two coherent light sources by illuminating a barrier containing two small openings (slits) S1 and S2 separated by a distance ‘d’ and kept at a distance ‘D’ from the screen.Waves originating from two coherent light sources S1 and S2 because maintain a constant phase relationship.In interference phenomenon, we have assumed that slits are point sources of light.BC
31DOUBLE SLIT INTERFERENCE When the light from S1 and S2 both arrive at a point on the screen such that constructive interference occurs at that location, a bright fringe appears.When the light from the two slits combine destructively at any location on the screen, a dark fringe results.BC
32DOUBLE SLIT INTERFERENCE At center of the screen O, we always get a bright fringe, because at this point the two waves from slit S1 and S2, interfere constructively without any path/phase difference.32
33DOUBLE SLIT INTERFERENCE - Analysis of Interference patterna-is the mid point of the slitWe consider waves from each slit that combine at an arbitrary point P on the screen C.The point P is at distances of r1 and r2 from the narrow slits S1 and S2, respectively.
34For D>>d, we can approximate rays r1 and r2 as being parallel. DOUBLE SLIT INTERFERENCEFor D>>d, we can approximate rays r1 and r2 as being parallel.The line S2b is drawn so that the lines PS2 and Pb have equal lengths.Path length (S1b) between the rays r1 and r2 reaching the point P decides the intensity at P.(i.e. maximum/minimum).
35DOUBLE SLIT INTERFERENCE Path difference between two waves from S1 & S2 (separated by a distance ‘d’) on reaching a point P on a screen at a distance ‘D’ from the sources is d sin .The path difference (S1b=d sin) determines whether the two waves are in phase or out of phase when they arrive at point P.If path length (S1b= d sinθ) is either zero or some integer multiple of the wavelength, constructive interference results at P.(d sinθ)
36d sin =m ……….(maxima) Constructive Interference: Maximum at P: The condition for constructive interference or Maxima at point P isPOd sin =m ……….(maxima)Where m= 0, ±1, ±2…….m- order number.Central maximum m=0Central maximum at O has order m=0.Each maximum above has a symmetrically located maximum below O; these correspond to m= -1, -2, -3……
37Destructive Interference: Minimum at P: OWhen path length (S1b=d sin ) is an odd multiple of λ/2, the two waves arriving at point P are (=π) out of phase and give rise to destructive interference.The condition for dark fringes, or destructive interference, at point P isCentral maximum m=0m=0m=1m=2m=3Negative values of m locate the minima on the lower half of the screen.
38DOUBLE SLIT INTERFERENCE Fringe width (Δy)The distance between two consecutive bright or dark fringes (for small θ)is known as fringe width Δy.ΔyThe spacing between the adjacent minima is same the spacing between adjacent maxima.
39For small value of , we can make following approximation. DOUBLE SLIT INTERFERENCEFor small value of , we can make following approximation.mQOym+1If d sin=m , mth order constructive interference will take place at P.
40What is the angular position of the first minimum? DOUBLE SLIT INTERFERENCESample 41-1:Problem 1:The double -slit arrangement is illuminated with light from a mercury vapor lamp filtered so that only the strong green line (λ=546 nm) is visible. The slits are 0.12 mm apart, and the screen on which the interference pattern appears is 55 cm away.What is the angular position of the first minimum?(b)What is the distance on the screen between the adjacent maxima?λ=546 nmD=0.55 md=0.12 mmθ=? for first minimum.Fringe width Δy=?
41Solve for First maximum : use d sinθ=mλ put m=1 DOUBLE SLIT INTERFERENCECHECK YOURSELFSolve for First maximum : use d sinθ=mλ put m=1
42(Assume that θ is small). d sin θ = mλ sinθ≈ θ (θ is very small) Problem 5:A double-slit arrangement produces interference fringes for sodium light (λ=589 nm) that are apart. For what wavelength would the angular separation be 10% greater?(Assume that θ is small).d sin θ = mλsinθ≈ θ (θ is very small)d θ=λ (for first maxima , m=1)Use: λ1/ λ2= θ 1/ θ 2, λ2= λ1 x θ 2/ θ 1,Given: θ 1 = (for 100%)θ 2 = x 1.1 = (10% more: for 110%)λ2= λ1 x θ 2/ θ 1 = (589 nm)(0.253/0.23)= nmDOUBLE SLIT INTERFERENCE
43Problem 11:DOUBLE SLIT INTERFERENCESketch the interference pattern expected from using two pinholes rather than narrow slits.In case two pinholes are used instead of slits, as in Young's original experiment, hyperbolic fringes are observed.If the two sources are placed on a line perpendicular to the screen, the shape of the interference fringes is circular as the individual paths travelled by light from the two sources are always equal for a given fringe.
44Solution: Use d sinθ=mλ d1 sin 150=λ gives the first maximum (m=1) DOUBLE SLIT INTERFERENCETutorial: Problem:2Monochromatic light illuminates two parallel slits a distance ‘d’ apart. The first maximum is observed at an angular position of 150. By what percentage should ‘d’ be increased or decreased so that the second maximum will instead be observed at 150?Solution: Use d sinθ=mλd1 sin 150=λ gives the first maximum (m=1)d2 sin 150 = 2λ put the second maximum (m=2) at the location of the first.Divide the second expression by the first and d2 = 2d1. This is a 100% increase in d1.
45To find λ: Use ym= λD/d gives λ=0.0108m Use: v= f x λ , f= 23 Hz. DOUBLE SLIT INTERFERENCETutorial: Problem 8In an interference experiment in a large ripple tank, the coherent vibrating sources are placed 120 mm apart. The distance between maxima 2.0 m away is 180 mm. If the speed of ripples is 25 cm/s, calculate the frequency of the vibrating sources.Given:d = 120 x m, λ= ? ym=180 x10-3 mD = 2 m, v= 25 x 10-2 m. f=?Use: v=f x λTo find λ: Use ym= λD/d gives λ=0.0108mUse: v= f x λ , f= 23 Hz.
46θ = ? in radians and in degrees. HRK:958 page: Exercise: 41-2Problem 1:Monochromatic green light of wavelength 554 nm, illuminates two parallel narrow slits 7.7 μm apart. Calculate the angular position of the third-order (m=3) bright fringe in radians and (b) in degrees.Given:d = 7.7 x m, m=3, λ= 554 nm = 554 x10-9 mθ = ? in radians and in degrees.Use: for the bright fringe: d sinθ = mλ (put m=3):Answers:θ=0.216 radians (b) θ =12.5 degreesDOUBLE SLIT INTERFERENCECHECK YOURSELF(Solve for third order dark fringe put m=2)
47Fringe width= Δy = λD/d = (512x 10-9m)(5.4 m)=(1.2 x 10-3m) Problem 3:A double-slit experiment is performed with blue-green light of wavelength 512 nm. The slits are 1.2mm apart and the screen is 5.4 m from the slits. How far apart are the bright fringes as seen on the screen?DOUBLE SLIT INTERFERENCEGiven: λ =512 nmd = 1.2 mmD= 5.4 mFringe width Δy=?Solution:Fringe width= Δy = λD/d =(512x 10-9m)(5.4 m)=(1.2 x 10-3m)=2.3 x 10-3 mCHECK YOURSELF: How far apart are the bright fringes as seen on the screen? Fringe width is the same 2.3 x 10-3 m.
48DOUBLE SLIT INTERFERENCE Problem 4:Find the slit separation of a double-slit arrangement that will produce bright interference fringes apart in angular separation. Assume a wavelength of 592 nm.Given: d=?θ=1.000λ= 592 nmUse:d =λ/sin θ = (592 x 10-9m)/sin(1.000) =3.39x 10-5mCHECK YOURSELFSolve this problem for dark interference fringes apart in angular separation.
49d sin θ0 =λ; and then find θ from d sin θ= λ/n. DOUBLE SLIT INTERFERENCEProblem 6A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are apart. What is the angular fringe separation if the entire arrangement is immersed in water (n=1.33)?Solution:Immersing the apparatus in water will shorten the wavelengths to λ/n. Start withd sin θ0 =λ; and then find θ from d sin θ= λ/n.Combining the two expressions,sin θ/sin θ0 =1/ngives θ=0.150
50The third-order fringe for a wavelength will be located at Problem 7In a double-slit experiment, the distance between slits is 5.22 mm and the slits are 1.36 m from the screen. Two interference patterns can be seen on the screen, one due to light with wavelength 480 nm and the other due to light with wavelength 612 nm. Find the separation on the screen between the third-order interference fringes of the two different patterns.Solution:The third-order fringe for a wavelength will be located atym = mλD/d= 3 λD/dwhere ym is measured from the central maximum.Then Δy is:y1 - y2 = 3(λ1 -λ2)D/d = 3(612x 10-9m – 480x10-9m)(1.36m)/(5.22x10-3m) = 1.03x10-4m:DOUBLE SLIT INTERFERENCE
51So with small angle approximation: d sin =(m+1/2 ) , DOUBLE SLIT INTERFERENCEProblem 9:If the distance between the first and tenth minima of a double slit pattern is 18 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits. What is the wavelength of the light used?mQOym+1So with small angle approximation:d sin =(m+1/2 ) ,sin tan = =ym/Dmth order minima will take place at P.d (ym/D)= =(m+1/2 ) ,ym = =(m+1/2 ) D/d
52We are given the distance (n the screen) between the first minima (m=0) and the tenth minima (m=9). Theny9-y0 = (9+1/2) D/d- (0+1/2) D/d = 9 ( D/d)Given:y9-y0 = 18 mm,d=50 cm=0.5 m,d= 0.15 mm= 0.15 x10-3 mSolving for λ = 18 x 10-3 x 0.15 x 10-3/ 9 x 0.5 = 600 nmSolve this problem with first and 10th maxima.(CHECK YOURSELF:Put first maxima m=1, and tenth maximum m=10 )
53CoherenceProblem 14:The coherence length of a wavetrain is the distance over which the phase constant is the same.If an individual atom emits coherent light for 1 x 10-8 s, what is the coherence length of the wavetrain?Suppose a partially reflecting mirror separates this wave train into two parts that are later reunited after one beam travels 5 m and other 10 m. do the waves produce interference fringes observable by a human eye?velocity = distance/ timex = c/t = (3.0 x108m/s)/1 x 108s) = 3 m.(b) No.
54YOUNG’S DOUBLE SLIT INTERFERENCE Thomas Young (1773–1829)Double slit experiment was first performed by Thomas Young in 1801.So double slit experiment is known as Young’s Experiment.He used sun light as source for the experiment.In young’s interference experiment, sunlight diffracted from pinhole S0 falls on pinholes S1 and S2 in screen B.Light diffracted from these two pinholes overlaps on screen C, producing the interference pattern.
55INTENSITY IN DOUBLE SLIT INTERFERENCE In this section we derive an expression for the intensity I at any point P located by the angle ‘θ’.We know, the intensity of light wave is proportional to square of its E-the electric vector.Intensity α Amplitude2I α E2Let us consider the electric components of two sinusoidal waves r1 and r2 from the two slits S1 and S2 have the same angular frequency ‘ω (=2πf)’ and a constant phase difference .E1= E0 sin ωt& E2= E0 sin (ωt + )POE0- amplitude of each wave.
56The resultant electric field : E = E1+E2 = Eo sin ωt + Eo sin (ωt+) INTENSITY IN DOUBLE SLIT INTERFERENCEThen the total magnitude of the electric field at point P on the screen is the superposition of the two waves.(Assumption: The slit separation d<<D, the electric vectors from the two interfering waves are nearly parallel, and we can replace the vector sum of the E- fields with the sum of their components).The resultant electric field :E = E1+E2 = Eo sin ωt + Eo sin (ωt+)
57INTENSITY IN DOUBLE SLIT INTERFERENCE Adding Wave disturbances: PhasorsThe combined electric field can be done algebraically, using a graphical method, which proves to be convenient in more complicated situations.The sinusoidal wave can be represented graphically by a Phasor of magnitude Eo rotating about the origin counterclockwise with an angular frequency .wPhasor diagramE1(=E0 sinwt)w
58Phasor representation of first wave disturbance E1(=E0sinwt) Phasor diagrams:Phasor representation of first wave disturbanceE1(=E0sinwt)is represented by the projection of the phasor on the vertical axis.w(b)The second sinusoidal wave isE2 = Eo sin (t + )It has the same amplitude and frequency as E1.Its phase is with respect to E1.
59(c) The sum E of wave disturbances E1 &E2 is the sum of the projections of the two phasors on the vertical axis, placing the tail of one arrow at the head of the other, maintaining the proper phase difference.E is the projection on the vertical axis of a phasor of length Eθ, which is the vector sum of the two phasors of magnitude E0.ωtE1E2EEE0
60From the right angled triangle ABC E1E2EEE0CABFrom the right angled triangle ABC
61β Eθ 2 β= φ φ or β =φ/2 E0 Where E- The resultant electric field PhaseshiftAmplitudeWave partWhere E- The resultant electric fieldEθ- Amplitude of the resultant waveβ – the phase difference between amplitude of the resultant wave and first wave.From the geometry of the right angled triangleβφEθE0The three phasors, E0,E0,Eθ form an isosceles triangle(An any triangle, an exterior angle (φ in this case) is equal to the sum of the opposite interior angles( β and β)). So,2 β= φor β =φ/2
62The amplitude ‘Eθ’ of the resultant wave disturbance is given by Eθ = 2 E0 cos βwhich determines the intensity of the interference fringes, depends on ‘β’, which in turn depends of the value of ‘θ’, that is , on the location of point P.θThen the intensity of the resultant wave Iθ at P is given byIθ α Eθ (for resultant wave) andI0 α E (Intensity of each single wave)Iθ = 4 E02 cos2 β = 4 E02 cos2 ф/2I θ = 4 I0 cos2 βNote: The intensity of the resultant wave at any point varies from zero (dark or minima) to four times (bright or maxima)the intensity I0 (i.e. 4I0=Im) of individual wave.
63Intensity distribution in double slit interference Iθ = 4 I0 cos2 ф/2Intensity Maxima occur wherecos2ф/2 =1, orФ =0 , 2π , 4π,…………..2m πф= 2mπ (maxima)Or the path difference:d sinθ= m λm= 0, ±1, ±2……………(maxima)Iθ = 4 I0 cos2 ф/2Intensity Minima occur wherecos2ф/2 =0, orФ =π , 3π, 5 π …………..(2m+1) πф= (2m+1)π (minima)Or the path difference:d sinθ= (m+1/2) λm= 0, ±1, ±2……… (minima)
64The horizontal solid line is I0: this describes the (uniform) intensity pattern on the screen if one slit is covered up.If the two sources were incoherent, the intensity would be uniform over the screen and would be 2 I0 indicated by the horizontal dashed line.For two coherent sources it would be 4I0.
65INTENSITY IN DOUBLE SLIT INTERFERENCE PHASE AND PATH DIFFERENCEThe phase difference (φ) between r1 and r2 associated with the path difference S1b (= d sinθ)
66Using Trigonometric Identity: E = E1+E2 = Eo (sin ωt + sin (ωt+))With = (t + ), = t, get:It can be written as, E= Eθ sin(wt+β)Where Eθ =2 E0 cos β: Amplitude of the resultant wave.‘β’ – the phase difference between resultant wave and first wave.β = ф/2
67INTENSITY IN DOUBLE SLIT INTERFERENCE Problem: SP 41-2Find graphically the resultant E(t) of the following wave disturbances.E1 = E0 sin tE2 = E0 sin (t + 15o)E3 = E0 sin (t + 30o)E4 = E0 sin (t + 45o)The phase angle φ between successive phasors is 150.E=E1+E2+E3+E4E=Eθ sin(wt+β)MIT-MANIPALBE-PHYSICS- INTERFERENCE
68To find β:In any closed n-sided polygon, the sum of the interior angles is (n-2)π.So, in the five sided polygon, abcdea:So, 2β= (1650)So, β = 22.50
69To find Eθ: ax1= ab cosβ x1x2= bc (cosβ-15) x3 x2x3= cd (cosβ-15) x2 x3e= de (cosβ)β-15cx1b
70Eθ = ax1+x1x2+x2x3+x3e= ab cosβ+ bc (cosβ-150)+ cd (cosβ-150)+de (cosβ)Substituteab=bc=cd=de= E0 and β=22.50Eθ=3.83 E0The resultant E(t) is the projection of E on the vertical axis:E= 3.83 E0 sin (wt+22.50)
71Alternate method: Verification: Standard equation: E(t)= Eθ sin (wt+β) Ey- sum of verical- componentsEθEh- sum of horizontal componentscos wtsin wtβWe can use the phasors to find sum and free to evaluate the phasors at any time t. We choose t=0.
72Sample Problem: Intensity in double-slit Interference Problem: E 41-15 Source A of long-range radio waves leads source B by 90 degrees. The distance rA to a detector is greater than the distance rB by 100m. What is the phase difference at the detector?Both sources have a wavelength of 400m.Initially, source A leads source B by 90◦, = 1/4 wavelength (100 m)However, source A also lags behind source B since rA is longer than rB by 100 m, which is 100m/400m= 1/4 wavelength.So, the net phase difference between A and B at the detector is zero.100 m=λ/4 corresponds to π/2 phaserADETCORrBThey are in phase and the reach the detector.
73Problem: E 41-16Find the phase difference between the waves from the two slits arriving at the mth dark fringe in a double-slit experiment.Answer: (2m + 1)π radians.
74Tutorial Problem: E 41-18Find the sum of the following quantities (a) graphically, using phasors; and (b) using trigonometry.y1=10 sin wt, y2=8.0 sin (wt+300)Similar to the equation:E1= E01 sinwt & E2= E02 sin (wt+φ)Resultant electric field E= Eθ sin (wt+β)Where Eθ- Amplitude of the resultant waveβ= phase difference between first electric field and the resultant wave.
75To find Eθ: The resultant electric field E= 17.39 sin (wt+13.3) (a) PhasorsTo find Eθ:βE01E02ABDCEθyxφEθ=17.39β= 13.30Substitute E01=10, E02=8, & φ=300The resultant electric field E= sin (wt+13.3)
76Alternate method: Trigonometry Standard equation: E(t)= Eθ sin (wt+β) Ey- sum of verical- componentsEθEh- sum of horizontal componentscos wtsin wtβWe can use the phasors to find sum and free to evaluate the phasors at any time t. We choose t=0.
77REFLECTION PHASE SHIFT Reflection causes a l/2 phase shiftIt has been observed that if the medium beyond the interface has a higher index of refraction, the reflected wave undergoes a phase change of (180o) ) or a path length λ/2.Incident waveReflected waveReflection causes no phase shiftIncident waveIf the medium beyond the interface has a lower index of refraction, there is no phase/path length change of the reflected wave.Reflected wave
78INTERFERENCE FROM THIN FILMS We see color when sunlight falls on a bubble, an oil slick, or a soap bubble - the interference of light waves reflected from the front and back surfaces of thin, transparent films.The film thickness is typically of the order of magnitude of the wavelength of light.A soapy water film on a vertical loop viewed by reflected light
79Thin films deposited on optical components- such as camera lenses-reduce reflection and enhance the intensity of the transmitted light.Thin coatings on windows can enhance the reflectivity for infrared radiation while having less effect on the visible radiation.It is possible to reduce the heating effect of sunlight on a building.
80INTERFERENCE FROM THIN FILMS dn1n2Source of lightA thin film (a soap film or thin film of air between two glass plates) is viewed by light reflected from a sourer S.Waves reflected from the front surface (r1) and back surface (r2) interfere and enter the eye.
81dn1n2Source of lightThe incident ray ‘i’ from the source enters the eye as ray r1 after reflection from the front surface of the film at ‘a’.The incident ray ‘i’ also enters the film at ‘a’ as refracted ray and is reflected from the back surface of the film at ‘b’.It then emerges from the front surface of the film at ‘c’ and also enters the eye, as ray r2.d- thickness of the film.
82Interference in light reflected from a thin film is due to a combination of rays r1 and r2. As the waves originated from the same source by division of amplitude, hence they are coherent and they are close together.dn1n2Source of lightThe region ac looks bright or dark for an observer depends on the phase difference between waves of rays r1 and r2.
83As r1 and r2 have travelled over paths of different lengths, have traversed different media, and have suffered different kinds of reflections at ‘a’ and ‘b’.The phase difference between two reflected rays r1 and r2 determine whether they interfere constructively or destructively.dn1n2Source of light
84EQUATIONS FOR THIN –FILM INTERFERENCE dn1n2Source of lightIf a wave travelling from a medium of index of refraction ‘n1’ toward a medium of index of refraction ‘n2’ undergoes a phase change upon reflection when n2>n1 and undergoes no phase change if n2<n1.
85The wavelength of light ‘λn’ in a medium whose index of refraction is ‘n’ λ - wavelength of the light in free space.
86To obtain Equations for thin film Interference, let us simplify by assuming near -normal incidence θi=0.The ‘r2’ travels a longer path (2d) than ‘r1’’ as ‘r2’ travels twice through the film before reaching the eye.dn1n2Source of light
87The path difference due to the travel of ray r2 through the film is approximately ‘2d’. Other possible contributions to the total path difference between r1 and r2 : the phase difference of π (or path difference of one-half wavelength) that might occur on reflection at the front/back surface of the film.dn1n2Source of light
88A general equation for Thin film Interference: The path difference between rays r1 and r2 is:Path difference = 2d + λn/2 + λn/2 …………(1)??Front surfaceBack surfaceNote: Depending on the relative index of refraction of the film in comparison with what is on either side of the film, we might need to include neither of the extra terms, or perhaps one of them, or perhaps both of them.
89(a) Interference in a thin soap film Examples:(a) Interference in a thin soap filmdAirn1800 phase changeNo phase changeFor the interference from a thin soap film of index of refraction ‘n’ surrounded by air, we must add the extra half wavelength for the front surface reflection, but not for the back surface .
90Interference in a thin soap film The total path difference=2d+λn/2=mλn ...(maxima)where m=1,2,3………Where we have dropped the m=0 solution because it is not physically meaningful.The total path difference:= 2d + λn/2 =(m+1/2) λn ……… (minima)where m=0,1,2,3…..Note: These equations apply when the index of refraction of the film is greater than the index of refraction of the material on either side.
91THIN FILM INTERFERENCE -WEDGE SHAPED FILM A thin wedge of air film can be formed by two glasses slides on each other at one edge and separated by a thin spacer (a thin wire or a thin sheet) at the opposite edge.A thin film having zero thickness at one end and progressively increasing to a particular thickness at the other end is called a wedge.
92The arrangement for observing interference of light in a wedge shaped film. The wedge angle is usually very small and of the order of a degree.When a parallel beam of monochromatic light falls normally on a wedge shaped film part of it is reflected from upper surface and some part from lower surface (division of amplitude).dwire
93Ray BC reflected from the top –NO phase change. Back surfaceirRay DE (the back surface reflection), undergoes a π phase change and λ/2 (half wave length) at the air to glass boundary due to reflection.These two coherent waves superpose-producing constructive and destructive interferences, the positions of which depend on the thickness of the film.
95Sample problem:41-3A soap film (n=1.33) in air is 320 nm thick. If it is illuminated with white light at normal incidence, what color will it appear to be in reflected light?No phase changedAirn=1.331800 phase changeNo phase changeSolution:The wavelengths which are maximally reflected are constructively interfered.
96Constructive interference maxima occur for the following wavelengths: 1702 nm (=1), nm(m=2),340 nm (m=3) and so on.Only the maximum corresponding to m=2 lies in the visible region (between about 400 nm and 700 nm); light of wavelength 567 nm appears yellow-green.
97Sample Problem: Page 951Lenses are often coated with thin films of transparent substances such as MgF2 (n=1.38) to reduce the reflection from the glass surface. How thick a coating is needed to produce a minimum reflection at the center of the visible spectrum (λ=550 nm)?Given: λ=550 nmn=1.38Minimum reflection:(Destructive interference)Thickness of coating: d=?
982nd+λn/2+λn/2=(m+1/2)λn…(minima) Solution:Light strikes the lens at near-normal incidence (θ).For both the front and back surfaces of the MgF2 film the reflection have additional path difference (λ/2).The path difference for destructive interference is therefore Path difference:2nd+λn/2+λn/2=(m+1/2)λn…(minima)
99Where m=1,2,3……………………………,Dropped m=0 solution : physically not meaningful.We seek the minimum thickness for destructive interference. For m=1, we obtain
100INTERFERENCE FROM THIN FILMS Problem: E 41-23A disabled tanker leaks kerosene (n=1.20) into the Persian Gulf, creating a large slick on top of water (n = 1.33).If you look straight down from aeroplane on to the region of slick where thickness is 460nm, for which wavelengths of visible light is the reflection is greatest?If you are scuba diving directly under this region of slick, for which wavelengths of visible light is the transmitted intensity is strongest?MIT-MANIPALBE-PHYSICS- INTERFERENCE
101White light (400 nm-700nm)Normal incidenceReflected light (λ=?)R.I=1.0R.I=1.33n=1.2Thin film of oilwaterSolution:The reflected light from the film is brightest at the wavelength (λ) for which the reflected rays are in phase with one another.(constructive interference).d=460nm(Both Front & back surface reflections have phase change) .
102The total path difference for maxima: Path difference = 2d + λn/2+ λn/2 = mλn , Where (m=1,2,3….)2d=(m-1) λ/n (λn=λ/n)λ=2nd/(m-1)Find λ for d=460 nm, n=1.2 & m=1,2,3…λ - for m=1 (not possible)For m=2: λ=1104 nm (IR region)For m=3: λ=552 nm (Green light-visible)For m=4: λ=368 nm (UV light)So, Green light appears in the reflected light
103The total path difference for minima: Path difference The wavelengths which are minimally reflected are maximally transmitted, and vice versa. Maximally transmitted wavelengths is the same as finding the minimally reflected wavelengths.Minimally reflectedlightMaximum Transmittedlight =?R.I=1.0R.I=1.33n=1.2Thin film of oilwaterThe total path difference for minima:Path difference= 2d + λn/2+ λn/2 = (m+1/2)λnwhere (m=1,2,3….)substitute : (λn=λ/n)λ=2nd/(m-1/2)
104Find λ for d=460 nm, n=1.2, & m=1,2,3…λ - for m=1: λ=2208 nm (not visible region)For m=2: λ=736 nm (IR region)For m=3: λ=442 nm(Blue light-visible)(maximum transmitted)
105Problem: E 41-25If the wavelength of the incident light is λ=572 nm, rays A and B are out of phase by 1.50 λ. Find the thickness d of the film.The total path difference for minima:Path difference =2d +λn/2= (m+1/2)λn2d= mλnm=0 not possibleTake m=1d=λ/2n=215nm
106OR In the given total phase of 1.50 λn. The top surface contributes a phase difference π=0.5λnSo, phase difference because of the thickness= 2d=λn=2π2d=λnd=λ/2n=572 nm/2 x 1.33=215 nm
107INTERFERENCE FROM THIN FILMS Problem: E 41-29A broad source of light (wavelength = 680nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.048mm in diameter at the other end. How many bright fringes appear over 120 mm distance?d=0.048 x 10-3 m)λ=680 x10- 9mθx=120 mmMIT-MANIPALBE-PHYSICS- INTERFERENCE
109INTERFERENCE FROM THIN FILMS Problem: E 41-27A thin film of acetone (n = 1.25) is coating a thick glass plate (n = 1.50). Plane light waves of variable wavelengths are incident normal to the film. When one views the reflected wave, it is noted that complete destructive interference occurs at 600nm and constructive interference at 700nm. Calculate the thickness of the acetone film?Put: λn=λ/nThin film n=1.25Glass plate 1.50MIT-MANIPALBE-PHYSICS- INTERFERENCE
111Problem 21:We wish to coat a flat slab of glass (n = 1 Problem 21:We wish to coat a flat slab of glass (n = 1.5) with a transparent material (n=1.25) so that light of wavelength 620nm (in vacuum) incident normally is not reflected. What should be the minimum thickness of the coating?Air=1Glass=1.5Film= 1.25
112Both the reflected rays r1(front surface reflection) and r2(back surface reflection) have additional path difference (λ/2).m=1
113Problem 22: A thin film in air is 410nm thick and is illuminated by white light normal to its surface. Its index of refraction is What wavelength in the visible spectrum will be intensified in the reflected beam?No phase changed=410 nmAirn=1.501800 phase change
114The result is only in the visible range when m = 3, so λ= 492 nm.
115Problem: 24:In costume jewelry, rhinestones (made of glass with n =1.50) are often coated with silicon monoxide (n = 2.0) to make them more reflective. How thick should the coating be to achieve strong reflection for 560 nm light incident normally?Air=n=1Film=n= 2.0Glass=n=1.5
117Problem 26:Light of wavelength 585nm is incident normally on a thin, soapy film (n=1.33) suspended in air. If the film is mm thick, determine whether it appears bright or dark when observed from point near the light source.No phase changed= mmAirn=1.331800 phase change
118So the interference is NOT dark. m should be an integer.So the interference is NOT dark.So the interference is bright.
119Problem 28:White light reflected at perpendicular incidence from a soap film in air has, in the visible spectrum, an interference maximum at 600nm and a minimum at 450nm with no minimum in between. If n = 1.33 for the film, what is the film thickness?No phase changed=?Airn=1.331800 phase change
120Use equation (1) or (2) to solve for d: With First source (λ=600nm):With second source (λ=450nm):Use equation (1) or (2) to solve for d:From equation (2)d= 2 x 450/2 x1.33=338 nm
121Problem 31: Light of wavelength 630nm is incident normally on a thin wedge shaped film with index of refraction There are ten bright and nine dark fringes over the length of the film. By how much does the film thickness changes over the length?d1d2xB10B1B2B3B4B5B6B7B8B9m1=1m2=10
122Number of bright bands in ‘x’ mm length. 10 bright & 9 dark bands Number of bright bands in ‘x’ mm length. 10 bright & 9 dark bands. Film thickness over the length d2-d1=?
123INTERFERENCE FROM THIN FILMS Instead of wedge shaped films, interference is possible even in curved films also.Circular interference fringes can be produced by enclosing a very thin film of air of varying thickness between a plane glass plate and a plano -convex lens of a large radius of curvature.
124If monochromatic light is allowed to fall normally and viewed, dark and bright circular fringes known as Newton’s Rings are produced.The fringes are circular because the air film has circular symmetry.
125Newton’s Rings:When the light is incident on the plano-convex lens part of the light incident on the system is reflected from glass-to-air boundary (say at point D).The reminder of the light is transmitted through the air film, and it is again reflected from the air-to-glass boundary (say from point J).The two rays are (1 and 2 ) reflected from the top and bottom of the air film interfere with each other to produce darkness and brightness .
126The interference effect is due to the combination of ray 1, reflected from the flat surface, with ray 2, reflected from the curved surface of the lens.Ray 1 undergoes a phase change of 1800 upon reflection (because it is reflected from a medium of higher index of refraction), whereas ray 2 undergoes no phase change (because it is reflected from a medium of lower refractive index).Air-Glass interfacePhase change (π)
127The condition for constructive interference remains unchanged. Path difference:
128The radius (r) of the bright ring: Consider the section AB of the lens, wherein it forms an air film of thickness d.Let mth order bright ring of radius ‘r‘ forms here.Let C be the center of curvature of the plano- convex lens.R(=CB=OC) be the radius of curvature of the plano-convex lens.OCBADR
129We can write: d= OD=OC-CD BADRWe can write: d= OD=OC-CD
130Substituting ‘d’ value in equation (1) and solving for ‘r’: …….(For bright ring)The radii/diameters of the bright rings.
131D- is the diameter of the mth bright ring. m=0 is physically not possible,So, m=1,2……………………..center ring must be the dark.Radii/diameters of bright rings are proportional to square root of odd numbers.
132Problem 33: In Newton’s ring experiment, the radius of curvature R of the lens is 5m and its diameter is 20mm.How many rings are produced?How many rings would be seen if the arrangement is immersed in water (n=1.33)?(Assume that wavelength = 589nm)Given: λ = 589 nmRadius of curvature R= 5 mdiameter of the ring=d=20mmTherefore, radius r= 10 mm=0.01 m(a) m=? (b) n=1.33, m=?
134Problem 34: The diameter of the tenth ring in a Newton’s rings apparatus changes from 1.42 to 1.27 cm as a liquid is introduced between the lens and the plate. Find the refraction of the liquid.Given: Radius of tenth bright ring in air: rair = 1.42 cmRadius of tenth bright ring in liquid: rliquid = 1.27 cm ringRefractive index of the liquid: nliquid = ?
136Problem 35: A Newton’s ring apparatus is used to determine the radius of curvature of a lens. The radii of the nth and (n+20)th bright rings are found to be 0.162cm and 0.368cm, respectively, in light of wavelength 546nm. Calculate the radius of curvature of the lower surface of the lens.Given: λ = 546 nmRadius of the nth ring=rn =0.162cmRadius of the (n+20)th=rn+20=0.368cmm=? Radius of curvature R= ?
138INTERFERENCE FROM THIN FILMS OPTICAL PATHDistance traveled by light in a medium in the time interval of ‘t’ is d = vtRefractive index n = c/vHence, ct = ndnd Optical path.dnMIT-MANIPALBE-PHYSICS- INTERFERENCE
139MICHELSON’S INTERFEROMETER PurposeInterferometers are basic optical tools used to precisely measure wavelength, distance, index of refraction of optical beams.It is a device working on the principle of interference of light and is used in precise measurements of length or changes in length.
140BE-PHYSICS- INTERFERENCE-2010-11 MICHELSON’S INTERFEROMETERLight from an extended monochromatic source P falls on a half-silvered mirror M.The incident beam is divided into reflected and transmitted beams of equal intensity (Division of amplitude).These two beams travel almost in perpendicular directions and will be reflected normally from movable mirror (M2) and fixed mirror (M1).BAGMIT-MANIPALBE-PHYSICS- INTERFERENCE
141MICHELSON’S INTERFEROMETER The two beams finally proceed towards a telescope (T) through which interference pattern of circular fringes will be seen.The interference occurs because the two light beams travel different paths between M and M1 or M2.Each beam travels its respective path twice. When the beams recombine, their path difference is 2 (d2 – d1).MIT-MANIPALBE-PHYSICS- INTERFERENCE
142MICHELSON’S INTERFEROMETER The path difference can be changed by moving mirror M2. As M2 is moved, the circular fringes appear to grow or shrink depending on the direction of motion of M2.New rings appear at the center of the interference pattern and grow outward or larger rings collapse disappear at the center as they shrink.MIT-MANIPALBE-PHYSICS- INTERFERENCE
143MICHELSON’S INTERFEROMETER Each fringe corresponds to a movement of the mirror M2 through one-half wavelength. The number of fringes is thus the same as the number of half wavelength.MIT-MANIPALBE-PHYSICS- INTERFERENCE
144If N fringes cross the field of view when mirror M2 is moved by d, then d = N (/2)d is measured by a micrometer attached to M2. Thus microscopic length measurements can be made by this interferometer.
145Michelson interferometer equation The interferometer is used to measure changes in length by counting the number of interference fringes that pass the field of view as mirror M2 is moved.Length measurements made in this way can be accurate if large numbers of fringes are counted.
146Applications Determination of wavelength The fact that whenever the movable mirror moves bya fringe originates or vanishes at the center is used to determine from the equation 2d = N , where d is the distance moved and N, the number of fringes originated or vanished.
147(nd–d)2= 2d(n-1) will be introduced. ApplicationsDetermination of refractive index (n)When a thin film (whose refractive index n is to be determined) of thickness ‘d’ is introduced on the path of one of the interfering beams, an additional path difference(nd–d)2= 2d(n-1) will be introduced.As a result there will be shift of fringes. If ‘m’ fringes shift, then,2d(n -1) = m from which ‘n’ can be determined.d
148MICHELSON’S INTERFEROMETER Problem: SP 41-6Yellow light (wavelength = 589nm) illuminates a Michelson interferometer. How many bright fringes will be counted as the mirror is moved through 1.0 cm?The number of fringes is the same as the number of half wavelengths in cm.Nλ= 2dN= 2 d/λ=2( x10-2 m)/(589 x 10-9m)= 33,956 fringesMIT-MANIPALBE-PHYSICS- INTERFERENCE
149MICHELSON’S INTERFEROMETER Problem: E41-39If mirror M2 in Michelson’s interferometer is moved through 0.233mm, 792 fringes are counted with a light meter. What is the wavelength of the light used?Nλ= 2dλ= 2 d/N=2(0.233 mm)/792= 588 nm= 588 nmMIT-MANIPALBE-PHYSICS- INTERFERENCE
150MICHELSON’S INTERFEROMETER Problem: E41-40An airtight chamber 5.0 cm long with glass windows is placed in one arm of a Michelson’s interferometer as indicated in Fig Light of wavelength λ = 500 nm is used. The air is slowly evacuated from the chamber using a vacuum pump. While the air is being removed, 60 fringes are observed to pass through the view. From these data find the index of refraction of air at atmospheric pressure.MIT-MANIPALBE-PHYSICS- INTERFERENCE
151Answer:The change in the optical path length is2(nd – d)2d(n-1)=mλn=mλ/2d+1=
152Problem HRK-41-38:A thin film with n=1.42 for light of wavelength 589nm is placed in one arm of a Michelson interferometer. If a shift of 7 fringes occurs, what is the film thickness?Solution:2d(n-1) = md = mλ/ 2(n-1)d = 7 x (589 x10-9 m)/ 2(1.41-1)= 4.9 x 10-6 m
153BE-PHYSICS- INTERFERENCE-2010-11 QUESTIONS – INTERFERENCEWhat is the necessary condition on the path length difference (and phase difference) between two waves that interfere (A) constructively and (B) destructively ? Obtain an expression for the fringe-width in the case of interference of light of wavelength λ, from a double-slit of slit-separation d. Explain the term coherence. Obtain an expression for the intensity of light in double-slit interference using phasor-diagram. MIT-MANIPALBE-PHYSICS- INTERFERENCE
154BE-PHYSICS- INTERFERENCE-2010-11 QUESTIONS – INTERFERENCE Draw a schematic plot of the intensity of light in a double-slit interference against phase-difference (and path-difference). Explain the term reflection phase-shift. Obtain the equations for thin-film interference. Explain the interference-pattern in the case of wedge-shaped thin-films. Obtain an expression for the radius of mTH order bright ring in the case of Newton’s rings. Explain Michelson’s interferometer. Explain how microscopic length measurements are made in this. MIT-MANIPALBE-PHYSICS- INTERFERENCE