Chapter 7 Interference of Light

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1 Chapter 7 Interference of Light Lecture Notes for Modern Optics based on Pedrotti & Pedrotti & Pedrotti Instructor: Nayer Eradat Spring 009 4/1/009 Interference of Light 1

2 Interference of light Interference phenomenon depends on superposition of two or more individual waves under strict conditions. Constructive interference leads to enhancement of the resulting amplitude with respect to that of the constituents. Destructive interference leads to diminution of the resulting amplitude with respect to that of the constituents. Interference produces an alternating spatial pattern of fringes. 4/1/009 Interference of Light

3 Two beam interference: vector treatment Interference of the plane waves of same frequency arriving at point P: E 1 = E 01 cos ( ks1 ω t + φ1 ) E = E0cos( ks ωt + φ) The combined disturbance at point P is: E= E1+ E 15 Ess are rapidly verying function s in time with frequencies of ~10 Hz and average to zero very quickly. Irradiance or radiant power density is a measure of time-average of the square of the field (detector reading) I = Ee ( W / cm ) 1-15 E in practice t averaging time for the eye >>T(period 10 s) t 30 s I = ε0c EE i = ε0c ( E1+ E) i( E1+ E) I = ε c E1iE1+ EiE + E1iE ( ) 0 I = I + I + I 1 1 1) if E1 and E are orthogonal no inte ) if E1 and E are parallel interference has rference happens. cosθ = 0 maximum effect cos θ = 1 3) if E1 and E have angle δ the interference term is: I = ε ce01ie0 cosδ 1 0 ( ) where δ = k s s + φ φ is the phase difference between the waves. 1 1 For a purely monochromatic wave δ is independent of time (a constant phase relationship between two waves). 4/1/009 Interference of Light 3

4 Two beam interference Interference of the plane waves of same frequency arriving at point P: I = I + I + I I cosδ I1 = ε0ce01; I1 = ε0ce0; I1 = ε0ce01ie0 cosδ = I1I cos δ where δ = k ( s s1 ) + φ φ1 For mutually incoherent fields and low quality laser sources the phases are random functions of time that very on time scales much larger than the E field oscillations but shorter than the detector averaging time. The interference term in this case will be: ( ( ) φ () φ ()) II cos k s s + t t = 0 for all incoherent sources and low quality lasers and I = I + I Light from independent sources even if they are same kind of lasers do not interfere. 1 ( ) For mutually coherent beams light from same lase split and recombined the phase difference at detector () t () t φ φ1 = 0 if the light paths are equal and δ = constant and the interference term is: II 1 cos ( k ( s s1 ) + φ ( t ) φ1 ( t ) ) = I1 I cos( ( k ( s s1 ) ) () ( ) Even if travel distance difference is nonzero φ t φ t+ δt 0 so long as δt << t 1 The coherence time of a source is inversely proportional to the range of the frequency components that 1 make up the electric field tc = τ 0 =. Δ ν Coherence length is the distance that electric field travels in coherence time: c L = l = cτ = ct c t 0 c For mutually coherent sources we assume the length traveled from the sources is much shorter than the coherence length, then the irradiance of the combined fields is: I = I + I + I I cos δ /1/009 Interference of Light 4

5 Two beam interference I = I + I + I I cosδ 1 1 δ is the total phase difference between the beam from the point of separation. It can be due to a) path length difference b) phase shift at reflecting beam splitters c) differing indecies es of refraction in the separate paths depanding on the sign of the cos δ the interference will be constructive (+) or destructive ( ). A pattern of alternating maxima and minima (fringes) will form at the plane of observation due to varying δ at different locations of the observation screen. ( ) Imax = I1 + I + I1I = E01 + E0 Imin = I1 + I I1 I = ( E01 E0 ) If I = I = I I = 4 I and I = max 0 min Fring visibility is a measure of fringe contrast I and is defined as: Fringe visibility = I max max I + I To ensure maximum visibility in experimental settings, we make sure the amplitudes are same in both arms when they arrive at the screen. min min 4/1/009 Interference of Light 5

6 Conservation of energy Rewriting the irradiance for the interference pattern of two equal amplitude components, we get: ( δ ) I = I + I + I cosδ = I 1+ cos δ Using 1+ cosδ = cos we get δ I = 4I0 cos The energy is not conserved at each point of the interference pattern I I but it is conserved over one spatial period I = I. = 0 So inteference causes redistribution of the energy of sources over the fringe pattern. av 0 The frine analysis performed ed here was for plane harminc waves but the results are general and hold for spherical and cylindrical waves. The only difference is that these waves do not have constant amplitudes and irradiances as they propagate. 4/1/009 Interference of Light 6

7 Young s (180) double slit experiment To demonstrate the interference pattern and fringes Young generated two coherent point sources by splitting the wavefront of a wave generated by a point source to two in-phase point sources. Today we uese a laser as the first source. We can assume that the amplitudes of the splitted waves are equal. Our goal is to develope an expression for the irradiance at an arbitrary point, P on a screen at a distance L from the sources. a is the separation of the sources. The irradiance at P is: ( δ ) I = 4I0 cos / where δ is the phase difference of the waves arriving at P In this geometry phase difference is caused by the optical path difference Δ= s s1 asinθ The phase difference associated with Δ is δ = k s = Δ ( s ) ( π / λ) 1 π Δ I = 4I0 cos λ πa sinθ I = 4I0 cos λ 4/1/009 Interference of Light 7

8 Young s (180) double slit experiment πa sinθ I = 4I0 cos λ constructive interference asin θ =Δ= mλ, m= 0, ± 1,... destructive interference asinθ =Δ= ( m+ 1/ ) λ, m= 0, ± 1,... y π ay For y << L sinθ = tan θ then I 4I0 cos L Lλ π ay mλl bright fringe = mπ ym,ma x =, m = 0, ± 1,... Lλλ a π ay 1 ( m+ 1/) λl dark fringe = m+ π ym,min =, Lλ a m = 0, ± 1,... Fringe separtion (peak to peak): λlλ L Δ ymaxima =Δ yminima = ym+ 1 ym = is constant for all fringes. a The minima are situated midway between the maxima. We can design experiments with this technique to measure the wavelength or width of the very tiny slits. 4/1/009 Interference of Light 8

9 Formation of the fringes with analysis of the crests and valleys of the spherical waves from two sources minima One valley and one crest maxima Crests of both sources 4/1/009 Interference of Light 9

10 The bright fringe surfaces for two coherent point source generated by rotating the pattern around the x axis 4/1/009 Interference of Light 10

11 Double slit experiments with virtual sources I Lloyd s mirror The interference pattern is generated by superposition of the light from the actual source S and the virtual source S that is image of the S on the mirror MM 4/1/009 Interference of Light 11

12 Double slit experiments with virtual sources II Fresnel s s mirror The interference pattern is generated by superposition of the light from the two virtual source S 1 and S that are images of the S on the mirror M 1 and M 4/1/009 Interference of Light 1

13 Double slit experiments with virtual sources II Fresnel s s biprisms The interference pattern is generated by superposition of the light from the two virtual source S 1 and S that are formed by refraction in the two halves of the biprism. 4/1/009 Interference of Light 13

14 Interference in dielectric films Dielectric here means index of refraction is real and positive. No considerable absorption hapens at the wavelength of interest. To create two coherent sources from one we can use two techniques: 1) Amplitude division. For example by partial reflection at two different interfaces. ) Wavefrot division. Fro example placing two slits in front of the waverfront. A transparent film bounded by parallel planes divides a beam of light into two parts. amplitude division some of the light is reflected from the first interface and some from ( ) the second interface. Two reflected portions from first and second planes can be brought together by a lens to interfer at a point on a screen. 4/1/009 Interference of Light 14

15 Thin film interference Δ= nf ( AB+ BC) n0 ( AD) ( ) ( ) ( ) The path difference is: Δ= nf AE+ FC n0 AD + nf EB+ BF = 0 it can be shown Snell's law: n sinθ = n sinθ 0 i f t ( ) AE = AG sin θ = AC / sinθ t AE = AC sinθt sinθt n AE AD AD AD = = = AC sinθ sinθ n n AD AEn n 0 i i f f t = = ( AE FC) n ( AE FC) n ( AD) ( ) f 0 + f + = 0 0 Δ= n EB+ BF = n EB Δ= n tcos θ this can be related to the angle of inceidence f f f t through the Snell's law. For normal incidence Δ = nt. The phase difference corresponding to the path difference Δ is: δ = kδ= π / λ. The net phase difference has to take into account the possible phase change by the reflection. Δ P optical path difference, Δ equivalent path difference arrising from the phase change. r Constructive interference: Δ P + Δ r = mλ where m= 0,1,,... ( ) Destructive ti interference: Δ P + Δ r = m+ 1/ λ where m= 0,1, 01,... What is the value of Δ for dirrerent types of interface? r 4/1/009 Interference of Light 15 f

16 Reflected wave amplitude Amplitude of the reflected electromagnetic wave at an interface E r E r n = n a a n + n b b E Three cases are recognized: i < n b ( ) i 1) If n external reflection then E and E have different a r i signs so the reflected wave has half-cycle phase difernce with respect to the incident wave. > n ( ) ) If n internal reflection then E and E have the same a b r i signs so the reflected wave and incident wave are in-phase. 3) If n = n then E = 0 no reflection happens. E i a b r n a n b

17 Amplitude of reflected EM wave from an interface f a b Er = na + nb n n E i

18 Condition for constructive and destructive interference by a thin film No relative phase shift at interfaces : Constructive: Δ +Δ = 0 + nt cos θ = m λ nt cos θ = m λ r p f t f t 1 Destructive: Δ r +Δ p = 0 + nt f cosθt = m+ λ m = 0, 1,, 3,... Half - cycle relative phase shift at interface : λ 1 Constructiv e: Δ r +Δ p = + nt f cosθt = mλ nt f cosθt = m+ λ 1 for normal incidence nt f = m+ λ λ 1 Destructive: Δ r +Δ p = + nt f cosθt = m+ λ nt f cos θt = mλ for normal incidence nt= mλ m = 0, 1,, 3,... f

19 Antireflecting coating For antireflecting coatings we want the reflected wave to vanish. 1 if Δ r +Δ p = m + λ then the reflected wave is out of phase and desctructive interference will happen. Usually n s > nf > na so both reflectons are external and no phase shift happens at the interfeces Δ r = 0. To minimize the film thickness and loss we take m = 1 and for one layer antireflection coating λ λ Δ p = t = t = this is the so-called quarter wavelength layer. 4 This works only for one wavelength. In order to make it for a brader spectrum designers use many layers. An antireflection coating is perfect if amplitudes of the waves coming from both surfaces are equal. To achieve that we need a special relationship between the indexes of the layers. The reflection coefficient 1 n / n1 of a surface is: r = the r at two interface have to be equal. 1 + n / n1 1 nf / n0 At air-film interface ra f = 1 + nf / n0 n f ns ra f = rf s = nf = nsn0 for n 0 = 1 nf = ns 1 ns / nf n0 nf At film-substrate interface For maximum air r f s = extinction 1 + ns / n f Example: for the ( ) Example: for the yellow-green 550nm that is eye's mostsensitive sensitive range n = 1. the clasest material to this is index is MgF with n = /1/009 Interference of Light 19 n f

20 Dielectric mirrors, filters, laser mirrors We can also design complete reflectors with alternating high-low λ/4 layers that create constructive interference for the reflected light ( here we have phase shift ). Advantage of these mirrors is that they don't heat up (no absorption) and are perfect reflectors for a specific wavelength. Good for making laser resonators. We can also design low-pass, high-pass, band-pass filters with very high precision. Optical thin film design is a branch of optics that deals with these applications. 4/1/009 Interference of Light 0

21 Localized and non localized finger Non localized fringes form everywhere and we can place a screen on their path to see them. For example fringes from a double slit are observable at every distance from the sources. Localized fringes form at a specific location in space and the screen has to be placed there to view them. For example fringes fi from a thin film are localized li at infinity i and we need a converging lens to project them on a screen. The fringes in the picture (on the right side) are fringes of equal inclination of Heidinger fringes formed by parallel rays not possible with point sources. 4/1/009 Interference of Light 1

22 Generating non localized real fringes with dielectric films using point sources 4/1/009 Interference of Light

Interference at a wedge Path difference between two rays is

interface. That is why the left corner appears dark.

23 Interference at a wedge Path difference between two rays is almost equal to t Wave 1 has zero phase shift due to reflection from the glass air interface. Wave has phase shift due to reflection from air glass interface. That is why the left corner appears dark. Light path difference is zero but phase difference due to reflection is or half a cycle. 1 E r = n n a a n + n b b E i

24 Fringes of equal thickness For a wedge with verying thickness the path difference Δ= ntcos varies even if the angle of incidence is f t constant. For a fixed direction of incident light bright or dark fringes will be associated with a particulat thickness (fringes of equal thickness). Fringes can be viewd with the arrangement in the figure and are called Fizeu Fringes. At normal incidence cosθ = 1 and the light path difference is Δ = nt mλ bright nt f +Δ r = 1 m + λ dark p f 4/1/009 Interference of Light 4

25 4/1/009 Interference of Light 5

26 Newton rings 4/1/009 Interference of Light 6

27 Film thickness measurements 4/1/009 Interference of Light 7

28 4/1/009 Interference of Light 8

29 Stokes relations In reality many internal reflections happen in the films. The Stokes relations offer a formalism to track the rflection and transmssion coefficients for the electric fields arriving at an interface. Let Ei = amplitude of the incident light, Et = amplitude of the tranmitted light, Er = amplitude of the reflected light E We define the reflection coefficient as: r Et r r = and the transmission coefficient t as: t = E i Ei Er = rei At the interface E i is devided th to parts Et = tei According to the principle of ray reversibility both cases in fig a and b are valid. When they happen together (figure c) we mark the reverse senario by primed parameters. But b and c are physically ( ) E= i r + tt ' E tt ' = 1 r equivalent. So 0 = ( rt ' + tr) Ei r = r ' for angles of incidence related through the Snell's law. There is a π phase difference between the rays incident from each side. i Stokes relations between the amplitude coefficients 4/1/009 Interference of Light 9

30 Multiple beam interference 4/1/009 Interference of Light 30

31 Multiple beam interference I We now consider interference between a narrow beam of amplitude E and angle of incidence θi and the multiple reflections from inside of a film of thickness t, index n f surrounded by air. We want to find superposition of the reflected waves from the top of the plate. The phase difference between the successive reflected dbeams is: δ = kδ= k ntcosθ Incident beam: E = ( ) iωt = ( ' ' ) E re e 1 0 E tt r E e E e iωt ( ωt δ) 3 i( t ) = ( tt' r' E ) e ω δ 0 5 i( ωt 3δ ) ( ' ' ) E = tt r E e 4 0 i ( N + 1) i( ωt ( N 1) δ) ( 0 )... E = tt' r' E e for N=,3,... N this form does not hold for E that never passes through the film. 1 ER = E = re e + tt E r e = E e r+ tt e r N 1 iδ x = 1 + x+ x +... = where x= r' e and x < 1 the series converges 1 x N = i t ( N 3) i t ( N 1) 0 ' i t 0 0 ' i ω ω δ ω δ ( N 4) i( N ) δ N e N= 1 N= N = 4/1/009 Interference of Light 31 0 f t

32 Multiple beam interference II i ω t tt ' r ' e ER = E0e r + 1 r' e iδ iδ using the Stokes relationships iδ iδ ( 1 r ) re r( 1 e ) iωt iωt ER = E0e r = E i 0e δ iδ 1 re 1 re ( 1 ) ( 1 ) e e e e 1 re 1 re iωt iδ iωt iδ R = R = 0 iδ iδ I E E r ( 1 cosδ) r ( 1 cosδ) iδ iδ I R using cos δ =e + e and = I I E E r I I 1+ r r cosδ 1+ r r cosδ R = R = 0 4 R = 4 i Using the conservation of energy equation I = I + I we find ( δ ) i R T ( ) 4 r 1 cos 1+ r r cosδ r 1 cosδ IT = ET = Ii IR = Ii 4 Ii I 4 1+ r r cosδ = 1+ r r cosδ I T ( 1 r ) = I 4 1+ r r cosδ i i E E R 0 i 4/1/009 Interference of Light 3

33 Multiple beam interference III r 1 cos 1 r IR = and 4 Ii IT I 4 i 1 r r cosδ = + 1+ r r cosδ A minimum for reflected irradiance occurs when 1 cosδ = 0 cosδ = 1 δ = πm ( δ ) ( 1 r ) ( ) Δ= mλ = n tcos θ condition for minimum reflectance f t ( ) 1 r cosδ = 1 This also provides the condition for maximum transmission I = I I = I r r cos δ All the secondary and higher order reflections are in-phase with each other and they all have π phase difference with the first reflection. So they all cancel out with the first reflection. T i T i tt ' r 'E E0 = = 1 r for interface of air and glass n1 = 1, n = 1.5, r = 0.04 and 96% of the light is re E E 1 0 cancelled in the first reflection so ignoring the higher order reflections is perfectly justified. When cos δ = 1 the reflection maximum occurs. 1 1 δ = m+ π Δ= m+ λ = nftcos θt condition for maximum reflectance 4r 1 r and in that case I = I and I I = ( 1+ r ) 1+ r R i T i 4/1/009 Interference of Light 33

Chapter 7. Interference of Light

Chapter 7. Interference of Light Last Lecture Superposition of waves Laser This Lecture Two-Beam Interference Young s Double Slit Experiment Virtual Sources Newton s Rings Film Thickness Measurement by