1 1. Light waves §24.1 Light waves and the coherent condition of waves Light wave is a small part of the whole electromagnetic wave spectrum. 2 §24.1 Light waves and the coherent condition of waves 2. The interference phenomena of waves 3. Coherence and the conditions of coherence 1A superposition of waves may give rise to variations in the resulting amplitude of the total wave disturbance, known as interference. 2the conditions of coherence ?The same physical type of waves, and same direction of the oscillation; ? The same frequency; ?A phase difference that is independent of time. The phase difference is the difference between the individual phases of the two waves. §24.1 Light waves and the coherent condition of waves 3 4. The phase difference Simple harmonic oscillation )cos()( φω += tAtx A---the amplitude ω---the angular frequency φ---the initial phase ω t + φ ---phase T π ω 2 = §24.1 Light waves and the coherent condition of waves Sinusoidal(harmonic) waves )cos(),( φωΨ +?= tkxAtx λ π2 =k ---angular wave number §24.1 Light waves and the coherent condition of waves 4 )cos(),( )cos(),( 22 11 φωΨ φωΨ +?= +?= tkxAtx tkxAtx If two waves The phase difference of the two waves 1212 )()( φφφωφωδ ?=+??+?= tkxtkx )cos(),( )cos(),( 22 11 φωΨ φωΨ +?= +?= tkxAtx tkxAtx If two waves The phase difference of the two waves xtkxtkx ? λ π φωφωδ 2 )()( 12path =+??+?= §24.1 Light waves and the coherent condition of waves 5. The principle of superposition and interference of waves If the amplitudes are not too large, the total wave disturbance at any point x and time t is the sum of the individual wave disturbance. L+++= ),(),(),(),( 321 txtxtxtx ΨΨΨΨ )cos()cos( ),(),(),( 2111 21 φωφω +?++?= += tkxAtkxA txΨtxΨtxΨ )cos(),( 111 φω +?= tkxAtxΨ )cos(),( 212 φω +?= tkxAtxΨ ?Case 1 §24.1 Light waves and the coherent condition of waves 5 ) 2 cos() 2 cos(2 ] 2 )()( cos[ ] 2 )()( cos[2),( 2112 1 21 21 1 φφ ω φφ φωφω φωφω + +? ? = +??+? ? +?++? = tkxA tkxtkx tkxtkx AtxΨ ) 2 cos() 2 cos(2),( 2112 1 φφ ω φφ + +? ? = tkxAtxΨ ),2,1,0(0,)12( ),2,1,0(2,2 1 L L ==+= === nAn nAAn πφ? πφ? constructive destructive §24.1 Light waves and the coherent condition of waves 0=?φ πφ =? 32πφ =? ),( and ),( 2 1 txΨ txΨ ),( 1 txΨ ),( 2 txΨ ),( 1 txΨ ),( 2 txΨ ),( txΨ ),( txΨ ),( txΨ Ψ Ψ Ψ Ψ ΨΨ §24.1 Light waves and the coherent condition of waves 6 if 21 AA ≠ then ),2,1,0(,)12( ),2,1,0(,2 21 21 L L =?=+= =+== nAAAn nAAAn πφ? πφ? )cos()cos( ),(),(),( 2111 21 φωφω +?++?= += tkrAtkrA trΨtrΨtrΨ ?Case 2 )cos(),( )cos(),( 212 111 φω φω +?=Ψ +?=Ψ tkrAtr tkrAtr o 2 r 2 r 1 p o 1 §24.1 Light waves and the coherent condition of waves ])( 2 cos[) 2 cos(2 ] 2 )()( cos[ ] 2 )()( cos[2),( 12 12 1 21 21 1 φω φωφω φωφω +?+ ? = +??+? ? +?++? = trr kkrkr A tkrtkr tkrtkr AtrΨ ])( 2 cos[)( 2 cos2),( 21121 φω +?+?= trr k rr k AtrΨ rrr k ?=? λ π )( 2 12 The total amplitude depends on §24.1 Light waves and the coherent condition of waves Path difference 7 ),2,1,0(0, 2 )12( ),2,1,0(2, 1 L L ==+=? ===? nAnr nAAnr λ λ constructive destructive In another words: rrrk ?=?= λ π δ 2 )( 12path The phase difference ),2,1,0()12(, 2 )12( ),2,1,0(2, path path L L =+=+=? ===? nnnr nnnr πδ λ πδλ out of phase in phase §24.1 Light waves and the coherent condition of waves )cos(2 1221 2 2 2 1 φφ ?++= AAAAA 2211 2211 coscos sinsin arctg φφ φφ φ AA AA + + = )cos( 21 φω += += tA xxx 21 AAA rrr += A r 1 A r 2 A r ω ω ω x 1 x 2 x φ 1 φ 2 φ x Same frequency and same direction of motion )cos( 111 φω += tAx )cos( 222 ?ω += tAx §24.1 Light waves and the coherent condition of waves 8 )cos( 111 φω += tAx )cos( 222 ?ω += tAx 2 cos4]cos1[2 )]( 2 cos1[2 )](cos[2 path 22 1path 2 1 2 1 1221 2 2 2 1 21 δ δ λ π AA rA rrkAAAAA AA =+= ?+= ?++= = )cos()cos( )sin()sin( arctg 2211 2211 φφ φφ ? +++ +++ = krAkrA krAkrA )cos(),( )cos(),( 222 111 φω φω +?=Ψ +?=Ψ tkrAtr tkrAtr §24.1 Light waves and the coherent condition of waves 6. Wave intensity The power of transmitted by the wave []zytkxAv t E P dd)(sin d d d 222 φωωρ +?== The intensity of a wave is the average power transmitted by the wave through one square meter oriented perpendicular to the direction the wave is propagating. vA zyttkxvA Tzyzy P I T 22 0 222av av 2 1 ddd])[(sin 1 dd 1 dd d ωρ φωωρ = +?== ∫ §24.1 Light waves and the coherent condition of waves 9 222 2 1 AvAI ∝= r r ωρ 6. The methods of obtaining the coherent waves 1wavefront division 2amplitude division §24.1 Light waves and the coherent condition of waves For light oscillation: )cos( 11 φω += tEE 2 1 2 2 1 2 1 EvEI ∝= r r ωρ §24.2 Young’s double slit experiment 1. Installation and the phenomena of experiment This is a typical method of wavefront division. 10 2. Theoretical analysis )cos( ),( )cos( ),( 20 2 10 1 φω φω +?= +?= tkrE trE tkrE trE Path length difference θ? sin 21 drrr =?= Two light oscillations at point P §24.2 Young’s double slit experiment Separation of the slits If D>>d r? The condition for constructive interference (bright fringe or maximum) and destructive interference (dark fringe or minimum)on the distant screen is Phase difference θ λ π ? λ π δ sin 22 path dr == §24.2 Young’s double slit experiment 11 Where m is an integer that can take on the values 0, ±1, ±2, ±3, nullnullnull. The absolute value of m is the order of interference. Bright fringes Dark fringes 2 )1( sin λ λ θ? + == m m dr Bright fringes Dark fringes π π θ λ π δ )1( 2 sin 2 path + == m m d §24.2 Young’s double slit experiment Min Min Min Min Min Min Max Max Max Max Max If θ is small, θθ ≈sin Then the multiple fringes will be uniformly spaced on the screen. §24.2 Young’s double slit experiment 12 3. the space of the adjacent fringes d D d D mmy d D m d m Dy md DDy λ λ ? λλ λθ θθ = ?+= == = ≈= ])1[( sin sintan ↑↓ yd ?, §24.2 Young’s double slit experiment 0=m 1+=m 1?=m 2+=m 3+=m 1+=m 2?=m 3?=m 4?=m 5?=m 5+=m §24.2 Young’s double slit experiment 13 S* Dy ∝? d y 1 ∝? For specified λ: λ? ∝y purplered yy ?>?For specified D and d: For white light: (a)m=0, fringe is white; (b)the other orders, fringes are colorful; (c)the fringes overlap each other. §24.2 Young’s double slit experiment Overlapping condition? Example 1: Suppose that Young’s experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits. How far apart are the bright fringes? Solution: mm25.2 1020.1 4.510500 ])1[( 3 9 = × ×× == ?+= ? ? d D d D mmy λ λ ? λθ θθ md DDy = ≈= sin sintan d D m d m Dy λλ == §24.2 Young’s double slit experiment 14 Example 2: In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.0 m from 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 600 nm. What is the separation on the screen between the third order (m =3 ) bright fringes of the two interference pattern? Solution: m102.7)()()( tan sin,sin 5 12 12 ? ×=?=?= =≈= ≈== d m d m Dyyy d m DDDy d m md λλ λλ? λ θθ θ λ θλθ §24.2 Young’s double slit experiment §24.2 Young’s double slit experiment Example 3: S 1 and S 2 are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d=4.00 m, and they emit at the same power. If a detector is moved to the right along the x axis from source S 1 , at what distances from S 1 are the first three interference maxima detected? 15 Solution: For constructive interference, the path difference λmxxd =?+ 22 Solve the equation gives λ λ m md x 2 222 ? = 1.17m3, 3m2,m;5.7,1 == ==== xm xmxm P x §24.2 Young’s double slit experiment §24.3 diffraction by a single slit and a circular aperture 1 Huygen’s principle Each point on a wavefront acts as a source of a wavelet that propagates outward from the point; The new position of the wavefront is determined by the envelope of the wavelet from the individual points on the original position of the wavefront. 1. Single slit diffraction 16 the smaller the obstacle is, the more evident the effect of diffraction is. 2 Huygens-Fresnel principle Huygen’s principle + secondary wavelets mutually interfere §24.3 diffraction by a single slit and a circular aperture 1Fresnel（close field diffraction）： 2Fraunholfer（far field diffraction）: 2. Classification of diffraction （or one of the distances is finite） source —— obstacle —— screen Finite distance Finite distance (That is parallel rays of light) source —— obstacle —— screen infinite distance infinite distance §24.3 diffraction by a single slit and a circular aperture 17 3. The installation of single slit diffraction §24.3 diffraction by a single slit and a circular aperture 4. Theoretical analysis (wavelet zone method) 1if 0sin == θ? ax Central bright fringe π? λ π δ == 1path 2 x 2 sin 2 1 λ θ? == a x 2if λθ? == sinax First dark fringe or 1 x 2 x x? 1 x? §24.3 diffraction by a single slit and a circular aperture 18 2/3sin λθ? == ax First bright fringe 3 Keep the a, increase θ λ a θ θ θ 2/λ 2/3λ 1 x 2 x 3/a 3/a 3/a x? 2 λ §24.3 diffraction by a single slit and a circular aperture 2/5sin λθ? == ax second bright fringe λθ? 3sin == ax third dark fringe λθ? 2sin == ax second dark fringe 2 sin 4 λ θ = a 4 Keep the a, increase θ x? 1 x 2 x 3 x 4 x 5 x §24.3 diffraction by a single slit and a circular aperture 19 ),2,1( 2 )12( 0 sin L±±= ? ? ? ? ? += m m ma λ λ θ central bright fringe bright fringes dark fringes I sinθ 0 aa 2 3 2 5 λλ ?? aa 2 5 2 3 λλ §24.3 diffraction by a single slit and a circular aperture )sin( 2 θ? λ π δ y= The phase difference of adjacent wavelets zone If θ=0, then 0)sin( 2 == θ? λ π δ y The the resultant amplitude is maximum, central maximum. y? y? y? y? 1 x 2 x 3 x 4 x 5 x 5. Phasor diagram and the intensity of diffraction §24.3 diffraction by a single slit and a circular aperture 20 If θ is very small 0)sin( 2 ≠= θ? λ π δ y §24.3 diffraction by a single slit and a circular aperture ↑= )sin( 2 θ? λ π δ y Increase θ, Continue to increase θ, the chain of phasors begins to wrap back itself, and the resulting coil begins to shrink. First maximum. The chain of phasors curls completely around so that the head of the last phasor just reaches the tail of the first phasor. The resultant amplitude is zero. First minimum. §24.3 diffraction by a single slit and a circular aperture 21 φ φ φ φ θ θ 2 1 sin )21( 22 1 sin m m E E R E R E = = = 2 2 2 ) sin ()( )( α α θ θ θ m mm II E E I I = = )sin( 2 2 θ λ π φ φ α a= = §24.3 diffraction by a single slit and a circular aperture 2 2 2 )sin( )sin(sin ) sin ()( θ λ π θ λ π α α θ a a III mm == I /Im sinθ 0 a λ 47.1 a λ 46.2 a λ 47.1? a λ 46.2? §24.3 diffraction by a single slit and a circular aperture 22 6.The angular width of the bright fringes L、、21 ±±=m central bright fringe bright fringes dark fringes a a m m 2 )12( 0 sin λ λ θ ? ? ? ? ? ? ? += a λ θθθ? 2 11 =?= ?+ Central bright fringe a mm λ θθθ? =?= +1 Other bright fringe §24.3 diffraction by a single slit and a circular aperture θtgfy= )tg(tg 12 θθ? ?= fy θ?θθ? ?=?= ffy )( 12 7. The line width of the bright fringes f a y ?= λ ? 2 Central bright fringe f a y ?= λ ? Other bright fringe f o y θ 2 L §24.3 diffraction by a single slit and a circular aperture 23 For specified λ: ↓a ↑θ? ↑a ↓θ? conspicuous diffraction unconspicuous diffraction For specified a: ↑λ ↓λ ↑θ? ↓θ? bright fringes appear purple close to center; red away from center for white light. The fringes become density if the installation immerged in water. §24.3 diffraction by a single slit and a circular aperture Passing white light through a single slit produces this diffraction patterns. §24.3 diffraction by a single slit and a circular aperture 24 θ?? sinsin aa += θ?? sinsin aa ?= a ? θ a ? θ L321 ±±±=、、m 0 λm 2 )12( λ +m=±= θ?? sinsin aa Central bright fringe bright fringes dark fringes §24.3 diffraction by a single slit and a circular aperture 8. Diffraction by a circular aperture and the resolution S * 1installation 2diffraction patterns Central circular bright spot surrounded by a series of increasingly dimmer concentric circle of light. §24.3 diffraction by a single slit and a circular aperture 25 3 Arrey spot λθ 220.1sin 1 =a The number 1.220 arises from a special class of functions, called Bessel function. The half angular width: a λ θ 220.1 1 = 1 θ §24.3 diffraction by a single slit and a circular aperture 4Rayleigh criterion Two incoherent, monochromatic, pointlike, distant sources each produce their own diffraction pattern. φφ λθ 220.1sin 1 =a The angular position of the first minimum of the diffraction pattern of a circular aperture §24.3 diffraction by a single slit and a circular aperture 26 What is the minimum separation angle to distinguish that there two diffraction patterns present? Rayleigh criterion: When two central peak of one diffraction pattern is located at the position of the first minimum of the other diffraction pattern. min φ min φ §24.3 diffraction by a single slit and a circular aperture That means λθθ θφφ 220.1sin 11 1min =≈ == aa Define the power of resolution λθ a ? ? = 221 11 1 when 1 θφ < , the two source can not be resolved. 1 θφ < §24.3 diffraction by a single slit and a circular aperture 27 §24.3 diffraction by a single slit and a circular aperture §24.4 The diffraction grating 1. The double slit revisited We must revisited the double slit to account for the simultaneous interference effect between the slit and the diffraction effects of each slit. When both slits are illuminated by monochromatic light, the slit sources are coherent and the light from one slit can also interfere with the light from the other slit. The double slit interference pattern, with equally spaced dark and bright fringes will be modulated by the envelope of the single slit diffraction pattern. 28 §24.4 The diffraction grating §24.4 The diffraction grating 29 §24.4 The diffraction grating §24.4 The diffraction grating 30 d fixed, a decreased: λθ = 1 sina §24.4 The diffraction grating a fixed, d decreased: λθ md = 1 sin §24.4 The diffraction grating 31 2. Multiple slits: the diffraction grating 1installation Many equally spaced and parallel slits are called diffraction grating. §24.4 The diffraction grating P 2grating equation §24.4 The diffraction grating 32 φ φ φ φ θ θ 2 1 sin 2 sin 2 sin2 2 1 sin2 N EE N RE RE = = = 2 2 2 ) sin sin )(( )( β β θ θ θ N II E E I I == )sin( 2 θ λ π φ d= §24.4 The diffraction grating let 2 φ β = θ E E φ φ φ φ R φ 2 2 2 )sin( )sin(sin ) sin ()( θ λ π θ λ π α α θ a a III mm == 22 ] )sinsin( )sinsin( [] )sin( )sinsin( [ θ λ π θ λ π θ λ π θ λ π d dN a a II m = §24.4 The diffraction grating 2 ) sin sin )(( β β θ N II = (a)Regardless the diffraction of the single slit λθ md =sin (1) ),2,1,0( L±±== mmπβ )( sin sin lim 2 θ β β πβ ININ N m == → )sin( 2 θ λ π φ d= 2 φ β = grating equation 33 (2) ),2,1,0( L±±==≠ kkNm πβπβ )sin( 2 θ λ π φ d= 2 φ β = )(sin)sin( mNk N k d N k d ≠=== λθπθ λ π β L LL ),12(),12( ),2(),1(),1(,,2,1 +±?± +±+±?±±±= NN NNNk Between two interference maxima there are N-1 interference minima. §24.4 The diffraction grating §24.4 The diffraction grating (3)Between the two minima is the secondary interference maximum, there are N-2 secondary interference maximum between two interference maxima. 34 §24.4 The diffraction grating §24.4 The diffraction grating Experiment result: Qualitative discussion: λθλθ 2sin,sin 21 == dd 35 λθ?θ λθ?θ λθ?θ λθ?θ λθ?θ 9.1)sin( 75.1)sin( 5.1)sin( 25.1)sin( 1.1)sin( 51 41 31 21 11 =+ =+ =+ =+ =+ d d d d d Between the θ 1 andθ 2 : Only the few slits left over from all this pairwise cancellation contribute a small amount of light to the maximum.Very narrow and bright fringes remain for the orders of the given wavelength. §24.4 The diffraction grating θ?θ + λ5.5 §24.4 The diffraction grating (b) Consider the single slit diffraction (1)The intensity of the fringes are not uniform; (2)Some interference maxima orders can be disappear. when )veconstructi,2,1,0(sin )edestructiv,3,2,1(sin L L = ′′ = == mmd mma λθ λθ m a d m m m a d = ′ ′ = orthen The will disappear.thm ′ 22 ] )sinsin( )sinsin( [] )sin( )sinsin( [ θ λ π θ λ π θ λ π θ λ π d dN a a II m = 36 §24.4 The diffraction grating 3the angular width of the interference maximum Similarly to a single slit λθ =sina NNd Nd 1 min min ∝= = λ θ λθ θ θ The angular width of the maximum is 2θ min . λθ = min sinNd For a grating, the central interference maximum §24.4 The diffraction grating 37 For mth interference maximum, we can project width of the illuminated grating, as the effective width of the aperture θcosNd θ λ θ cos min Nd = ④resolution and angular dispersion of a diffraction grating If the illuminating light include wavelength λ 1 and λ 2 , the two interference patterns may superimpose each other, how to distinguish the two diffraction patterns? §24.4 The diffraction grating λ?θ? θλ θ λ θ θ d m d m d m md =?≈= = cosd d d d cos Derivative θ with respect to λ, we can get the angular dispersion According to the Raleigh criterion λ? λ λ θθ? d m Nd Nd ≈ == min λθ md =sinRecalling the grating equation §24.4 The diffraction grating 38 White light §24.4 The diffraction grating The resolving power of the diffraction grating Nm= λ? λ The zeroth, first, second and fourth orders of the visible emission lines from Hydrogen. §24.4 The diffraction grating 39 §24.4 The diffraction grating 3. X-ray diffraction Bragg’s law )3,2,1(sin2 L== mmd λθ Example 1: two slit, each of width 0.20 mm, are separated by 0.70 mm and illuminated with light of wavelength 633 nm in a Young’s double slit experiment. How many bright interference fringes appear under the central diffraction envelope? Solution: λθλθ == 11 sin,sin amd 5.3 20.0 70.0 , === mm a d The seven bright fringes appear under the central diffraction envelope, m=0 ,±1, ±2 ±3. §24.4 The diffraction grating 40 §24.4 The diffraction grating Example 2: A three-slit grating has separation d between adjacent slits. If the middle slit is covered up, will the half-width of the intensity maxima become broader or narrower and by what factor? dNd 3 1min λλ θ?θ === Solution: 4 3 1 2 = θ? θ? ddN NNdd 4)2( 23,2 2 λλ θ? ==∴ =→=→Q §24.4 The diffraction grating Example 3: A grating with 2000 slits spaced equally over a 1.000 cm distance is used to analyze the spectrum of the element mercury. Among other wavelengths, mercury emits light at 576.959 nm and 579.065 nm. What is the angular separation of the two wavelengths in the second-order spectrum? Solution: λ?θ? θλ θ λ θ θ d m d m d m md =?≈= = cosd d d d cos λθ md =sin Grating equation 41 o 04961.0 (rad)10658.8 10)959.576065.579( )2000/10000.1( 2 4 9 2 = ×= ×? × = = ? ? ? λ?θ? d m §24.4 The diffraction grating Another way: λθ md =sin d m d m λ? θθθ? θ λ θ λλ =?= ≈= 12 sin §24. 5 thin-film interference 1. The index of refraction and the speed of light The index of refraction of a medium is the ratio of the speed of light in vacuum to its speed in the transparent medium v c n = The frequency of the light is the same in both medium. 11 11 , λ λ νλ =∴= v c vQ 42 then 1 1 n λ λ = §24. 5 thin-film interference 2. Equivalent path difference and phase difference xnx ?=?= λ π λ π δ 22 1 path πδ 4 path = πδ 4 path = x? Equivalent path = geometrical path × index of refraction 3. thin-film interference--amplitude division Medium n 1 , n 3 Reflect light 2 and 3 is coherent Transmission light 4 and 5 is coherent p p′ d s λ A B C D 1 n 2 n 3 n b d h f 1 2 3 5 4 i a c γ Thin-film n 2 , d Incident light λ, i ,γ §24. 5 thin-film interference 43 2 )( differencepathequivelent 12 λ +?+= adnbcabn 2 sin2 22 1 2 2 λ +?= innd differencepathequivelent 2 path λ π δ = Phase difference ? ? ? =+ = = ve)(destructi)2,1,0()12( ive)(construct)2,1,0(2 L L mm mm π π Phase changes upon reflection must be done on a case-by-case basis §24. 5 thin-film interference 2 sin2 2 )( differencepath 22 3 2 2 32 λ λ +?= +?+= innd bhncfbcn p p′ d s λ A B C D 1 n 2 n 3 n b d h f 1 2 3 5 4 i a c γ differencepathequivelent 2 path λ π δ = ? ? ? =+ = = ve)(destructi)2,1,0()12( ive)(construct)2,1,0(2 L L mm mm π π Phase difference §24. 5 thin-film interference 44 1 n 2 n 3 n 1phase changes upon reflection 2321 , nnnn << 2321 , nnnn >> Reflection light has phase change 321 nnn >> 321 nnn << transmission light has phase change 2if i=0, 2 2 2 λ += dn Path difference ) 2 (2 2 2path λ λ π δ += dnPhase difference §24. 5 thin-film interference 4. wedge thin-film θ n λ 2 2 2 sin2 22 1 2 2 λ λ += +?= nd innd Path difference ? ? ? ? ? + = 2 )12( λ λ m m ? ? ? + = fringesdark)12( fringesbright2 path π π δ m m θ l d m d m+1 ?d §24. 5 thin-film interference 45 θ l d m d m+1 ?d n ddd mm 2 1 λ ? =?= + The thickness difference between adjacent bright(dark) fringes θ λ θ λ θ nn d l 2sin2sin ≈= ? = The fringes width (space between adjacent dark fringes ) §24. 5 thin-film interference The interference fringes trace out contours of constant thickness of the wedge thin-film. §24. 5 thin-film interference Soap film—white light 46 Example 1: A broad beam of red light, with wavelength λ=632.8 nm, incidences on an air wedge as shown in figure. An observer looking down on the the block with the air wedge sees an interference pattern consisting of six dark fringes and five bright red fringes along the wedge. What is the change in thickness ?T(=T R -T L )along the wedge? §24. 5 thin-film interference Solution: 2 )12( 2 2 λλ +=+ mnT At the left and right ends, dark fringe implies the destructive interference n m T n m T R R L L 2 , 2 λλ == nn m n m TTT LL LR λλλ 2 5 22 )5( =? + =?=? (m)1058.1 00.1 108.632 2 5 6 9 ? ? ×= × =?T §24. 5 thin-film interference 47 5. Newton’s rings ? ? ? ? ? + =+= 2 )12( 2 2d.p. λ λ λ m m nd Phase difference ) 2 (2 2 path λ λ π δ += nd ? ? ? =+ = = L L ,2,1,0)12( ,2,1,02 mm mm π π Bright rings Dark rings Newton’s rings traced out contours of constant thickness of the air gap between the surfaces. d Glass r R Glass §24. 5 thin-film interference 222 222 2 )( dRdRr dRrR +?+= ?+= R r d 2 2 ≈ therefore ? ? ? ? ? ? ? ? = ringsdark ringsbright 2 )12( n mR n Rm r λ λ The radius of the Newton’s rings: d Glass r R Glass §24. 5 thin-film interference 48 §24. 5 thin-film interference 6. Iridescence of feather and butterfly wing §24. 5 thin-film interference 49 Iridescence of a morpho butterfly wing Thin teraces of transparent cuticle-like material §24. 5 thin-film interference Example 2: a glass lens is coated on one side with a thin film of magnetium fluoride(MgF 2 ) to reduce reflection from the lens surface. The index of refraction of MgF 2 is 1.38; that of the glass is 1.50. What is the least coating thickness that eliminates the reflections at the middle of the visible spectrum(λ=550nm)? Assume that the light is approximately perpendicular to the lens surface. §24. 5 thin-film interference 50 Glass n 3 =1.50 MgF 2 n 2 =1.38 n 1 =1.00 2 2 2 sin2 2 22 1 2 2 λ λ += +?= dn innd ? ? ? ? ? + = 2 )12( λ λ m m Path difference )nm(6.99 38.14 550 4 2 )12(2 2 min 2 = × == += n d mdn λ λ m=0, d=d min §24. 5 thin-film interference Glass n 3 =1.50 MgF 2 n 2 =1.38 n 1 =1.00 m=1 )nm(9.298 38.14 5503 4 3 2 )12(2 2 2 = × × == += n d mdn λ λ )nm( 825108.29838.122 9 2 mmm dn = ××× == ′ ? λ Constructive interference m=2 )nm(5.412 2 825 ==′λ Blue-purple §24. 5 thin-film interference 51 Only transverse waves can exhibit polarization. 1. Polarization phenomena and Polarized light §24.6 Polarization of light x y z 2. Five polarized states of light (1) Unpolarized light The light wave emitted during a very short interval from an individual atom is polarized. Atoms in a typical light source all act independently of each other. §24.6 Polarization of light 52 Any plane polarized wave can be resolved into two components along two mutually perpendicular directions chosen. We can model unpolarized light by two mutually perpendicular, collective, incoherent oscillations. How can the unpolarized light from a typical light source be polarized? §24.6 Polarization of light ?????? (2)Polarize by dichroic Some special material can absorb the one oscillation of light. If the oscillations of the transverse wave at a particular location are confined to a line, the wave is plane polarized. §24.6 Polarization of light 53 §24.6 Polarization of light (3)Partial polarized light If the part of one oscillation is absorbed, this kind of light is called partial polarized light. ????????? ? ? ? (4) Circular polarized light Other types of polarized light can be created by judicious superposition of plane polarized light. jtkxEE ? )cos( 0 ω?= r Light vector: jtkxEE ? )cos( 01 ω?= r ktkxEE ? ) 2 cos( 02 π ω +?= r Such light is said to be right circularly polarized §24.6 Polarization of light ] ? )(sin ? )[(cos 0 21 ktjtE EEE ωω += += rrr If we look at x=0, then (4)Elliptic polarized light 54 If jtkxEE ? )cos( 1101 ?ω +?= r ktkxEE ? )cos( 2202 ?ω +?= r Then according to different we will get different results. 12 ??? ?=? Generally the result is a ellipse. j ? j ? j ? k ? k ? k ? §24.6 Polarization of light jtkxEE ? )cos( 01 ω?= r ktkxEE ? )cos( 02 πω ??= r if ) ?? )(cos( 0 21 kjtkxE EEE ??= += ω rrr then Why the two waves do not interfere? This is a plane polarized light. §24.6 Polarization of light j ? k ? 55 3. Polarized by absorption and Malus’s law Unpolarized light can be polarized by selective absorption in passing through a polarizing sheet such as a Polaroid. Half the light intensity is absorbed by the polarizing material and half is transmitted with a plane polarization parallel to the axis of the material. §24.6 Polarization of light θcosEE y = θ 2 2 2 0 cos== E E I I y θ 2 0 cosII = Malus’s law analyzer §24.6 Polarization of light 56 4. Polarization by reflection: Brewster’s law ....... 1 2 tan n n B =θ Brewster’s law n 1 n 2 ? ? ? ? ? ? ? ? ? ? ? θ θ 2 θ n 1 n 2 ? ? ? ? ? B θ B θ ? ? ? ? ? ? 2 θ o 90 §24.6 Polarization of light The special angle of incidence that results in a reflected beam that is completely plane polarized is called Brewster angle. The refracted beam makes a 90o angle with this reflected beam. o 90 2 =+θθ B The Brewster angle for an air-water interface: 00.1 33.1 tan 1 2 == n n B θ o 1.53= B θ The Brewster angle for an glass-air interface: 50.1 00.1 tan 1 2 == n n B θ o 3.56= B θ §24.6 Polarization of light 57 ? . Natural light i 0 ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? Polarized light Polarized light Stack of glass plates §24.6 Polarization of light §24.6 Polarization of light 58 5. Polarization by double refraction(birefringence) When an unpolarized light ray is incident on the crystal in certain directions, two refracted rays result! This phenomenon is called double refraction. Ordinary ray(O) follows the law of refraction; Extraordinary ray(E) does not follow the law of refraction. The two refraction rays are completely plane polarized along two perpendicular directions. The advantages: P1132-1133 §24.6 Polarization of light