824.3 diffraction by a single slit and a circular aperture Diffracted the smaller the obstacle is the more evident the effect of diffraction is Huygens-Fresnel principle Huygens principle+ secondary wavelets mutually interfere 824.3 diffraction by a single slit and a circular aperture 2. Classification of diffraction O Fresnel (close field diffraction) Finite Finite distance Source obstacle distance screen Cor one of the distances is finite) 2 Fraunholfer (far field diffraction) infinite infinit distance distance source-obstacle screen (That is parallel rays of light 16
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
824.3 diffraction by a single slit and a circular aperture 3. The installation of single slit diffraction 824.3 diffraction by a single slit and a circular aperture 4. Theoretical analysis(wavelet zone method) Oif x=asine=0 Central bright fringe Totally destructive nterference ②ifAx1=S62 2 2 2 Ar;=兀 Mx, dextral axis--no or c=asin e= = First dark fringe Incident wave
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 ∆x = a sinθ = 0 Central bright fringe ∆ π λ π δ path = 1 = 2 x 2 sin 2 1 λ ∆ = θ = a 2if x ∆x = a sinθ = λ First dark fringe or 1 x 2 x ∆x ∆x1 §24.3 diffraction by a single slit and a circular aperture
824.3 diffraction by a single slit and a circular aperture Keep the a, increase 0 4r=asin 6=32/2 First bright fringe a/3 /3 、A a/3 31 difference between 824.3 diffraction by a single slit and a circular aperture ④ Keep the a, increase 6 sin 8 dx=asin 6=21 second dark fringe dr=asin 6=52/2 second bright fringe Ar= asing=3元 Incident third dark fringe
18 ∆x = a sinθ = 3λ / 2 First bright fringe 3 Keep the a, increase θ λ a θ θ θ λ / 2 3λ / 2 1 x 2 x a / 3 a / 3 a / 3 ∆x 2 λ §24.3 diffraction by a single slit and a circular aperture ∆x = a sinθ = 5λ / 2 second bright fringe ∆x = a sinθ = 3λ third dark fringe ∆x = a sinθ = 2λ second dark fringe 2 sin 4 λ θ = a 4 Keep the a, increase θ ∆x 1 x 2 x 3 x x4 x5 §24.3 diffraction by a single slit and a circular aperture
824.3 diffraction by a single slit and a circular aperture central bright fringe asin 0=(2m+1) 2 bright fringes m dark fringes (m=±1,±2,) 3元 32 52 sine 2 2a 2a 824.3 diffraction by a single slit and a circular aperture 5. Phasor diagram and the intensity of diffraction The phase difference of P adjacent wavelets zone 2兀 6=-(4ysin) If 60. then 元 6=-(4ysin6)=0 The the resultant amplitude is maximum central maximum Incident 19
19 ( 1, 2, ) 2 (2 1) 0 sin = ± ± L ⎪ ⎩ ⎪ ⎨ ⎧ = + m m a m λ λ θ central bright fringe bright fringes dark fringes I 0 sinθ a 2a 3 2 5λ λ − − a 2a 5 2 3λ λ §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 ( sin ) 0 2 = ∆ θ = λ π δ y The the resultant amplitude is maximum, central maximum. ∆y ∆y ∆y ∆y 1 x 2 x x3 x4 x5 5. Phasor diagram and the intensity of diffraction §24.3 diffraction by a single slit and a circular aperture
824.3 diffraction by a single slit and a circular aperture Eg(Em) Phasor for Phasor for op I bottom ray 2 If 0is very smallδ=-(4ysin6)≠0 824.3 diffraction by a single slit and a circular aperture 6=2(4ysin)↑ 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 Continue to increase 0, the chain of phasors begins to wrap back itself, and the resulting coil to shrink. b> First maximum
20 If θ is very small ( sin ) 0 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