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潭明
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1 波的干涉 ( Interference of Waves )

2 線性相加原理 ( Linear Superposition Principle ) 線性相加原理 ( Linear Superposition Principle ) y t ( total wave function ) = y 1 + y y N For scalar wave functions ( e.g. gas pressure in sound waves ) algebraic sum For vector wave functions vector sum 當兩個波相遇時產生干涉 ( interference ) constructive interference destructive interference 當它們分開時保持它們各自原有的波形 ( 兩個波動之間並不會相互造成彼此的改變 )

4 當兩個相同波速與波長的波同向前進時

5 駐波 ( Standing Waves ) Y 1 = A sin (kx wt) ( +x direction ) Y 2 = A sin (kx + wt) ( -x direction ) Y(x,t) = A sin (kx wt) + A sin (kx + wt) sina + sinb = 2 sin ( A + B ) cos ( A B ) 2 2 Y(x,t) = 2A cos wt sin kx - a stationary harmonic standing wave when kx = 0, p, 2p.., or x = 0, l/2, l, 3 / 2 l,. ( k = 2p/l ) y always = 0 節點 ( nodes ) when kx = p/2, 3 / 2 p, 5 / 2 p,.., or x = l/4, 3 / 4 l, 5 / 4 l,. y = 2A coswt 反節點 ( antinodes )

7 繩波的共振駐波 ( resonant standing wave on a string ) v ( 若繩子兩端被固定了 ) l 1 = (2/1) L, 1 st harmonic fundamental frequency : u 1 = (1/2) v/l (u = v / l ) l 2 = (2/2) L, 2 nd harmonic u 2 = (2/2) v/l = 2u 1 l 3 = (2/3) L, 3 rd harmonic u 3 = (3/2) v/l = 3u 1

8 n th harmonic of a resonant standing wave : l n = (2/n) L u n = (n/2) v/l l n 只能是某些固定的值, 但是只要改變波速 ( v ) 就能將頻率 ( u n ) 調整至任何的值 e.g. 改變吉他弦上的張力 ( tension ) 就能改變弦上的波速 ( v tension ) ( v = [ F / (M/L) ] 1/2 ) 就能改變頻率 u

10 displacement pressure displacement pressure

11 displacement

1016 = 3373 Hz Human ear responds to 3000 Hz

15 L = 2.54 cm L = l / 4 l = 4 L = m u = 343 / = 3373 Hz Human ear responds to 3000 Hz more efficiently Displacement pressure

16 兩個波源的干涉 ( Interference From Two Sources ) 當我們說兩個波源發出的兩個波同相 ( in phase ) 時 Y 1 = A 1 sin (kx wt + F 1 ) Y 2 = A 2 sin (kx wt + F 2 ) F 1 = F 2 假設有兩個點波源 ( point sources ), 發出的波具有相同的 w, k, A, 而且同相 x 1 x2 Y 1 ( from s 1 ) = A sin (kx 1 wt ) Y 2 ( from s 2 ) = A sin (kx 2 wt ) s 1 s 2

17 when x 2 x 1 ( path difference ) = nl, n = 0, 1, 2,.. kx 2 kx 1 = (2p/l) nl = 2np kx 2 = kx 1 + 2np y 2 ( from s 2 ) = A sin ( kx 1 + 2np wt ) = A sin ( kx 1 wt ) = y 1 ( from s 1 ) y = y 1 + y 2 = 2y 1 = 2A sin ( kx 1 wt ) or = 2y 2 = 2A sin ( kx 2 wt ) when x 2 x 1 ( 路徑差 ) = nl, n = 0, 1, 2,.. constructive interference

18 when x 2 x 1 ( path difference ) = ( n + 1/2 )l, n = 0, 1, 2,.. kx 2 kx 1 = (2p/l) ( n +1/2 ) l = 2np + p kx 2 = kx 1 + 2np + p y 2 ( from s 2 ) = A sin ( kx 1 + 2np + p wt ) = - A sin ( kx 1 wt ) = - y 1 ( from s 1 ) y = y 1 + y 2 = 0 when x 2 x 1 ( 路徑差 ) = ( n +1/2 )l, n = 0, 1, 2,.. destructive interference

21 光的雙狹縫干涉 ( Double Slit Interference of Light ) 每一個狹縫形同一個點波源 ( point source )

22 D D >> d s 1 p // s 2 p

23 D >> d s 1 p // s 2 p s 1 p s 2 p ( 路徑差, path difference ), d = d sinq d q Constructive interference when : d sinq = nl, n = 0, 1, 2,.. Destructive interference when : d sinq = (n + ½ ) l, n = 0, 1, 2,..

24 D d = d sinq constructive interference when : d sinq = nl, n = 0, 1, 2, d 相鄰建設型干涉亮點之間的夾角減小

25 d 相鄰建設型干涉亮點之間的夾角減小 If d >> l 一定夾角中會夾有太多干涉亮點很難再觀察到干涉條紋的存在 Ex. d = 1000 l, q = 1 o 1000 l sin1 o = nl n = 1000 sin1 o = 1000 x = intensity maxima within an angle of 1 o 1 o

26 西元 1801 年, Thomas Young 是如何證明光具有波動性? Thomas Young 利用單頻光 ( single-wavelength light, monochromatic light ) 通過雙狹縫, 觀察到干涉條紋, 因此證明了光具有波動性 Ex. l of visible light : 4000 Å ~ 7000 Å Å, Å ngstron = m d ~ a few m ( micron, micrometer, 10-6 m ) to enable observation of interference pattern on the screen

$Diffraction ) 惠更斯原則$ (Huygens' Principle) : 每一個波前

27 單狹縫繞射 ( Single Slit Diffraction ) 惠更斯原則 (Huygens' Principle) : 每一個波前 ( wavefront ) 的點形同另一個點波源 ( point source ) 當狹縫寬度 ( d ) 波長 ( l ) 當 d >> l what will happen when d > l? ~

29 what will happen when d > l? ~

30 D d A B C D E to point P 假設 d << D screen d 當滿足以下條件時, 屏幕上會出現暗帶 : CP EP ( path difference ) = l/2 AP EP = l d sinq = l 當滿足以下條件時, 屏幕上會出現另一個暗帶 : DP EP ( path difference ) = l/2 AP EP = 2l d sinq = 2l 繞射弱點發生在 : d sinq = nl, n = 1, 2, 3,.. sin q 1 = l/d, sin q 3 = 2l / d

31 what will happen when d > l? ~ 繞射弱點發生在 : d sinq = nl, n = 1, 2, 3,.. sin q 1 = l/d, sin q 3 = 2l / d

32 If d = l sin q 1 = 1, q 1 = 90 o 屏幕上沒有繞射弱點, 中央的繞射強點擴展至整個屏幕 If d < l 同樣的, 屏幕上沒有繞射弱點觀察不到繞射圖樣 If d < l 狹縫形同一個點波源 If d >> l 屏幕上, 在一定夾角內會有太多繞射條紋, 使得繞射條紋難以觀察, 與雙狹縫干涉的情況相似

34 解析力 ( Resolving Power ) 當 Dx ( Dq ) 什麼時候會無法分辨出這兩個繞射亮點? q 一個定義何時無法再分辨 S1 和 S2 的指標當其中一個波源在屏幕上的中心亮點與另一個波源在屏幕上的第一個繞射弱點重疊時 q Rayleigh Criterion of the limit of resolution s1 s2 Dx

35 一個定義何時無法再分辨 S1 和 S2 的指標當其中一個波源在屏幕上的中心亮點與另一個波源在屏幕上的第一個繞射弱點重疊時 ( Rayleigh Criterion of the limit of resolution ) sin q 1 = l / d q 1 ( in radians ) Dq = q 1 ( in radians ) = limiting angle of resolution l / d for high resolution need l and d

36 晶体的繞射現象 ( X-Ray Diffraction by Crystals ) : - Bragg Scattering scattered wave 當前進中的波遇到與其波長相當或稍大的物体時產生散射 ( scattering ), 形同一個新的點波源 Atomic size ~ 1-3 Å ( m ) inter-atomic separation in a crystal ~ a few Å to 10 Å l of x-ray : 0.1 ~ 10 Å ( 10-9 ~ m ) x-ray 會被晶体中的原子散射 ( visible light l : 4000 ~ 7000 Å )

37 晶体中的原子擁有週期性的規則排列晶格結構 ( lattice structure ) 當兩組反射波的路徑差剛好等於波長的整數倍時 : 2 d sinq = nl, n = 1, 2,.. ( Bragg condition ) 在反射角為 q 的方向上會觀察到建設型干涉的亮點假如我們將入射角 q 由 0 o 開始逐漸變大, 直到在反射角為 q 的方向上觀察到建設型干涉的亮點, 此時 d ( inter-atomic distance ) = nl / ( 2 sinq ), 原子間距就可以求得了

38 QUIZ #6 姓名學號

39 QUIZ #6 姓名學號 (1) Light of wavelength 6000 A is sent through a single slit. If the angular separation between adjacent diffraction minima is radians, what is the width of the slit? ( For small angles, sinq = q (in radians ))

40 QUIZ #6 姓名學號 (2) 西元 1801 年, Thomas Young 是如何證明光具有波動性? 他的實驗中最困難的部分是甚麼?

41 QUIZ #5 姓名學號 A 有一個脈衝 ( pulse ) 在 t = 0 的時候, 可以 y(x) = x 3 + B 來描述其波形, 若此一脈衝是以 5 m/s 的速率向正 x ( + x ) 方向移動, 請寫出其波函數 y ( x, t )

42 A f k, f s B ( f k or f s moves block A to the right also )

43 D A to point P screen d B C D E 假設 d << D d (1) 請利用上圖推導出單狹縫繞射弱點發生的條件 (2) 請解釋何謂 Rayleigh Criterion of the limit of resolution (3) 請利用 Rayleigh Criterion of the limit of resolution 解釋如何能夠提高單眼相機的解析度

投影片 1

投影片 1 Coherence ( ) Temporal Coherence Michelson Interferometer Spatial Coherence Young s Interference Spatiotemporal Coherence 參料 [1] Eugene Hecht, Optics, Addison Wesley Co., New York 2001 [2] W. Lauterborn,