824.2 Youngs double slit experiment m bright fringes Ar=dsin= (m+1)Dark fringes 2 2丌 2m兀 Bright fringes d sine (m+I)T Dark fringes Where m is an integer that can take on the values0,士1,士2,士3,“ The absolute value of m is the order of terfe 824.2 Youngs double slit experiment If @is small sin6≈6 Max Then the multiple fringes Min will be a.+Max uniformly Min aced on the screen
11 Where m is an integer that can take on the values 0, ±1, ±2, ±3, xxx. The absolute value of m is the order of interference. Bright fringes Dark fringes 2 ( 1) sin λ λ ∆ θ + = = m m r d Bright fringes π Dark fringes 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, sinθ ≈ θ Then the multiple fringes will be uniformly spaced on the screen. §24.2 Young’s double slit experiment
824.2 Youngs double slit experiment 3. the space of the adjacent fringes y=Dtan6≈Dsin6 d sin b=ma D ncident y=D 4=I(m+1)-m D D V,4个 824.2 Youngs double slit experiment m=+ n m=-5
12 3. the space of the adjacent fringes d D d D y m m d D m d m y D d m y D D λ λ ∆ λ λ θ λ θ θ = = + − = = = = ≈ [( 1) ] sin tan sin d ↓, ∆y ↑ §24.2 Young’s double slit experiment m = 0 m = +1 m = −1 m = +2 m = +3 m = +1 m = −2 m = −3 m = −4 m = −5 m = +5 §24.2 Young’s double slit experiment
824.2 Youngs double slit experiment For specified a:!y∝D4 For specified D and d: Ay oc AAyred >Ay purple For white light: a)0, fringe is white (b)the other orders fringes are colorful (c)the fringes overlap each other. Overlapping condition? 824.2 Youngs double slit experiment Example 1: Suppose that Youngs 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: y=Dtan 8 a Dsin 8 Ay=[(m+1) d sin=m1 D500×10-×54 J=D m2 D d1.20×10-3 =2.25mm
13 S* ∆y ∝ D d y 1 For specified λ: ∆ ∝ ∆y ∝ λ red purple For specified D and d: ∆y > ∆y 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: 2.25mm 1.20 10 500 10 5.4 [( 1) ] 3 9 = × × × = = = + − − − d D d D y m m λ λ ∆ θ λ θ θ d m y D D = = ≈ sin tan sin d D m d m y D λ λ = = §24.2 Young’s double slit experiment
824.2 Youngs double slit experiment 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 the separation on the screen between the third order(m=3)bright fringes of the two interference pattern Solution: dsin 0=mn, sin 8 m≈6 y= Dane≈De=Dm 4y=y(4)-m(n)=D(m2-m)=72×10m 824.2 Youngs double slit experiment Example 3: S, 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,, at what distances from S, are the first three interference maxima detected?
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: ( ) ( ) ( ) 7.2 10 m tan sin , sin 2 1 5 2 1 − = − = − = × = ≈ = = = ≈ d m d m y y y D d m y D D D d m d m λ λ ∆ λ λ λ θ θ θ λ θ λ θ §24.2 Young’s double slit experiment §24.2 Young’s double slit experiment Example 3: S1 and S2 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 S1, at what distances from S1 are the first three interference maxima detected?
824.2 Youngs double slit experiment Solution: For constructive interference, the path difference d+x-x=mn Solve the equation gives d2-m222 2ma 1,x=7.5m;m=2,x=31 m=3,x=1.17m 824.3 diffraction by a single slit and a circular aperture 1. Single slit diffraction O 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 at time f original position of the wavefront
15 Solution: For constructive interference, the path difference d + x − x = mλ 2 2 Solve the equation gives λ λ m d m x 2 2 2 2 − = 3, 1.17m 1, 7.5m; 2, 3m = = = = = = m x m x m x 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