Effects 6.2 Two-Slit Thin film is a general property of waves. A condition for is that the wave source is coherent. between two waves gives characteristic patterns due to constructive and destructive. For two waves to show they must have coherence. Two waves are coherent if one wave has a constant phase relation to the other coherent incoherent Light from two separate light bulbs is Incoherent Light from a single light bulb passing through a small slit is coherent x 2 x φ= π phase shift Laser light is coherent E δ=0 Constructive E δ= 2 Destructive Sum Distance -> distance 1
δ= Constructive Coherent waves barrier r 1 path difference path difference =δ =r 2 r 1 A In phase r 2 Superposition of waves at A shows due to path differences Condition for constructive Condition for destructive δ = m 1 δ = ( m + ) 2 Order number m m = 0 + 1, + 2,. pattern due to two spherical waves Amplitude on screen Destructive Constructive δ= δ=- δ=0 m=1 m=0 m=-1 pattern of water waves Coherent waves pattern Young s two slit experiment Is light a wave (Huygens) or a particle (Newton)? Path difference from the two slits Not to scale In the limit L>>d, the rays are nearly parallel Thomas Young perpendicular to r 1 and r 2 Light shows wave properties path difference δ= dsinθ 2
d Bright constructive Dark destructive pattern maxima m=2 m=1 θ m=0 Central maximum m= -1 m= -2 dsinθ = m bright dsin θ dark = (m + 1/ 2) m = 0, + 1, + 2,... Wavelength of light Light from a laser is passed through two slits a distance of 0.10 mm apart and is hits a screen 5.0 m away. The separation between the central maximum and the first bright fringe is 2.6 cm. Find the wavelength of the light. First maxima y m=1 d θ 0.10mm L 5.0 m 2.5cm for small dsinθ=m sinθ θ y y angles d m L L = solve for for m= 1 = yd 2 3 (2.6x10 m)(0.1x10 m) 7 = = 5.2x10 m = 520nm ml (1)(5.0m) Thin film Thin film In thin film is between light reflected from front an back surfaces of a thin film. The phase difference is due to two factors: Path difference through the film (corrected for the change in speed of light in the material) Phase shift due to reflection at the interface Phase shift due to reflection n 1 < n 2 phase shift=180 o Reflection with inversion phase shift = 180 o Phase shift due to reflection n 1 > n 2 Phase shift = zero Reflection without inversion Phase shift = zero 3
Thin film between light reflected from Top and bottom surfaces. For a film in air the phase difference due to reflection is 180 o. If the path difference (2t) is negligible then there is destructive. Destructive occurs when the path length difference equals integral multiples of the wavelength. Condition for destructive δ=2t= m film = m The wavelength in the film is shorter than in air. n m=0, 1, 2, 3. Thin film for a soap film in air For constructive the path difference (2t) must be half integral multiples of the wavelength to make up for the phase shift on reflection. Condition for constructive 1 1 δ=2t= (m + ) film = (m + ) 2 2 n M=0, 1, 2, 3 Soap film Question A vertical soap film displays a series of colored band due to reflected light. Find the thickness of the film at the position of the 5 th green band from the top (=550 nm, n =1.33) Constructive The 5th band has m=4 (the first is m=0) 1 2t = (m + ) 2 n 1 1 550nm t = (m + ) = (4 + ) = 930nm 2 2n 2 2(1.33) no coating anti-reflective coating Anti-reflective Coating Anti-reflective coatings are used to reduce reflections at the air-glass interface. n 1 Anti-reflective Coating n2 n3 t Anti-reflective coatings consists of a thin-layer of material with a refractive index in between that of air and glass. Destructive between light reflected at the two surfaces reduces the intensity of reflected light. What is the condition for destructive? n 1 =1.00 < n 2 < n 3 There is a phase shift of 180 o 1 2 t = ( m+ ) at both interfaces. 2 n2 The phase difference due to reflection is zero The path difference must be a half-integral number of wavelengths. 4
Question An anti-reflective coating of MgF 2 (n=1.38) is used on a glass surface to reduce reflections. Find the minimum thickness of the coating that can be used for green light (=550 nm). Optical compact disc For destructive 1 2 t = ( m+ ) minimum 2 n2 at m=0 Solve for t 550nm t = = = 100nm 4n 4(1.38) 1 2t = 2 n Quarter wavelength (in coating) thickness A CD stores information in a series of pits and bumps in the plastic. The information is read by a reflected laser beam. The intensity of the beam is changed by destructive of the reflected light t = 4n destructive 5