Physics 1161 PreLecture 20 Interference textbook sections 28
Physics 1161: Pre-Lecture 20 Interference • textbook sections 28 -1 -- 28 -3
Superposition Constructive Interference +1 -1 t + +1 -1 In Phase t +2 t -2
Superposition Constructive Interference +1 -1 t + +1 -1 In Phase t +2 t -2
Superposition Destructive Interference +1 -1 t + +1 -1 Out of Phase t +2 t -2 180 degrees
Superposition Destructive Interference +1 -1 t + +1 -1 Out of Phase t +2 t -2 180 degrees
Interference for Light … • Can’t produce in phase (coherent) light from separate sources. (f 1014 Hz) • Need two waves from single source taking two different paths – Two slits – Reflection (thin films) – Diffraction* Single source Two different paths Interference possible here
Young’s Double Slit Concept At points where the difference in path length is 0, , 2 , …, the screen is bright. (constructive) d Single source of monochromatic light At points where the difference in path length is L 2 slitsseparated by d the screen is dark. (destructive) Screen a distance L from slits
Young’s Double Slit Key Idea L Two rays travel almost exactly the same distance. (screen must be very far away: L >> d) Bottom ray travels a little further. Key for interference is this small extra distance.
Young’s Double Slit Quantitative d d Path length difference = d sin q Constructive interference Destructive interference where m = 0, or 1, or 2, . . . Need l < d
Young’s Double Slit Quantitative L y d A little geometry… sin(q) tan(q) = y/L Constructive interference Destructive interference where m = 0, or 1, or 2, . . .
Thin Film Interference 1 2 t n 0=1. 0 (air) n 1 (thin film) n 2 Get two waves by reflection off of two different interfaces. Ray 2 travels approximately 2 t further than ray 1.
Reflection + Phase Shifts Incident wave Reflected wave n 1 n 2 Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change. • If n 1 > n 2 - no phase change upon reflection. • If n 1 < n 2 - phase change of 180º upon reflection. (equivalent to the wave shifting by /2. )
Thin Film Summary Determine d, number of extra wavelengths for each ray. 1 2 n = 1. 0 (air) n 1 (thin film) t This is important! n 2 Reflection Distance Ray 1: d 1 = 0 or ½ Ray 2: d 2 = 0 or ½ + 2 t/ lfilm If |(d 2 – d 1)| = 0, 1, 2, 3 …. If |(d 2 – d 1)| = ½ , 1 ½, 2 ½ …. Note: this is wavelength in film! (lfilm= lo/n 1) (m) constructive (m + ½) destructive
Thin Film Practice 1 2 t n = 1. 0 (air) nglass = 1. 5 nwater= 1. 3 Blue light (lo = 500 nm) incident on a glass (nglass = 1. 5) cover slip (t = 167 nm) floating on top of water (nwater = 1. 3). Is the interference constructive or destructive or neither? d 1 = d 2 = Phase shift = d 2 – d 1 =
Thin Film Practice 1 2 t n = 1. 0 (air) nglass = 1. 5 nwater= 1. 3 Blue light (lo = 500 nm) incident on a glass (nglass = 1. 5) cover slip (t = 167 nm) floating on top of water (nwater = 1. 3). Is the interference constructive or destructive or neither? d 1 = ½ Reflection at air-film interface only d 2 = 0 + 2 t / lglass = 2 t nglass/ l 0= 1 Phase shift = d 2 – d 1 = ½ wavelength
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