1. textbook sections 28-1 -- 28-3 Physics 1161: Pre-Lecture 26 Interference

2. Superposition t +1 t +1 t +2 -2 + Constructive Interference In Phase

3. Superposition t +1 t +1 t +2 -2 + Constructive Interference In Phase

4. Superposition t +1 t +1 t +2 -2 + Destructive Interference Out of Phase 180 degrees

5. Superposition t +1 t +1 t +2 -2 + Destructive Interference Out of Phase 180 degrees

6. Interference for Light … Can’t produce in phase (coherent) light from separate sources. (f  10 14 Hz) Single source Two different paths Interference possible here Need two waves from single source taking two different paths –Two slits –Reflection (thin films) –Diffraction *

7. Young’s Double Slit Concept Screen a distance L from slits Single source of monochromatic light d 2 slits- separated by d L At points where the difference in path length is 0,,2, …, the screen is bright. (constructive) At points where the difference in path length is the screen is dark. (destructive)

8. 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.

9. d Path length difference = d Young’s Double Slit Quantitative Destructive interference Constructive interference where m = 0, or 1, or 2,... d sin  Need  < d

10. d Destructive interference Constructive interference where m = 0, or 1, or 2,... Young’s Double Slit Quantitative y sin(  )  tan(  ) = y/L L A little geometry…

11. Thin Film Interference n 1 (thin film) n2n2 n 0 =1.0 (air) t 1 2 Get two waves by reflection off of two different interfaces. Ray 2 travels approximately 2t further than ray 1.

12. Reflection + Phase Shifts n1n1 n2n2 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.) Incident wave Reflected wave

13. Thin Film Summary n 1 (thin film) n2n2 n = 1.0 (air) t 1 2 Ray 1:  1 = 0 or ½ Determine  number of extra wavelengths for each ray. If |(  2 –  1 )| = ½, 1 ½, 2 ½ …. (m + ½) destructive If |(  2 –  1 )| = 0, 1, 2, 3 …. (m) constructive Note: this is wavelength in film! ( film = o /n 1 ) + 2 t/ film ReflectionDistance Ray 2:  2 = 0 or ½ This is important!

14. Thin Film Practice n glass = 1.5 n water = 1.3 n = 1.0 (air) t 1 2  1 =  2 = Blue light (  = 500 nm) incident on a glass (n glass = 1.5) cover slip (t = 167 nm) floating on top of water (n water = 1.3). Is the interference constructive or destructive or neither? Phase shift =  2 –  1 =

15. Thin Film Practice n glass = 1.5 n water = 1.3 n = 1.0 (air) t 1 2  1 = ½  2 = 0 + 2t / glass = 2t n glass / 0 = 1 Blue light (  = 500 nm) incident on a glass (n glass = 1.5) cover slip (t = 167 nm) floating on top of water (n water = 1.3). Is the interference constructive or destructive or neither? Phase shift =  2 –  1 = ½ wavelength Reflection at air-film interface only