Chapter 37 Interference

1. Chapter 37 Interference PHYS 2326-34

2. Concepts to Know • Interference • Principle of Superposition • Monochromatic Light • Coherent Light • Antinodal Curves (Constructive Interference) • Nodal Curves (Destructive Interference) • Interference Fringes • Phase Shift (Reflection off Slow Medium) • Thin Films

3. Inteference • Like sound waves, light waves can produce interference • Nodes exist where there are minima (dark areas) from destructive interference • Antinodes exist where there are bright areas from constructive interference

4. With sound, we discussed the interference using examples having only one frequency and phase such as a pipe that resonates. • To see such effects with light, we need to have one frequency as well. Otherwise the effect is hidden by too much going on.

5. Monochromatic Light • Monochromatic light simply means one color – which is 1 frequency or wavelength • Lasers produce monochromatic light very well • Sodium vapor lights do a fair job although they are producing bichromatic light but there are a variety of arc lamps which are monochromatic • Lightbulbs produce a wide range of wavelengths

6. Coherent Light • Coherent light refers to light having a consistent phase between sources. • From our study of sound, we determined that phasing between PA system speakers could have negative effects on one’s hearing at some locations • In order to see interference effects we must have consistent phase.

7. Conditions for Interference • Sources must be coherent – have a constant phase • Sources must be monochromatic, a single wavelength

8. Young’s Double-Slit Experiment • Performed by Thomas Young in 1800 • It showed Huygens’s principle of subdividing wave fronts into new wavelets or wave fronts to be correct

9. S1 and S2 are narrow slits separated by distance d. Distance δ is the difference in path lengths r1 and r2 and θ is the angle from the normal at the slit center line and y is the offset distance at the screen • When δ is a multiple of wavelength there is constructive interference r1 y S1 r2 θ d S2 δ L

10. d sin θbright = mλ • d sin θdark = (m+ ½) λ • since tan θ = y/L • ybright = L tan θbright • ydark = L tan θdark • for small angles where θ ~ sin θ • ybright = L (m λ /d)

11. Phase Shift from Reflection • Section 37.5 refers to Lloyd’s mirror where our two sources are one source plus a reflected image of the source • When an electromagnetic wave is reflected from the interface to a medium with a higher index of refraction than the one in which it is traveling, it undergoes a 180 degree phase reversal.

12. When a reflection occurs at a surface where the index of refraction is less than that of the medium the wave is traveling in, there is NO phase reversal at the reflection • This corresponds to a ½ wavelength displacement shift or a ½ integer shift

13. Thin Films • Thin films such as oil on water or soap bubble surfaces exhibit interference effects by showing varied colors when white light is incident on the films. • This occurs because the reflection from the top surface has a 180 degree phase shift and the second surface at some distance t does not have this phase shift and the total effective path difference becomes a multiple of certain wavelengths • 2t= (m+1/2)λn where λn = λ /n (index of refraction) • 2nt = (m+1/2) λ for constructive interference and • 2nt = m λ for destructive interference (eqn 37.15-17)

14. Color & Wavelength 500nm 600nm 400nm 700nm blue green red • Primary colors are red green and blue and have a range of wavelengths for each. Adding any combination creates other colors. Often called RGB • Cyan magenta and yellow are combinations of two primary colors • Wavelengths are shown in nanometers (10-9 meters) sometimes these are shown in Angstroms which is 10-10 yellow cyan magenta

15. Example Problem 1 In a double slit interference experiment, the slits are 10 micron (10-6 meters) apart and the screen is 2 meters away. If 500nm wavelength light is used, find a) the location of the first dark fringe, b) the location of the 3rd bright fringe, c) the spacing between fringes, d) the theoretical number of bright fringes possible.

16. a) d sin θ = (m+1/2)λ, 1E-5 sin θ =(1/2) 5E-7 θ = 1.43° y = L tan θ = 2.0 tan(1.43°) = 0.05m b) d sin θ = (m)λ, 1E-5 sin θ = (3) 5E-7 θ = 8.63° y = L tan θ = 2.0 tan(8.63°) = 0.30m c) d sin θ = (1)λ, θ = 2.86° y = 0.10m d) let maximum θ = 90°, d sin 90 = (m)λ, m=20. This is for 1 side and there is a middle fringe total = 41

17. Example Problem 2 • What is the minimum thickness of a soap bubble film with index of refraction 1.33 that would reflect 650nm most brightly? b) What is the minimum thickness for an anti-reflecting coating of index of refraction 1.4 or a glass of index 1.5 which would reflect no green light of wavelength 550nm? c) what would be the color of the light that is reflected off the lens

18. n= 1.33, λo = 650nm, λ = λo /n = 488nm, Δm = 0 = m2-m1 m2 = 2d/ λ, since m1 = ½ (due to 180 deg. inversion not present at m2) our path difference can be ½ wavelength. Get wavelength inside material and determine d = m2 * λ/2 = ½ * 488 / 2=122nm b) coating n=1.4, glass = 1.5 at 550nm λ = λo /n = 393nm, m1 = ½, m2 = 2d/ λ + ½ since both m1 and m2 reflect from greater index of refraction mediums Δm = ½ = m2-m1 = m2 – ½, m2 = 1 λ = 550/1.4 = 393, hence m2 =1= 2d/ λ + ½, 2d/ λ = 1/2, d = λ/4 = 393/4 = 98nm c) green transmitted, blue & red reflected, = magenta