1. Physics 102: Lecture 20
Interference 1

2. But first…More on the Eye
Recall the lens formula:
Normal eyes
Far point = 
Near point = 25 cm
Nearsightedness
Far point dfar < 
To correct, produce virtual image of far object d0 =  at the far point (di = dfar)

3. Nearsightedness Far point dfar < 
To correct, produce virtual image of far object d0 =  at the far point (di = dfar)
flens = - dfar
1/flens = -1/dfar
Example:
My prescription reads dipoters
flens = -1/6.5 = m = cm (a diverging lens)
dfar = 15.4 cm (!)

4. Farsightedness Near point dnear > 25 cm
To correct, produce virtual image of object at d0 = 25 cm to the near point (di = dnear)
Example:
My near prescription reads +2.5 dipoters
flens = +1/2.5 = 0.4 m = 40 cm
therefore dnear = 67 cm (with my far correction)

5. Magnifying Glasses Will not cover in class Read on your own
Lecture 19 at the very end
Text Ch. 24.4

6. Constructive Interference
Superposition
Constructive Interference t +1 -1 + t +1 -1
In Phase t +2 -2 5

7. Destructive Interference
Superposition
Destructive Interference +1 t -1 + +1
Out of Phase
180 degrees t -1 t +2 -2 7

8. Superposition ACT + Different f
1) Constructive 2) Destructive 3) Neither 10

9. Interference Requirements
Need two (or more) waves
Must have same frequency
Must be coherent (i.e. waves must have definite phase relation) 12

10. Interference for Sound …
For example, a pair of speakers, driven in phase, producing a tone of a single f and l:
hmmm… I’m just far enough away that l2-l1=l/2, and I hear no sound at all! l1 l2
Demo 440 Sound Interference
But this won’t work for light--can’t get coherent sources 15

11. Interference for Light …
Can’t produce 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 18

12. ACT: Young’s Double Slit
Light waves from a single source travel through 2 slits before meeting on a screen. The interference will be:
Constructive
Destructive
Depends on L d
The rays start in phase, and travel the same distance, so they will arrive in phase.
Single source of monochromatic light  L
2 slits-separated by d
Screen a distance L from slits 23

13. Preflight 20.1
The experiment is modified so that one of the waves has its phase shifted by ½ l. Now, the interference will be:
Constructive
Destructive
Depends on L
½ l shift d
The rays start out of phase, and travel the same distance, so they will arrive out of phase.
60% got this correct
Single source of monochromatic light  L
2 slits-separated by d
Screen a distance L from slits 25

14. Young’s Double Slit Concept
At points where the difference in path length is 0, l,2l, …, the screen is bright. (constructive)
2 slits-separated by d d
At points where the difference in path
length is
the screen is dark. (destructive)
Demo slit
Single source of monochromatic light  L
Screen a distance L from slits 27

15. Young’s Double Slit Key Idea
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. 30

16. Young’s Double Slit Quantitative
Path length difference =
d sin q
Constructive interference
where m = 0, or 1, or 2, ...
Destructive interference
Need l < d 32

17. Young’s Double Slit Quantitative
A little geometry…
sin(q)  tan(q) = y/L
Constructive interference
Destructive interference
where m = 0, or 1, or 2, ... 33

18. Preflight 20.3 L y d
When this Young’s double slit experiment is placed under water. The separation y between minima and maxima
1) increases 2) same 3) decreases
21% 23% 53%
Under water l decreases so y decreases 35

19. Preflight 20.2 Need: d sin q = m l => sin q = m l / d
In the Young double slit experiment, is it possible to see interference maxima when the distance between slits is smaller than the wavelength of light?
1) Yes 2) No
Need: d sin q = m l
=> sin q = m l / d
If l > d then l / d > 1
so sin q > 1
Not possible! 35

20. Thin Film Interference1 2
n0=1.0 (air)
n1 (thin film) t n2
Demo 70 soap film
Get two waves by reflection off two different interfaces.
Ray 2 travels approximately 2t further than ray 1. 37

21. Reflection + Phase Shifts
Incident wave
Reflected wave n1 n2
Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change.
If n1 > n2 - no phase change upon reflection.
If n1 < n2 - phase change of 180º upon reflection. (equivalent to the wave shifting by l/2.) 39

22. Note: this is wavelength in film! (lfilm= lo/n1)
Thin Film Summary
Determine d, number of extra wavelengths for each ray. 1 2
n = 1.0 (air)
n1 (thin film) t n2
This is important!
Note: this is wavelength in film! (lfilm= lo/n1)
Reflection
Distance
Ray 1: d1 = 0 or ½
+ 0
Ray 2: d2 = 0 or ½
+ 2 t/ lfilm
If |(d2 – d1)| = 0, 1, 2, 3 … (m) constructive
If |(d2 – d1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructive 42

23. Thin Film Practice (ACT)
Example 1 2
n = 1.0 (air)
nglass = 1.5 t
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?
A) d1 = 0 B) d1 = ½ C) d1 = 1
What is d1, the total phase shift for ray 1 45

24. Thin Film Practice Example 1 2 n = 1.0 (air) nglass = 1.5 t
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?
d1 = ½
Reflection at air-film interface only
d2 = 0 + 2t / lglass = 2t nglass/ l0= (2)(167)(1.5)/500) =1
Phase shift = d2 – d1 = ½ wavelength 45

25. ACT: Thin Film
Blue light l = 500 nm incident on a thin film (t = 167 nm) of glass on top of plastic. The interference is:
(A) constructive
(B) destructive
(C) neither 1 2
n=1 (air)
nglass =1.5 t
nplastic=1.8
Brian Dvorkin has had good luck with the following presentation. 1) Look for the ½ wavelength shifts on reflection. Then ask for constructive interference what do I need ray 2 to do? (0 or ½ wavelength more?) If choose 0, increase to 1.
d1 = ½
Reflection at both interfaces!
d2 = ½ + 2t / lglass = ½ + 2t nglass/ l0= ½ + 1
Phase shift = d2 – d1 = 1 wavelength 48

26. Preflights 20.4, 20.5
t = l
nwater=1.3
ngas=1.20
nair=1.0
noil=1.45
A thin film of gasoline (ngas=1.20) and a thin film of oil (noil=1.45) are floating on water (nwater=1.33). When the thickness of the two films is exactly one wavelength…
The gas looks:
bright
dark
The oil looks:
bright
dark
d1,gas = ½
d2,gas = ½ + 2
d1,oil = ½
d2,oil = 2
| d2,gas – d1,gas | = 2
| d2,oil – d1,oil | = 3/2
constructive
destructive 50

27. See you Wednesday!