1. Supplementary Material
This set of slides contains material dealing with thin films and with the Michelson Interferometer. Both of these phenomena can be explained nicely by the interference of waves. Also included are slides about the telescope.

2. Interference: Thin Films
Before, we had several different parts of a wide beam interfering with one another.
Can we find other ways of having parts of a beam interfere with other parts?

3. Interference: Thin Films
We can also use reflection and refraction to get different parts of a beam to interfere with one another by using a thin film.
reflected red interferes with
refracted/reflected/refracted blue.
air
film
water

4. Interference: Thin Films
Blue travels an extra distance of 2t in the film.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

5. Interference: Thin Films
And, blue undergoes two refractions and reflects off of a different surface.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

6. Interference: Thin Films
When a wave encounters a new medium:
the phase of the refracted wave is NOT affected.
the phase of the reflected wave MAY BE affected.

7. Interference: Thin Films
When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

8. Interference: Thin Films
When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

9. Interference: Thin Films
When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

10. Interference: Thin Films
When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

11. Interference: Thin Films
When light is incident on a SLOWER medium (one of index of refraction higher than the one it is in), the reflected wave is 180 degrees out of phase with the incident wave.
When light is incident on a FASTER medium, the reflected wave does NOT undergo a 180 degree phase shift.

12. Interference: Thin Films
If na < nf < nw, BOTH red and blue reflected rays will be going from fast to slow, and no difference in phase will be due to reflection.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

13. Interference: Thin Films
If na < nf > nw, there WILL be a 180 degree phase difference (/2) due to reflection.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

14. Interference: Thin Films
There will ALWAYS be a phase difference due to the extra distance of 2t/.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

15. Interference: Thin Films
When t=/2 the phase difference due to path is 360 degrees (equivalent to no difference)
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

16. Interference: Thin Films
When t=/4 the phase difference due to path is 180 degrees.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

17. Interference: Thin Films
Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf.
reflected red interferes with
refracted/reflected/refracted blue.
air
film t
water

18. Interference: Thin Films
reflection: no difference if nf < nw;
180 degree difference if nf > nw.
distance: no difference if t = a/2nf
180 degree difference if t = a/4nf
Total phase difference is sum of the above two effects.

19. Interference: Thin Films
Total phase difference is sum of the two effects of distance and reflection
For minimum reflection, need total to be 180 degrees.
anti-reflective coating on lens
For maximum reflection, need total to be 0 degrees.
colors on oil slick

20. Thin Films: an example
An oil slick preferentially reflects green light. The index of refraction of the oil is 1.65, that of water is 1.33, and of course that of air is
What is the thickness of the oil slick?

21. Thin Films: an example
Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).

22. Thin Films: an example
Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).
Since we have nf > nw, we have 180 degrees due to reflection.

23. Thin Films: an example
Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).
Since we have nf > nw, we have 180 degrees due to reflection.
Therefore, we need 180 degrees due to extra distance, so need t = a/4nf where a = 500 nm, nf = 1.65, and so: t = 500 nm / 4(1.65) = 76 nm.

24. Michelson Interferometer
Split a beam with a Half Mirror, the use mirrors to recombine the two beams.
Mirror
Half Mirror
Light
source
Mirror
Screen

25. Michelson Interferometer
If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen.
Mirror
Half Mirror
Light
source
Mirror
Screen

26. Michelson Interferometer
If the blue beam goes a little extra distance, s, the the screen will show a different interference pattern.
Mirror
Half Mirror
Light
source
Mirror s
Screen

27. Michelson Interferometer
If s = /4, then the interference pattern changes from bright to dark.
Mirror
Half Mirror
Light
source
Mirror s
Screen

28. Michelson Interferometer
If s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift).
Mirror
Half Mirror
Light
source
Mirror s
Screen

29. Michelson Interferometer
By counting the number of fringe shifts, we can determine how far s is!
Mirror
Half Mirror
Light
source
Mirror s
Screen

30. Michelson Interferometer
If we use the red laser (=632 nm), then each fringe shift corresponds to a distance the mirror moves of 316 nm (about 1/3 of a micron)!
Mirror
Half Mirror
Light
source
Mirror s
Screen

31. Limits on Resolution: further examples: telescope
Light from far away is almost parallel.
eyepiece
objective
lens fo fe

32. Limits on Resolution: further examples: telescope
The telescope collects and concentrates light.
eyepiece
objective
lens fo fe

33. Limits on Resolution: further examples: telescope
Light coming in at an angle, in is magnified to out .
eyepiece
objective
lens x fo fe

34. Limits on Resolution: further examples: telescope
in = x/fo, out = x/fe; M = out/in = fo/fe
eyepiece
objective
lens x fo fe

35. Limits on Resolution: further examples: telescopes
magnification: M = out/in = fo /fe
light gathering: Amt D2
resolution:  = D sin(limit) so
in = limit and out = 5 arc minutes
so limit  1/D implies Museful = 60 * D
where D is in inches
surface must be smooth on order of 

36. Limits on Resolution: Telescope
Mmax useful = out/in = eye/limit
= 5 arc min / (1.22 *  / D) radians
= (5/60)*(/180) / (1.22 * 5.5 x 10-7 m / D)
= (2167 / m) * D * (1 m / 100 cm) * (2.54 cm / 1 in)
= (55 / in) * D

37. Limits on Resolution: Telescope
What diameter telescope would you need to read letters the size of license plate numbers from a spy satellite?

38. Limits on Resolution: Telescope
need to resolve an “x” size of about 1 cm
“s” is on order of 100 miles or 150 km
limit then must be (in radians)
= 1 cm / 150 km = 7 x 10-8
limit = 1.22 x 5.5 x 10-7 m / D = 7 x 10-8
so D = 10 m (Hubble has a 2.4 m diameter)