1. Waves (in general) sine waves are nice
other types of waves (such as square waves,
sawtooth waves, etc.) can be formed by a
superposition of sine waves - this is called
Fourier Series . This means that sine
waves can be considered as fundamental.

2. Waves (in general) E = Eo sin() where  is a phase angle
which describes the location along the wave
 = 90 degrees is the crest
= 270 degrees is the trough

3. Waves (in general) E = Eo sin() where  is a phase angle
in a moving wave,  changes with both
time (goes 2 radians in time T) and
distance (goes 2 radians in distance )
so = (2/)*x +/- (2/T)*t
where 2/T =  and
where 2/ = k and so
phase speed: v = distance/time = /T = f = /k

4. Waves (in general) For nice sine waves: E = Eo sin(kx +/- t)
For waves in general, can break
into component sine waves; this is
called spectral analysis

5. Light and Shadows Consider what we would expect from
particle theory: sharp shadows
dark
dark
light

6. Light and Shadows Consider what we would expect from
wave theory: shadows NOT sharp
crest
crest
crest
dark
dark
dim
light
dim

7. Light and Shadows What DOES happen? look at a very bright laser beam
going through a vertical slit.
(A laser has one frequency
unlike white light.)

8. Double Slit Experiment
We will consider this situation
but only after we consider another:
the DOUBLE SLIT experiment:

9. Double Slit Experiment
Note that along the green lines
are places where crests meets crests
and troughs meet troughs.
crest on crest
followed by
trough on trough

10. Double Slit Experiment
Note that along the dotted lines
are places where crests meets troughs
and troughs meet crests.
crest on trough
followed by trough on crest
crest on crest
followed by
trough on trough

11. Double Slit Experiment
Further explanations are in the
Introduction to the Computer Homework
Assignment on Young’s Double Slit, Vol 5, #3.
crest on trough
followed by trough on crest
crest on crest followed by trough on trough

12. Double Slit Experiment
Our question now is: How is the pattern
of bright and dark areas related to the
parameters of the situation: , d, x and L?
bright  x d
bright L
SCREEN

13. Young’s Double Slit Formula
λ/d = sin() ≈ tan() = x/L
The two (black) lines from the two slits to the first bright spot are almost parallel, so the two angles are almost 90 degrees, so the two ’s are almost equal.
bright x d
bright L λ

14. Double slit: an example
n = d sin() = d x / L
d = 0.15 mm = 1.5 x 10-4 m
x = ??? measured in class
L = ??? measured in class
n = 1 (if x measured between adjacent bright spots)
 = d x / L = (you do the calculation)

15. Interference: Diffraction Grating
The same Young’s formula works for multiple slits as it did for 2 slits.
lens
bright s1 s2 s3
s2 = s1 + 
s3 = s2 +  = s1 + 2
s4 = s3 +  = s1 + 3
s5 = s4 +  = s1 + 4 d
bright s4 s5 λ

16. Interference: Diffraction Grating
With multiple slits, get MORE LIGHT and get sharper bright spots.
lens
bright s1 s2 s3
s2 = s1 + 
s3 = s2 +  = s1 + 2
s4 = s3 +  = s1 + 3
s5 = s4 +  = s1 + 4 d
bright s4 s5

17. Interference: Diffraction Grating
With 5 slits, get cancellation when s = 0.8; with two slits, only get complete cancellation when s = 0.5 .
lens
bright
dark s1 s2 s3
s2 = s1 + .8
s3 = s2 + .8 = s 
s4 = s3 + .8 = s 
s5 = s4 + .8 = s  d
bright s4 s5

18. Diffraction Grating: demonstrations
look at the white light source
(incandescent light due to hot filament)
look at each of the gas excited sources
(one is Helium, one is Mercury)

19. 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?

20. 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

21. 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

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

23. 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.

24. 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.

25. 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.

26. 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.

27. 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.

28. 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.

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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.

36. 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

37. 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 or course that of air is
What is the thickness of the oil slick?

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

39. 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.

40. 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.

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

42. 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

43. 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

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

45. 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

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

47. 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

48. Michelson Interferometer
We can also use the Michelson interferometer to determine the index of refraction of a gas (such as air).
Put a cylinder with transparent ends into one of the beams.

49. Michelson Interferometer
Evacuate the cylinder with a vacuum pump
Slowly allow the gas to seep back into the cylinder and count the fringes.
Mirror
Half Mirror
Light
source
Mirror
cylinder
Screen

50. Michelson Interferometer
In vacuum, #vv = 2L .
In the air, #aa = 2L .
Since va < c, a < v and #a > #v .
Knowing v and L, can calculate #v .

51. Michelson Interferometer
Knowing v and L, can calculate #v .
By counting the number of fringe shifts, we can determine #.
Since # = #a - #v , we can calculate #a .
Now knowing L and #a, we can calculate a .

52. Michelson Interferometer
We now know v and a, so:
with vf = c and af = va , we can use
na = c/va = vf / af = v / a .

53. Michelson Interferometer an example
If L = 6 cm, and if v = 632 nm, and if 50 fringes are counted when air is let back into the cylinder, then:
#v = 2L/v = 2 * .06 m / 632 x 10-9 m = 189,873
#a = #v + # = 189, = 189,923
a = 2L/#a = 2 * .06 m / 189,923 = nm
na = v / a = nm / nm =