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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.
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
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
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
Light and Shadows
• Consider what we would expect from
particle theory: sharp shadows
lightdark dark
Light and Shadows
• Consider what we would expect from
wave theory: shadows NOT sharp
lightdark darkdimdim
crest
crest
crest
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.)
Double Slit Experiment
• We will consider this situation
but only after we consider another:
the DOUBLE SLIT experiment:
Double Slit Experiment
• Note that along the green lines
are places where crests meets crests
and troughs meet troughs.
crest on crestfollowed bytrough on trough
Double Slit Experiment
• Note that along the dotted lines
are places where crests meets troughs
and troughs meet crests.
crest on crestfollowed bytrough on trough
crest on troughfollowed by trough on crest
Double Slit Experiment
• Further explanations are in the
Introduction to the Computer Homework
Assignment on Young’s Double Slit, Vol 5, #3.
crest on crest followed by trough on trough
crest on troughfollowed by trough on crest
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?
d
SCREEN
L
xbright
bright
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.
d
λ L
x
bright
bright
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)
Interference: Diffraction Grating
• The same Young’s formula works for multiple slits as it did for 2 slits.
d
lens bright
bright
s1
s2
s3
s4
s5
s2 = s1 + s3 = s2 + = s1 + 2s4 = s3 + = s1 + 3s5 = s4 + = s1 + 4
λ
Interference: Diffraction Grating
• With multiple slits, get MORE LIGHT and get sharper bright spots.
d
lens bright
bright
s1
s2
s3
s4
s5
s2 = s1 + s3 = s2 + = s1 + 2s4 = s3 + = s1 + 3s5 = s4 + = s1 + 4
Interference: Diffraction Grating
• With 5 slits, get cancellation when s = 0.8; with two slits, only get complete cancellation when s = 0.5 .
d
lens bright
bright
s1
s2
s3
s4
s5
s2 = s1 + .8s3 = s2 + .8 = s1 + 1.6s4 = s3 + .8 = s1 + 2.4s5 = s4 + .8 = s1 + 3.2
dark
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)
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?
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.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
Interference: Thin Films
• Blue travels an extra distance of 2t in the film.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
Interference: Thin Films
• Also, blue undergoes two refractions and reflects off of a different surface.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
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.
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.
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.
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.
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.
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.
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.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
Interference: Thin Films
• If na < nf > nw, there WILL be a 180 degree phase difference (/2)due to reflection.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
Interference: Thin Films
• There will ALWAYS be a phase difference due to the extra distance of 2t/.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
Interference: Thin Films
• When t=/2 the phase difference due to path is 360 degrees (equivalent to no difference)
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
Interference: Thin Films
• When t=/4 the phase difference due to path is 180 degrees.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
Interference: Thin Films
• Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf.
air
film
water
reflected red interferes withrefracted/reflected/refracted blue.
t
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.
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
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 1.00 .
What is the thickness of the oil slick?
Thin Films: an example
• Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).
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.
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.
Michelson Interferometer
Split a beam with a Half Mirror, the use mirrors to recombine the two beams.
Mirror
Mirror
Half Mirror
Screen
Lightsource
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
Mirror
Half Mirror
Screen
Lightsource
Michelson Interferometer
If the blue beam goes a little extra distance, s, the the screen will show a different interference pattern.
Mirror
Mirror
Half Mirror
Screen
Lightsource
s
Michelson Interferometer
If s = /4, then the interference pattern changes from bright to dark.
Mirror
Mirror
Half Mirror
Screen
Lightsource
s
Michelson Interferometer
If s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift).
Mirror
Mirror
Half Mirror
Screen
Lightsource
s
Michelson Interferometer
By counting the number of fringe shifts, we can determine how far s is!
Mirror
Mirror
Half Mirror
Screen
Lightsource
s
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
Mirror
Half Mirror
Screen
Lightsource
s
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.
Michelson Interferometer
• Evacuate the cylinder with a vacuum pump• Slowly allow the gas to seep back into the cylinder
and count the fringes.Mirror
Mirror
Half Mirror
Screen
Lightsource cylinder
Michelson Interferometer
• In vacuum, #vv = 2L .
• In the air, #aa = 2L .
• Since va < c, a < v and #a > #v .
• Knowing v and L, can calculate #v .
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 .
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 .
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 + # = 189,873 + 50 = 189,923
a = 2L/#a = 2 * .06 m / 189,923 = 631.83 nm
na = v / a = 632.00nm / 631.83nm = 1.00026