Waves in general
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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).

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Waves in general
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 general1
Waves (in general)

  • E = Eosin()where  is a phase angle

    which describes the location along the wave

     = 90 degrees is the crest

    =270 degrees is the trough


Waves in general2
Waves (in general)

E = Eosin() where  is a phase angle

in a moving wave,  changes with both

  • time (goes2radians in time T) and

  • distance (goes 2radians in distance )

    so = (2/)*x +/- (2/T)*t

  • where2/T =  and

  • where 2/ = k and so

    phase speed: v =distance/time = /T = f =/k


Waves in general3
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
Light and Shadows

  • Consider what we would expect from

    particle theory: sharp shadows

dark

dark

light


Light and shadows1
Light and Shadows

  • Consider what we would expect from

    wave theory:shadows NOT sharp

crest

crest

crest

dark

dark

dim

light

dim


Light and shadows2
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
Double Slit Experiment

  • We will consider this situation

    but only after we consider another:

    the DOUBLE SLIT experiment:


Double slit experiment1
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


Double slit experiment2
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


Double slit experiment3
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


Double slit experiment4
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


Young s double slit formula
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


Double slit an example
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
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


Interference diffraction grating1
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


Interference diffraction grating2
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 = s1 + 1.6

s4 = s3 + .8 = s1 + 2.4

s5 = s4 + .8 = s1 + 3.2

d

bright

s4

s5


Diffraction grating demonstrations
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
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 films1
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


Interference thin films2
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


Interference thin films3
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


Interference thin films4
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 films5
Interference: Thin Films

  • When a wave on a string encounters a fixed end, the reflected wavemust 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 films6
Interference: Thin Films

  • When a wave on a string encounters a fixed end, the reflected wavemust 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 films7
Interference: Thin Films

  • When a wave on a string encounters a free end, the reflected wavedoes NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.


Interference thin films8
Interference: Thin Films

  • When a wave on a string encounters a free end, the reflected wavedoes NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.


Interference thin films9
Interference: Thin Films

  • When light is incident on a SLOWERmedium (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 films10
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


Interference thin films11
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


Interference thin films12
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


Interference thin films13
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


Interference thin films14
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


Interference thin films15
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


Interference thin films16
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 films17
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
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 example1
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 example2
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 example3
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
Michelson Interferometer

Split a beam with a Half Mirror, the use mirrors to recombine the two beams.

Mirror

Half Mirror

Light

source

Mirror

Screen


Michelson interferometer1
Michelson Interferometer

If the red beamgoes the same length as the bluebeam, then the two beams will constructively interfere and a bright spot will appear on screen.

Mirror

Half Mirror

Light

source

Mirror

Screen


Michelson interferometer2
Michelson Interferometer

If the blue beamgoes a little extra distance, s, the the screen will show a different interference pattern.

Mirror

Half Mirror

Light

source

Mirror

s

Screen


Michelson interferometer3
Michelson Interferometer

If s = /4, then the interference pattern changes from bright to dark.

Mirror

Half Mirror

Light

source

Mirror

s

Screen


Michelson interferometer4
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


Michelson interferometer5
Michelson Interferometer

By counting the number of fringe shifts, we can determine how far s is!

Mirror

Half Mirror

Light

source

Mirror

s

Screen


Michelson interferometer6
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


Michelson interferometer7
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 interferometer8
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


Michelson interferometer9
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 .


Michelson interferometer10
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 interferometer11
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
Michelson Interferometeran 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


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