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

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:


• Note that along the green lines

are places where crests meets crests

and troughs meet troughs.

crest on crestfollowed bytrough on trough


• 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


• 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


• 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


• 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


• 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?


• 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.


• Blue travels an extra distance of 2t in the film.

air

film

water


t


• Also, blue undergoes two refractions and reflects off of a different surface.

air

film

water


t


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.


• 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.


• 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.


• 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.


• 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.


• 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.


• 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


t


• If na < nf > nw, there WILL be a 180 degree phase difference (/2)due to reflection.

air

film

water


t


• There will ALWAYS be a phase difference due to the extra distance of 2t/.

air

film

water


t


• When t=/2 the phase difference due to path is 360 degrees (equivalent to no difference)

air

film

water


t


• When t=/4 the phase difference due to path is 180 degrees.

air

film

water


t


• Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a/nf.

air

film

water


t


• 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.


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?


• 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.



• 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


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


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


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

Mirror

Mirror

Half Mirror

Screen

Lightsource

s


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


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

Mirror

Mirror

Half Mirror

Screen

Lightsource

s


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


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.


• 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


• In vacuum, #vv = 2L .

• In the air, #aa = 2L .

• Since va < c, a < v and #a > #v .

• Knowing v and L, can calculate #v .


• 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 .


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

Documents

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