Thin-Film Interference

Light is both reflected and refracted.

The refracted light ray has to travel farther before returning to your eye.

If this distance is equal to a whole wavelength integer, then the film would seem uniformly bright (constructive interference).

If it differs by an odd integer number of half wavelengths, then the film would appear dark.

Fig 27.10

The wavelength

difference between ray

1 and ray 2 occurs

inside the thin film,

the wavelength of the

wave while in the film

is the wavelength that

must be considered.

In this case, the

wavelength while in the

gasoline must be

considered (not in the

air).

Fig 27.10

Equation… you knew it was coming!

The wavelength within the film can be

calculated by using the index of refraction

for the thin film:

n=c/v and λ=v/f, so

n=c/v = (c/f) / (v/f) = λvacuum / λfilm

Or… n = λvacuum / λfilm ,

and λfilm = λvacuum /n But… there’s one more thing to consider…

Whenever a wave reflects at a boundary, it is possible for them to change phase.

Ex: Wave on string hitting wall (dense material compared to air)—the wave inverts when reflecting. http://phet.colorado.edu/simulations/sims.php?sim=Wave_on_a_String

This inversion is equal to 1/2λ.

In contrast, a phase change does NOT occur when a wave on a string reflects from the end of a string that is hanging free.

Fig 27.11

When light reflects off of a

medium which has a larger

refractive index, it changes

phase by 1/2λ.

Reflecting off of a smaller

refractive index medium does

NOT result in a phase

change.

In figure, light changes phase

by 1/2 λ when it reflects off of

the gasoline, but does not

change when it reflects at the

gasoline/H2O boundary.

In this case, you would want

the refracted light wave to be

in a 1/2 λ increment for

constructive interference!

Fig 27.10reflection

refraction

Gasoline film appears yellow because when white

light hits it, the blue part of the spectrum

destructively cancels out! (Recall: red and green

light make yellow light! i.e., The absence of blue

light makes yellow light.)

The colors you see in thin films of gasoline or

soap bubbles occur because of constructive and

destructive interference in conjunction with

fluctuating film thickness or the interference

changes for specific λ because the film thickness

is changing.

Camera lenses and binoculars are made to try

and reduce reflection of light by coating them with

magnesium fluoride, n=1.38.

DiffractionDiffraction is the

bending of waves

around obstacles or

edges of an opening.

Sound waves leaving a

room through an open

doorway diffract;

person around corner

hears sound.

Diffraction is an

interference effect.

Fig 27.17

Huygens’ PrincipleEvery point on a wave front acts as a source of tiny wavelets that move forward w/the same speed as the wave.

The wave front at a later point is a surface which is tangent to the wavelets.

Each dot acts as a sound or light source which produces wavelets.

In the center they link together, but on the edges they curve.

Reason why you can hear music around corners.

Light also bends at openings, but only by a small degree.

Fig 27.17

The amount of bending is dependent on the ratio of λ/W where W=opening width.

The smaller λ/W is, the less the diffraction (can’t see around corners because doorways, W, are usually large compared to the λ of light)

The greater λ/W is, the greater the diffraction.

Big λ and small W results in maximized diffraction.

Fig 27.18

http://www.phys.hawaii.edu/~teb/op

tics/java/slitdiffr/index.html

Destructive interference leading to the first dark

fringe on either side of the central bright fringe.

(Only one dark fringe is being shown.)

Double slit: sin=mλ/(d)

for bright fringes

Single slit: sin=m(λ)/(W)

for dark fringes

m = 0, 1, 2, 3, 4…

W = width of slit

The distance 2y is the width of the central bright

fringe. What is this distance for (a) red light of

λ=690nm, and (b) violet light of λ=410nm?

W = 4.0 X10-6m

Fig 27.24

2y = ? For Red Light (first find = ?)

m=1, W = 4.0 X10-6 m, L = 0.40 m a) λ = 690 nm,

= sin-1 [m (λ) / (W)]

= sin-1[1 (690 X 10-9 m) / (4.0 X10-6 m)]

= 9.9o

tan = y/L

y = L (tan ) = (0.40 m) (tan 9.9o)

y = 0.070 m

2y = 0.14 m

b) Blue light: when λ=410 nm, then 2y = 0.083 m

Resolving Power

Resolving power is the ability to

distinguish between 2 closely spaced

points. Ex: A car’s lights at night, is it a

motorcycle or a car on-coming?

Diffraction occurs

when light passes

through circular

openings. Ex: Eyes,

cameras,

telescopes. The

diffraction pattern

places a limit on the

resolving power of

the device.

The Math

The first dark fringe relative

to the central bright fringe

can be found using:

sin = 1.22λ/D

D represents the diameter of

the opening (compared to d

which is distance between

slits)

The smaller D, The bigger

sin.

For two objects off

in the distance, as

they get farther

away from the

optical device the

images get closer

together. Throw

in diffraction and

the images get

blurry.

Rayleigh Criterion for Resolution

Two points are just

resolved when the

first dark fringe of one

falls on the central

bright fringe of the

other.

More Resolution

The minimum angle between the two

objects shown can be found using:

sin = 1.22λ/D or if the angle is small and

expressed in radians where sinmin = min

then we can say min = 1.22λ/D in radians.

For an optical device, the best resolution is

with small λ and large D.

Hubble can resolve 2 objects 1cm apart 62

miles away.

Tail lights blur in my vision at night.

X-rays diffracted

when passed

through a crystalline

structure will

diffract. The

amount of

diffraction can be

used to determined

spacing between

atoms.