thin-film interference...interference). if it differs by an odd integer number of half wavelengths,...
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![Page 1: Thin-Film Interference...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](https://reader035.vdocuments.us/reader035/viewer/2022081403/60b14dbb6115ab6d2a4aede3/html5/thumbnails/1.jpg)
Thin-Film Interference
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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
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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
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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…
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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
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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
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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.
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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
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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
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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
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http://www.phys.hawaii.edu/~teb/op
tics/java/slitdiffr/index.html
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Destructive interference leading to the first dark
fringe on either side of the central bright fringe.
(Only one dark fringe is being shown.)
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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
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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
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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
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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?
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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.
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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.
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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.
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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.
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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.
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X-rays diffracted
when passed
through a crystalline
structure will
diffract. The
amount of
diffraction can be
used to determined
spacing between
atoms.