Videos
SP212
Chapter 35 - Interference
Double Slit Experiment and Explain
http://youtu.be/Iuv6hY6zsd0
Maj Jeremy Best USMC
Physics Department, U.S. Naval Academy
April 14, 2016
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
April 14, 2016
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Interference
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SP212
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Find the Physics
We have talked about light, and electromagnetic waves.
We have also determined that light is a wave. Therefore,
like other waves, it can interfere. Light waves interfere
when there is a phase difference between two waves
which meet at the same point.
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SP212
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Huygens’ Principle
Wavelength and Refraction
All points on a wave front serve as
point sources of spherical secondary
wavelets. After a time t, the new
position of the wavefront will be that
surface tangent to all of the secondary
wavelets .
The index of refraction indicates the speed of light in a
material: n = c/v . But the speed of a wave is the
frequency times the wavelength. For light in a vacuum,
that means c = λν . When light enters another
medium, the wavelength of the wave changes:
λn =
λ
n
The frequency remains the same.
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SP212
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SP212
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Solution:
Problem: Example
How much faster, in meters per second, does light travel
in a crystal with refraction index 1.72 than in another
with refraction index 2.12?
c
c
∆v = v1 − v2 =
−
=c
n1 n2
1
1
−
n1 n2
=
32887231.2 m/s
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SP212
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SP212
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Phase Shift
This change in wavelength means that light waves
initially in phase may be out of phase after passing
through different media:
Count the number of waves in each
medium.
The Law of Refraction again, is:
n1 sin θ1 = n2 sin θ2
from chapter 33.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
L
Ln1
=
λn1
λ
L
Ln2
N2 =
=
λn2
λ
L
N2 − N1 = (n2 − n1 )
λ
N1 =
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Phase Differences
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Phase Shift
It is only fractional phase differences that matter. If
two waves are 5 wavelengths out of phase, they are back
in phase. But if they are 5.5 wavelengths out of phase,
they are 0.5 wavelengths out of phase - totally
destructive interference.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
April 14, 2016
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Diffraction
Young’s Double Slit Experiment
We will look at diffraction in more detail next chapter,
but for now, we need to understand that diffraction
occurs when waves encounter a small opening
approximately the same size as the wavelength of the
wave.
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Young’s Double Slit Experiment
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Young’s Double Slit Experiment
If the path length difference in a double slit experiment is
an integer number of wavelengths, we will see
constructive interference (a bright spot)
∆L = d sin θ = mλ m = 0, 1, 2, ...
mλ
θ = sin−1
d
Passing through two different slits results in each wave
having a different path length ∆L = d sin θ
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
April 14, 2016
For example, we would refer to the m = 2 maximum as
the second-order maximum
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SP212
April 14, 2016
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Young’s Double Slit Experiment
Thin-film Interference
We will see destructive interference (a dark spot) if the
path-length difference is a half-integer number of
wavelengths.
We’ve all seen the rainbow patterns on soap bubbles or
oil slicks. These arise due to the interference of light
reflecting off the front and back surfaces of a thin-film.
∆L = d sin θ = (m + 1/2) λ m = 0, 1, 2, ...
1/2)λ
(m
+
θ = sin−1
d
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Thin-Films: The Tricky Part
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Reflection Phase Shifts
Constructive/destructive interference still occurs for the
same reason: constructive: waves in phase, destructive:
waves out of phase. But now, to determine that, we
need to take into account three effects:
The two rays travel different lengths
The two rays travel through different media, and
have different wavelengths
The process of reflection off the back surface can
induce a π-radian phase shift on its own
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
April 14, 2016
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When light reflects off a material with a higher index of
refraction than it is traveling through, the phase shifts by
π radians. When light reflects off a surface of lower
index of refraction, there is no phase shift
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
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Solution:
Air to acetone to glass in increasing refractive index. To
see very little or no reflection,
1 λ
2L = m +
where m = 0, 1, 2, ......
2 n2
Problem: Example
A thin film of acetone (n = 1.25) coats a thick glass
plate (n = 1.50). White light is incident normal to the
film. In the reflections, fully destructive interference
occurs at 620 nm and fully constructive interference at
682 nm. Calculate the thickness of the acetone film.
MUST give us the minima (no light) because of the π
phase shift. and
2L = m
λ
n2
for maxima (again due to the phase shift)
Giving us: 1364 nm where they are both the same.
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SP212
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Thin film Interference
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Thin-Film Interference
You must trace through the “history” of the rays, and
determine if the reflections leave them in phase or out of
phase. Then decide if you want to preserve this
condition, or reverse it.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
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2L = (m + 1/2)
2L = m
λ
n2
λ
n2
Reverses reflection phase shifts
Preserves reflection phase shifts
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SP212
April 14, 2016
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Problem:
If mirror M2 in a Michelson interferometer (the figure
below) is moved through 0.244 mm, a shift of 802 bright
fringes occurs. What is the wavelength of the light
producing the fringe pattern?
Thin-Film Interference
You must THINK about when phase shifts occur, and
what leads to a phase difference between two rays.
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Solution:
A shift of one fringe corresponds to a change in the
optical path length of one wavelength. When the mirror
moves a distance d, the path length changes by 2 d
since the light traverses the mirror arm twice. Let N be
the number of fringes shifted. Then, 2d = Nλ and
2d
2(.244 × 10−3 m
λ=
=
) = 608.5 nm
N
802
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