Transcript 24.3 Interference – Young`s Double-Slit Experiment

Chapter 24
The Wave Nature of Light
How many times does the
incident light reflect from this
set of three mirrors?
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Definition: Refraction
Refraction is the movement of light
from one medium into another
medium.
Refraction cause a change in speed
of
light as it moves from one medium to
another.
Refraction can cause bending of the
light at the interface between media.
Index of Refraction
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚
𝑛=
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑚𝑒𝑑𝑖𝑢𝑚
𝑛=
𝑐
𝑣
When light slows down…
…it bends. Let’s take a look at a
simulation.
http://interactagram.com/physics/optic
s/refraction/
The Lawnmower model
fast medium
slow medium
fast medium
slow medium
Speed changes
but
not direction
Speed and direction change
Ray is bent toward normal
Refraction occurs when there
is a change in wave
Frequency
Amplitude
Speed
All of these
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n1 < n 2
When the index of
refraction increases,
light bends toward
the normal. FST
SFA
Problem
Light enters an oil from the air at an angle of
50o with the normal, and the refracted beam
makes an angle of 33o with the normal.
a) Draw this situation.
b) Calculate the index of refraction of the oil.
c) Calculate the speed of light in the oil
Problem
Light enters water from a layer of oil at an
angle of 50o with the normal. The oil has a
refractive index of 1.65, and the water has a
refractive index of 1.33.
a) Draw this situation.
b) Calculate the angle of refraction.
c) Calculate the speed of light in the oil, and
in the water
Critical Angle of Incidence
• If light passes into a medium with a
greater refractive index than the original
medium, it bends away from the normal and
the angle of refraction is greater than the
angle of incidence.
• If the angle of refraction is > 90o, the light
cannot leave the medium.
• The smallest angle of incidence for which
light cannot leave a medium is called the
critical angle of incidence.
Calculating Critical Angle
n1sin(θ1) = n2sin(90o)
n1sin(θc) = n2sin(90o)
sin(θc) = n2/n1
θc = sin-1(n2/n1)
Total Internal Reflection Video: prism
and fiber optic
Total Internal Reflection Video: tank
of water
Problem
A. What is the critical angle of incidence for
a gemstone with refractive index 2.45 if it
is in air?
B. If you immerse the gemstone in water
(refractive index 1.33), what does this do
to the critical angle of incidence?
Why is the sky blue?
The Visible Spectrum and Dispersion
Wavelengths of visible light: 400 nm to 750 nm
Shorter wavelengths are ultraviolet; longer are
infrared
The Visible Spectrum and Dispersion
The index of refraction of a material varies
somewhat with the wavelength of the light.
The Visible Spectrum and Dispersion
This variation in refractive index is why a prism
will split visible light into a rainbow of colors.
Formation of Rainbows
Since each color
Ray of sunlight (all colors)
has a different
wavelength, they
Red light
each refract slightly
Normal
differently into and
Blue light
out of the water
droplet thus
separating from
white to individual
colors
Water Droplet
Close-up
The Visible Spectrum and Dispersion
Actual rainbows are created by dispersion in tiny
drops of water.
Diffraction
The bending of a wave around a barrier.
Diffraction of light combined with
interference of diffracted waves causes
“diffraction patterns”.
The phenomenon of diffraction
involves the spreading out of waves
past openings which are on the order
of the wavelength of the wave.
Ripple Tank
• Diffraction around obstacles in a
ripple tank.
• Diffraction and interference in a
ripple tank.
Interference of waves
 Two sound waves interfere each other
destructive
constructive
d1
d2
d 1  d 2  n
 ( n  1 / 2 )
n  0,1,2,....
(constructive)
(destructive)
Interference – Young’s Double-Slit
Experiment
If light is a wave, interference effects will be
seen, where one part of wavefront can interact
with another part.
One way to study this is to do a double-slit
experiment:
Interference – Young’s Double-Slit
Experiment
If light is a wave,
there should be
an interference
pattern.
Interference – Young’s Double-Slit
Experiment
The interference occurs because each point on
the screen is not the same distance from both
slits. Depending on the path length difference,
the wave can interfere constructively (bright
spot) or destructively (dark spot).
Interference – Young’s Double-Slit
Experiment
We can use geometry to find the conditions for
constructive and destructive interference:
(24-2a)
(24-2b)
24.3 Interference – Young’s Double-Slit
Experiment
Condition for a maximum: d sinθ = mλ
For small angles, we may assume sin θ ≈ θ
THEN tanθ ≈ sinθ ≈ θ ≈ x/L (where x is the distance between the
Central maximum and the next bright fringe (one λ difference)
Therefore: d(x/L) ≈ mλ → x ≈ mλL / d
(for small angles, screen is far away, L>>d)
Double slit interference
Between the maxima and the minima, the
interference varies smoothly as shown in the intensity
profile:
**How does interference pattern change with λ or with d? **
Physics Classroom two-slit diffraction
two-slit diffraction
Interference – Young’s Double-Slit
Experiment
Since the position of the maxima (except the
central one) depends on wavelength, the firstand higher-order fringes contain a spectrum of
colors.
Interference – Young’s
Double-Slit Experiment:
Remember, as λ increases, θ
increases as well (if all else
remains the same).
Therefore, red light is
diffracted at a larger θ than
violet light.
Diffraction Grating
The maxima of the diffraction pattern are
defined by
(24-4)
single slit diffraction
Single slit diffraction applet
single slit diffraction
http://hyperphysics.phyastr.gsu.edu/hbase/phyopt/sinsl
it.html
Monochromatic light passed through a double slit produces
an interference pattern on a screen a distance 2.0m away.
The third-order maximum is located 1.5cm away from the
central maximum. Which of the following adjustments would
cause the third-order maximum instead to be located 3.0cm
from the central maximum?
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Doubling the distance between slits
Tripling the wavelength of the light
Using a screen 1.0m away from
the slits
Using a screen 3.0m away from
the slits
Halving the distance between slits
D
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FRQ
Laser light is passed through a diffraction grating with
7000 lines per cm. Light is projected onto a screen far
away. An observer by the diffraction grating observes the
first order maximum 25° away from the central maximum.
a) What is the wavelength of the laser?
b) If the first order maximum is 40cm away from the
central maximum on the screen, how far away is the
screen from the diffraction grating?
c) How far, measured along the screen, from the central
maximum will the second-order maximum be?
Interference by Thin Films
Another way path lengths can differ, and
waves interfere, is if the travel through
different media.
If there is a very thin film of material – a few
wavelengths thick – light will reflect from both
the bottom and the top of the layer, causing
interference.
This can be seen in soap bubbles and oil
slicks, for example.
Interference by Thin Films
The wavelength of the
light will be different
in the oil and the air,
and the reflections at
points A and B may or
may not involve
phase change.
Thin Films
Why do you see a rainbow pattern on a soap bubble?
It’s a result of thin film interference.
The colored pattern is a result of
reflection from both the front and back
surfaces of the bubble and different
bubble thickness
There are two effects:
1) An inversion of the wave incident on
the first surface of the bubble
2) A path length difference between
wave reflected from 1st surface and
wave reflected from 2nd surface
Thin Films
There is a path length difference ABC:
Results in constructive interference if
ABC = 2t = mλn
Results in destructive interference if
ABC = 2t = (m+1/2) λn
t
λn = λ/n wavelength of light material
In addition, depending on index of refraction of the materials, an
additional phase change may exist:
•
If n2 > n1 the reflected wave has a phase change of  (½ cycle,
equivalent to a ½ path length difference)
(like reflection from a string with a fixed end, medium 2 is denser)
•
If n2 < n1 the reflected wave has a phase change of 0 (equivalent
to a 0 path length difference)
(like reflection from a string with a free end, medium 2 is less dense)
Thin Films
 Intuitive
explanation of phase change at a interface
(picture below is for a soap bubble)
gravity
Thin Films
1) Colored pattern is result of
varied thickness of the bubble:
Gravity causes the bubble to be
thinner at the top and thicker at
the bottom.
2) The thinnest part of film (top) is
dark (destructive interference),
Thin film interference
Air
Soap film, thickness t
Water
Extra path length = 2t
• Film thickness:
t ~ 
(0.5 – 1 m)
• Interference between light reflected off the top and bottom surfaces
can either be constructive (bright band) or destructive (dark band)
• Constructive interference occurs for a certain wavelegth depending
on the local film thickness
 bright color appears at that value of 
Problem solving strategy:
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Count the phase changes. A phase change occurs for
every reflection from low to high index of refraction
The extra distance traveled by the wave in the thin
film is twice the thickness of the film.
The wavelength in the film is λn = λ/n where n is the
index of refraction of the film’s material.
Use 2t = mλn . If the light undergoes zero or two phase
changes, then m is a whole number for constructive
interference (half integers for m give destructive
interference). For one phase change, the conditions
are reversed – whole numbers give destructive
interference, half integers , constructive.
Practice Problem
A soap bubble appears green (λ = 540nm) at the
point on its front surface nearest the viewer. What is
the smallest thickness the soap bubble film could
have? Assume n=1.35.
Anti-reflective / Nonreflective coating
Nonreflective coatings are commonly found on high
quality optical equipment including camera lenses,
telescopes, and eye glasses
Less light reflection means more light transmission
A thin film is placed on the surfaces of the lens
The film thickness is such that light (designed to
apply to just wavelength) reflected from it
destructively interferes. No coating can apply to all
wavelengths.
The amount of reflection depends on the difference
in index of refraction between the two materials
Other wavelengths will have partial destructive
interference. Wavelengths farthest from the center
will have the most reflection (why the camera lens
looks purplelish)
Anti-reflective / Nonreflective coating
Practice Problem
What is the thickness of an optical
coating of MgF2 whose index of
refraction is n=1.38 and is designed
to eliminate reflected light at
wavelengths (in air) around 550 nm
when incident normally on glass for
which n=1.5?
2Michelson Interferometer
The Michelson interferometer is centered
around a beam splitter, which transmits about
half the light hitting it and reflects the rest. It
can be a very sensitive measure of length.