PHYS 202
Lecture 17
Professor Stephen Thornton
March 30, 2005
Reading Quiz:
Which one of the following experiments was
best known as the justification for the wave
theory of light?
1) Huygens’ experiment
2) Young’s double slit experiment
3) Newton’s rings experiment
4) Soap bubbles experiment
Answer: 2
Young first performed the experiment
in 1801, and it clearly required a
wave explanation. Newton believed
in the particle nature, and he was such
a dominant figure that it was difficult
for the scientific community to accept
the wave concept.
Last Time
The human eye
Vision correction
Camera
Magnifier
Microscopes, telescopes
Today
Superposition of waves
Interference of waves
Young’s two-slit experiment
Phase change due to reflection
Interference effects – Newton’s rings,
thin films, etc.
Single-slit diffraction
Exam 3 has been postponed from
April 13 until April 20!
Constructive and Destructive Interference
waves cancel
Computer interference
http://www.colorado.edu/physics/200
0/applets/fourier.html
http://id.mind.net/~zona/mstm/physic
s/waves/interference/intrfrnc.html
Two Radio Antennas Transmitting the Same Signal
Min
Max
Min
Max
The light sources
are coherent – in
relative phase with
each other.
A 2 − A1 = mλ constructive interference
waves add
1⎞
⎛
A 2 − A1 = ⎜ m − ⎟ λ destructive interference
2⎠
⎝
waves cancel
Young’s Two-Slit
Experiment
Makes light coherent
http://www.phy.ntnu.edu.tw
/java/doubleSlit/doubleSlit.
html
Do demo
Huygens’s Principle
Each point acts as a new
source of waves.
Path Difference in the
Two-Slit Experiment
∆A
∆A = d sin θ = mλ ***
m = integer
for constructive
interference
1
d sin θ = (m − )λ ***
2
m = 0, ±1, ±2,...
for destructive
inteference
Two-Slit
Pattern
Two slit
interference
Linear Distance in an Interference Pattern
y = L tan θ
Phase Change with Reflection
or solid
surface
An Air Wedge
Phase change with reflection, not refraction.
A System for Generating Newton’s Rings
Do demo
Test for Flatness
Dent in Newton’s
ring flat plate
Interference in
Thin Films
λn =
2t
2nt
=
=m
λvac / n λvac
λvac
n
m = 0, ± 1, ± 2,... destructive
2t
2nt
1
=
= m+
2
λvac / n λvac
m = 0, ± 1, ± 2,... constructive
Thickness and Color in a Thin Film
Do demo
A Thin Film with One Phase Change
A Thin Film with Two Phase Changes
Oil film interference
Oil has higher n
than water.
Soap bubbles
Antireflection
Look at Active
Example 28-1.
Quarter wave
coatings.
Antireflective coatings
t = λ /4
Antireflective coatings
http://www.zeiss.de/de/ophtalmic/co
mp/home_e.nsf/6f2a76c25f0237fbc1
2566fe003b25ff/26007f9aa67f3dee41
25688f003dd3c0?OpenDocument
Conceptual Quiz:
The colors observed on a soap bubble tend to
change over a short time period, but the colors
of a gas spill on water do not. Why?
1. The soap bubble is colored, but gas is
colorless. The water evaporates.
2. They are due to different phenomena.
3. The colors for soap bubbles are due to
reflection only, but not for gas on water.
4. Both colors are due to interference, but the
soap film changes thickness with time.
Answer: 4
The soap bubbles change thickness mostly due
to gravity. This changes the interference effect
due to different wavelengths. The oil thickness
stays more constant with time.
Conceptual Quiz:
Sometimes we want to put coatings on
mirrors to increase the reflection. Keeping
in mind what we just learned for
antireflective coatings, the thickness of the
mirror coating would be for increased
reflection
1)
2)
3)
4)
λ/8
λ/4
λ/2
λ
Answer: 3
We want the waves to constructively
interfere after reflection, so we want
λ/2 so that when included with the
two reflections we will have a total of
λ phase change and we have
constructive interference. Look at
Active Example 28-1, step 3. Set that
to 1 for constructive interference
instead of ½ for destructive
interference.
The iridescent tipped feathers
of the hummingbird's gorget
are unique to their avian
family.Hummingbirds can
precisely control the
direction of the iridescent
light by moving the feather
tips relative to the angle of
the sun. Light waves
refracted by air bubbles in a
thin melanin pigment coating
of the feather barbules negate
or reinforce one another to
result in the bright coloration.
Work Problems
28-15
28-32
Diffraction of Water Waves
Single-Slit Diffraction
Do demo
Locating the First Dark Fringe
in Single-Slit Diffraction
Locating the
Second Dark
Fringe in SingleSlit Diffraction
Single slit diffraction
W = slit width
λ = wavelength
θ = angle
W sin θ = mλ
m = ±1, ± 2, ...
This is when destructive interference
occurs. Dark spots. The bright spots
will be about halfway between.