Transcript PHY2054_03-31-11

Announcements
• HW set 9 due this week; covers Ch 23 and Ch 24.1-4
• Office hours:
• My office hours Th 2 -3 pm
• or make an appointment
• Come to class April 19
• course and instructor evaluation
• 8 bonus HITT points
• Always check out http://www.phys.ufl.edu/courses/phy2054/spring11/
for more announcements
QUESTIONS? PLEASE ASK!
Converging lenses
From last time…

Thin Lenses


Converging, diverging
Magnification
M =

h'
q
=h
p
Lens equation
1 1
1
+
=
p q
f


Remember the sign conventions!
(Table 23.3)
Lens maker’s equation
æ1
ö
1
1
= (n - 1) çç - ÷÷
f
è R1 R2 ø

Ray tracing
Diverging lenses
http://upload.wikimedia.org/wikipedia/commons/5/5d/
Michelson_Interferometer_Green_Laser_Interference.jpg
Chapter 24
http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html
Wave Optics
http://www.ligo.org/
http://fuff.org/interference/two_sources_interference.gif
Wave Optics: Coherence and Interference
http://www.black-holes.org/gwa4.html

The wave nature of light 


To produce interference –



Interference, diffraction,
polarization
sources must be coherent
(maintain a constant phase
w.r.t. each other)
(waves should have identical
wavelengths)
Coherent sources

Old days: Single Slit +
Double slit


Young’s double slit
Today: Lasers!
sciencecity.oupchina.com.hk
Young’s Double Slit Experiment



Light incident on a screen with a narrow
slit, So
Light emerging from So arrive at a second
screen that contains two narrow, parallel
slits, S1 and S2
Light waves emerging from S1 and S2
originate from the same wave front and
therefore are always in phase

The light from the two slits form a visible
pattern on a screen

The pattern consists of a series of bright
and dark parallel bands called fringes


Constructive interference occurs where a
bright fringe appears
Destructive interference results in a dark
fringe
DEMO
Interference Patterns
Constructive
interference
Destructive
interference
Constructive
interference
(again)
Question: What is the difference in the path length
(distance) that the two waves travel of the above
pictures ?
Interference Equations

δ = r2 – r1 = d sin θ



Bright fringe (constructive
interference)  δ must be
either zero or some integral
multiple of the wavelength, λ
δ = d sin θbright = m λ




assumes the paths are
parallel, a very good
approximation since L >> d
m = 0, ±1, ±2, …
m is called the order number
Dark fringe (destructive
interference)  δ must be an
odd half wavelength
δ = d sin θdark = (m + ½) λ

m = 0, ±1, ±2, …
Interference Equations


The positions of the fringes
can be measured vertically
from the zeroth order
maximum
y = L tan θ  L sin θ



L>>d>>λ
θ << 1
For bright fringes
(constructive interference)
sin  bright 

m
d
y bright
lL
=
m
d
m = 0, ± 1, ± 2 …
For dark fringes
(destructive interference)
sin  dark

1
 m  
d
2
y dark
lL æ
1ö
çç m + ÷÷ m = 0, ± 1, ± 2 …
=
d è
2ø
Phase Changes Due To Reflection

An electromagnetic wave
undergoes a 180° phase
change upon reflection from
a medium of higher index of
refraction than the one in
which it was traveling


Analogous to a reflected
pulse on a string
There is no phase change
when the wave is reflected
from a boundary leading to
a medium of lower index of
refraction

Analogous to a pulse in a
string reflecting from a free
support
Problem 23.13, p 818

Radio waves from a star, of
wavelength 250 m, reach a
radio telescope by two
separate paths as shown in
the figure. One is a direct
path to the receiver, which is
situated on the edge of a
cliff. The second is by
reflection off the water. The
first minimum of destructive
interference occurs when the
star is is 25° above the
horizon. Find the height of
the cliff. (Assume no phase
change on reflection.)
Interference in Thin Films




Interference is due to the
interaction of the waves reflected
from both surfaces of the film
Ray 1 - phase change of 180°
with respect to the incident ray
Ray 2 - no phase change with
respect to the incident wave
Ray 2 travels an additional
physical distance of 2t in the film


The wavelength λ is reduced by n
in the film  the optical path
length is 2 n t
Constructive interference

2 n t = (m + ½ ) λ


m = 0, 1, 2 …
takes into account both the
difference in optical path length for
the two rays and the 180° phase
change
Destructive interference

2nt=mλ
m = 0, 1, 2 …
DEMO
Handling thin films problems

Identify the thin film causing the interference

Determine the indices of refraction in the film and the media on
either side of it

Determine the number of phase reversals: zero, one or two

Interference is constructive if the path difference is an integral
multiple of λ and destructive if the path difference is an odd half
multiple of λ

NOTE: The conditions are reversed if one of the waves undergoes a phase
change on reflection
Equation
1 phase reversal
0 or 2 phase
reversals
2nt = (m + ½) 
constructive
destructive
destructive
constructive
2nt = m 
Problem 24.26, p 819

A plano-convex lens with a radius of curvature R = 3.0 m is
in contact with a flat plate of glass. A light source and the
observer’s eye are both close to normal, as shown below.
The radius of the 50th bright Newton’s ring is found to be 9.8
mm from the center. What is the wavelength of the light
produced by the source?
Anti-reflection coatings on solar cells
Answer to 23.13
Answer to 23.26