3/30/11
Converging lenses
Announcements
From last time…
•  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
Thin Lenses
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Converging, diverging
Magnification
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Lens equation
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M =
•  Come to class April 19
•  course and instructor evaluation
•  8 bonus HITT points
h'
q
=!
h
p
1 1
1
+
=
p q
f
•  Always check out http://www.phys.ufl.edu/courses/phy2054/spring11/
for more announcements
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Diverging lenses
Remember the sign conventions!
(Table 23.3)
Lens maker’s equation
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QUESTIONS? PLEASE ASK!
"1
1
1 %'
= (n ! 1) $$ !
'
f
# R1 R2 &
Ray tracing
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Wave Optics: Coherence and Interference
http://www.black-holes.org/gwa4.html
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http://upload.wikimedia.org/wikipedia/commons/5/5d/
Michelson_Interferometer_Green_Laser_Interference.jpg
Chapter 24
The wave nature of light à
http://www.exploratorium.edu/ronh/bubbles/bubble_colors.html
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To produce interference –
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Wave Optics
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Interference, diffraction,
polarization
sources must be coherent
(maintain a constant phase
w.r.t. each other)
(waves should have identical
wavelengths)
Coherent sources
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Old days: Single Slit +
Double slit
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Today: Lasers!
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http://www.ligo.org/
sciencecity.oupchina.com.hk
Young s double slit
http://fuff.org/interference/two_sources_interference.gif
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3/30/11
Interference Patterns
Young s Double Slit Experiment
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Constructive
interference
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
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
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Constructive interference occurs where a
bright fringe appears
Destructive interference results in a dark
fringe
Question: What is the difference in the path length
(distance) that the two waves travel of the above
pictures ?
DEMO
Interference Equations
Interference Equations
δ = r2 – r1 = d sin θ
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assumes the paths are
parallel, a very good
approximation since L >> d
Bright fringe (constructive
interference) à δ must be
either zero or some integral
multiple of the wavelength, λ
δ = d sin θbright = m λ
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Constructive
interference
(again)
Light waves emerging from S1 and S2
originate from the same wave front and
therefore are always in phase
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Destructive
interference
m = 0, ±1, ±2, …
m is called the order number
Dark fringe (destructive
interference) à δ must be an
odd half wavelength
δ = d sin θdark = (m + ½) λ
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m = 0, ±1, ±2, …
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The positions of the fringes
can be measured vertically
from the zeroth order
maximum
y = L tan θ ≈ L sin θ
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L>>d>>λ
θ << 1
For bright fringes
(constructive interference)
sin θbright =
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mλ
d
y bright =
For dark fringes
(destructive interference)
sin θ dark =
1 ⎞
⎜ m + ⎟
d ⎝
2 ⎠
λ ⎛
!L
m
d
y dark =
m = 0, ± 1, ± 2 …
!L !
1$
## m + && m = 0, ± 1, ± 2 …
d "
2%
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3/30/11
Problem 23.13, p 818
Phase Changes Due To Reflection
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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
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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.)
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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
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Analogous to a pulse in a
string reflecting from a free
support
Interference in Thin Films
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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
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The wavelength λ is reduced by n
in the film à the optical path
length is 2 n t
2 n t = (m + ½ ) λ
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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
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2nt=mλ
m = 0, 1, 2 …
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
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Constructive interference
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Handling thin films problems
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DEMO
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) λ
constructive
destructive
destructive
constructive
2nt = m λ
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3/30/11
Problem 24.26, p 819
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Anti-reflection coatings on solar cells
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?
Answer to 23.13
Answer to 23.26
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