Download PHYS 1030 MidTerm Week of March 16-20 Week of March 23

PHYS 1030 MidTerm
Average: 12.2/20
Marks by student number are on the web
Friday, March 13, 2009
37
Week of March 16-20
Tutorial and Test 3: ch 22, 24, 25, 26
Week of March 23-27
Experiment 5: Spectroscopy
Week of March 30 - April 3
Tutorial and Test 4: ch. 27, 28
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38
Thin film interference
2 internal or 2 external reflections:
(refractive indices in increasing or decreasing order)
2t = mλf ilm
1
2t = (m + )λf ilm
2
→
constructive interference
→
destructive interference
1 internal and 1 external reflection:
(refractive indices not in increasing or decreasing order)
1
2t = (m + )λf ilm
2
2t = mλf ilm
→ constructive interference
→ destructive interference
λf ilm =
λair
nf ilm
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Young’s double slits and diffraction grating
Bright fringes when: d sin ! = m", m = 0, 1, 2 . . .
Dark fringes when:
1
d sin ! = (m + )", m = 0, 1, 2 . . .
2
d = distance between slits
Diffraction from a single slit
Dark fringes when:
W sin ! = m", m = 1, 2, 3 . . .
(no simple formula for bright fringes)
W = width of slit
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m = 1 dark fringe
m = 1 dark fringe
m=2
m=2
0o
θ
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Prob. 27.-/20: A slit of width W = 4.3!10-5 m is located 1.32 m from
a flat screen. Light shines through the slit and falls on the screen.
Find the width of the central fringe of the diffraction pattern when
the wavelength of the light is 635 nm.
• At what angle is the first minimum found?
m = 1 dark fringe
m = 1 dark fringe
m=2
m=2
θ!
00
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Clicker Question
A diffraction pattern is produced on a viewing screen by using a single
slit with blue light. Does the pattern broaden or contract (become
narrower) when
(a) the blue light is replaced by red light and,
(b) the slit width is increased?
A) broaden, broaden
B) broaden, contract
C) contract, broaden
D) contract, contract
m = 1 dark fringe
m = 1 dark fringe
m=2
Minima: W sin! = m"
m=2
!
Answer: B) broaden (" increases), contract (W increases)
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Diffraction by a
small opaque disk
Because of diffraction of light
from the edge of the circular
disk, the shadow cast by the
disk has:
• a central bright spot
(constructive interference of
light waves from around edge
of disk)
• circular bright fringes inside
the shadow
• bright and dark fringes
outside the shadow
Shadow of the disk
Bright spot at
centre of shadow!
(Parallel beam, that is, a plane wave)
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Diffraction by a
circular aperture
The bright and dark fringes are
circular and concentric with the
aperture.
The first dark fringe from the
centre is at the angle shown in
the diagram:
! = 1.22
"
radians
D
D = diameter of aperture
The smaller the aperture, the
larger the angle of diffraction.
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circular hole of diameter D
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Position of images in absence of diffraction
Resolving power
Light passing through any
aperture, such as a camera lens,
or an eye lens, is diffracted.
If two objects are close
together, the diffraction
patterns from them may
overlap.
If the overlap of the patterns
is large enough, it may not be
possible to tell there are two
objects – “resolution” is limited
by diffraction.
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Resolving Power
Headlights not resolved
Headlights resolved
Objects well-resolved
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Resolving power – Rayleigh criterion
Two objects can be resolved when their angular separation is greater than:
!min = 1.22
"
radians (Rayleigh criterion)
D
Objects well-resolved
Diameter, D
At !min, the first
minimum of one
object overlaps the
central maximum
of the other
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Resolving Power
For green light, " = 555 nm, and the diameter of the pupil of the eye,
D = 2.5 mm, the angular resolution is:
θmin = 1.22
λ
= 2.7 × 10−4 rad
D
At 120 m altitude, the minimum
resolvable distance on the ground is:
s = Hθmin = 0.033 m
Eagle/owl: D = 6.2 mm
θmin = 1.1 × 10−4 rad,
s = 0.013 m
# better ability to find small
furry animals on the ground
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Prob. 27.30: Two stars are 3.7!1011 m apart and are equally distant
from the earth. A telescope has an objective lens with a diameter of
1.02 m and just detects these stars as separate objects.
Assume that light of wavelength 550 nm is being observed and that
diffraction effects rather than atmospheric turbulence limit the
resolving power of the telescope.
Find the maximum distance that these stars could be from earth.
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Diffraction grating
Pairs of slits act as two Young’s slits. The bright fringes (“principal
maxima”) are at the same angles as for Young’s double slits.
Interference also occurs between more distant slits # sharper
peaks
Bright fringes (principal maxima): d sinθ = m"
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Diffraction grating
Much sharper fringes than
Young’s double slits
d sin ! = m"
Principal
maxima
d sin ! = m"
Principal
maxima
$ much more precise
measurement of
wavelength
The small peaks are
“subsidiary maxima”
Young’s double slits
with same distance
between slits, d
θ!
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A grating spectrometer
Parallel rays fall
on the grating
Grating equation: d sin ! = m" for the bright fringes
!=
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d sin "
→ wavelength, !
m
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Angstroms: 10 A = 1 nm
http://astro.u-strasbg.fr/~koppen/discharge/helium.html
Only the lines are present in the spectrum. A faint continuum was added to
place the lines within the spectrum of visible light.
The spectrum of light from a star allows elements in the star’s “atmosphere”
to be identified.
Helium was observed in the sun’s atmosphere before it had been discovered on
earth. Named after the Greek word for the sun.
Friday, March 13, 2009
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Prob. 27.42: A diffraction grating has 2604 lines per centimetre
and produces a principal maximum at ! = 30º. The grating is used
with light that contains all wavelengths between 410 and 660 nm.
What wavelengths of the incident light that could have produced
this maximum?
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http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html#c2
CD as a “reflection grating”
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CDs, DVDs
Information is stored along spiral tracks
1600 nm apart as binary 0’s and 1’s that
are read by how much a laser beam is
reflected from the surface.
No pit (land)
A pit
t
1
Laser beam
2t =
2
Destructive
interference,
weak reflection
!
2
1
2
Constructive
interference,
strong reflection
For " = 520 nm, t = "/4 = 130 nm
for destructive interference.
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Keeping the laser
beam on track
Two tracking beams are the first
order maxima produced by a
diffraction grating.
Land
If the laser drifts off track, one of
the tracking beams will start to hit
pits and will be reflected less strongly.
A feedback system steers the beam to
centre it so that both tracking beams
are once again strongly reflected.
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Summary of Chapter 27
• Principle of Linear Superposition ! interference and
diffraction
• Diffraction by Young’s double slits, by a single slit, and the
diffraction grating
• Interference in thin films, phase change on reflection (for
external reflections only)
• Michelson interferometer
• Interference by a circular aperture ! Rayleigh criterion and
resolving power
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Chapter 28: Special Relativity
• Events, inertial reference frames
• Postulates of special relativity
• Time dilation, length contraction
• Relativistic momentum
• Equivalence of mass and energy
• Omit 28.7, relativistic addition of velocities
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Reference frames
A reference frame is a coordinate system in which positions
and times may be measured: (x, y, z, t).
Inertial reference frame - is not accelerated – Newton’s laws
of motion apply so an object remains at rest when zero net
force acts on it.
Non-inertial reference frame - is accelerated. Example, car
going around corner, objects move if not held in place.
Special relativity applies to inertial reference frames.
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Events
An event is something that happens (surprise!).
Observers record the position and time (x, y, z, t) of events
according to their own distance scales and clocks.
L = vt
x
y
An event
occurs
here
y%
x%
y
Event occurs at (x, y, z, t)
in O frame
x
O
y%
Event occurs at (x%, y%, z%, t%)
in O% frame
x%
O%
Special relativity deals with events observed in inertial frames
that are in relative motion
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Postulates of Special Relativity
1) Relativity Postulate: The laws of physics are the same in all
inertial reference frames.
$ All inertial frames are equally good. There is no way to determine
whether an inertial frame is absolutely at rest.
2) Speed of Light Postulate: The speed of light in vacuum, measured in
any inertial reference frame, has the same value, independent of the
motion of the source of light or of the observer.
# Implies that light does not need a material medium to travel
through, with respect to which its speed would be fixed.
Compare sound, travelling at a fixed speed relative to air.
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Prob. 28.C1: The speed of light in water is c/n, where n = 1.33 is the
refractive index of water. Thus, the speed of light in water is less
than c.
Why doesn’t this violate the speed of light postulate?
The speed of light postulate deals with the speed of light in
vacuum, and says nothing about the speed in a transparent medium.
The speed of light in vacuum has special significance. Its constancy
leads to special consequences.
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