Interference, Diffraction

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Interference, Diffraction Polarization
PHY232 Spring 2007 Jon Pumplin http//www.pa.msu
.edu/pumplin/PHY232 (Ppt courtesy of Remco
Zegers)
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light as waves

• so far, light has been treated as if it travels
  in straight lines
• ray diagrams
• refraction, reflection
• To describe many optical phenomena, we have to
  treat light as waves.
• Just like waves in water, or sound
• waves, light waves can interact
• and form interference patterns.
• Remember c f ?

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interference
constructive interference
destructive interference
at any point in time one can construct the total
amplitude by adding the individual components
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demo interference
Interference III


constructive interference waves in phase
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Interference in spherical waves
maximum of wave
minimum of wave
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(No Transcript)
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light as waves
it works the same for light waves, sound waves,
and small water waves
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double slit experiment

• the light from the two sources is
  incoherent (fixed phase with respect to each
  other
• in this case, there is
• no phase shift between
• the two sources
• the two sources of light must have identical wave
  lengths

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Youngs interference experiment
there is a path difference depending on its size
the waves coming from S1 or S2 are in or out of
phase
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Youngs interference experiment
If the difference in distance between the screen
and each of the two slits is such that the waves
are in phase, constructive interference occurs
bright spot difference in distance must be a
integer multiple of the wavelength d sin?
m?, m0,1,2,3 m 0 zeroth order, m1 first
order, etc. if the difference in distance is off
by half a wavelength (or one and a half etc.),
destructive interference occurs (d sin?
m1/2?, m0,1,2,3)
path difference
demo
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distance between bright spots
tan?y/L
L
if ? is small, then sin ? ? ? ? tan ? so d sin?
m?, m0,1,2,3 converts to dy/L m?
difference between maximum m and maximum
m1 ym1-ym (m1)?L/d-m?L/d ?L/d ymm?L/d
demo
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question

• two light sources are put at a distance d from a
  screen. Each source produces light of the same
  wavelength, but the sources are out of phase by
  half a wavelength. On the screen exactly midway
  between the two sources will occur
• a) constructive interference
• b) destructive interference

1/2?
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question

• two narrow slits are illuminated by a laser with
  a wavelength of 600 nm. the distance between the
  two slits is 1 cm. a) At what angle from the beam
  axis does the 3rd order maximum occur? b) If a
  screen is put 5 meter away from the slits, what
  is the distance between the 0th order and 3rd
  order maximum?

• use d sin? m? with m3
• ?sin-1(m?/d)sin-1(3x600x10-9/0.01)0.01030
• b) Ym m?L/d
• m0 y0 0
• m3 y3 3x600x10-9x5/0.01 9x10-4 m 0.9
  mm

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other ways of causing interference

• remember

equivalent to
n1gtn2
n1ltn2
1
2
2
1
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phase changes at boundaries
If a light ray travels from medium 1 to medium 2
with n1ltn2, the phase of the light ray will
change by 1/2?. This will not happen if n1gtn2.
n1gtn2
1
2
1
2
n1ltn2
1/2? phase change
no phase change
In a medium with index of refraction n, the
wavelength changes (relative to vacuum) to ?/n
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thin film interference
n1
The two reflected rays can interfere. To analyze
this system, 4 steps are needed
n1.5
n1

1. Is there phase inversion at the top surface?
2. Is there phase inversion at the bottom surface
3. What are the conditions for constructive/destructi
   ve interference?
4. what should the thickness d be for 3) to happen?

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thin film analysis

1. top surface?
2. bottom surface?
3. conditions?
4. d?

1
2
n1
n1.5
n1

• top surface n1ltn2 so phase inversion 1/2?
• bottom surface n1gtn2 so no phase inversion
• conditions
• constructive ray 1 and 2 must be in phase
• destructive ray 1 and 2 must be out of phase by
  1/2?
• if phase inversion would not take place at any of
  the surfaces
• constructive
• 2dm? (difference in path lengthinteger
  number of wavelengths)
• due to phase inversion at top surface
  2d(m1/2)?
• since the ray travels through film
  2d(m1/2)?film (m1/2)?/nfilm
• destructive 2dm?film m?/nfilm

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Note
The interference is different for light of
different wavelengths
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question

• Phase inversion will occur at
• top surface
• bottom surface
• top and bottom surface
• neither surface

na1
nb1.5
nc2

• constructive interference will occur if
• 2d(m1/2)?/nb
• 2dm?/nb
• 2d(m1/2)?/nc
• 2dm?/nc

note if destructive 2d(m1/2)?/nb this is used
e.g. on sunglasses to reduce reflections
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another case
1
2
The air gap in between the plates has varying
thickness. Ray 1 is not inverted (n1gtn2) Ray 2 is
inverted (n1ltn2) where the two glasses touch no
path length difference dark fringe. if
2t(m1/2)? constructive interference if 2tm?
destructive interference.
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question
Given h1x10-5 m 30 bright fringes are seen, with
a dark fringe at the left and the right. What is
the wavelength of the light?
2tm? destructive interference. m goes from 0
(left) to 30 (right). ?2t/m2h/m2x1x10-5/306.67
x10-7 m667 nm
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newtons rings
demo
spacing not equal
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quiz (extra credit)

• Two beams of coherent light travel different
  paths arriving
• at point P. If constructive interference occurs
  at point P,
• the two beams must
• travel paths that differ by a whole number of
• wavelengths
• b)travel paths that differ by an odd number of
  half
• wavelengths

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question

• why is it not possible to produce an interference
  pattern
• in a double-slit experiment if the separation of
  the slits
• is less than the wavelength of the light used?
• the very narrow slits required would generate
  different
• wavelength, thereby washing out the
  interference pattern
• the two slits would not emit coherent light
• the fringes would be too close together
• in no direction could a path difference as large
  as one wavelength be obtained

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diffraction
In Youngs experiment, two slits were used to
produce an interference pattern. However,
interference effects can already occur with a
single slit.
This is due to diffraction the capability of
light to be deflected by edges/small openings.

In fact, every point in the slit opening acts as
the source of a new wave front
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(No Transcript)
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interference pattern from a single slit
pick two points, 1 and 2, one in the top top half
of the slit, one in the bottom half of the
slit. Light from these two points
interferes destructively if ?x(a/2)sin??/2
so sin??/a we could also have divided up the
slit into 4 pieces ?x(a/4)sin??/2 so
sin?2?/a 6 pieces ?x(a/6)sin??/2 so
sin?3?/a Minima occur if sin? m?/a m1,2,3
In between the minima, are maxima sin?
(m1/2)?/a m1,2,3
AND sin?0
or ?0
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slit width
a
a
if ?gta sin??/a gt 1 Not possible, so no patterns
if ?ltlta sin?m?/a is very small diffraction
hardly seen
?lta interference pattern is seen
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the diffraction pattern
The intensity is not uniform II0sin2(?)/?2
??a(sin?)/ ?
a
a
a
a
a
a
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question
light with a wavelength of 500 nm is used to
illuminate a slit of 5?m. At which angle is the
5th minimum in the diffraction pattern seen?
sin? m?/a ? sin-1(5x500x10-9/(5x10-6))300
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diffraction from a single hair
instead of an slit, we can also use an
inverse image, for example a hair! demo
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double slit interference revisited
The total response from a double slit system is a
combination of two single-source slits, combined
with a diffraction pattern from each of the slit
due to diffraction
minima asin?m?, m1,2,3 maxima asin?(m1/2)?,
m1,2,3 and ?0 a width of
individual slit
due to 2-slit interference
maxima dsin?m?, m0,1,2,3 minima
dsin?(m1/2)?, m0,1,2,3 d distance between
two slits
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double-slit experiment
a
d
if ?gtd, each slit acts as a single source of
light and we get a more or less prefect
double-slit interference spectrum
if ?ltd the interference spectrum is folded with
the diffraction pattern.
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question
A person has a double slit plate. He measures the
distance between the two slits to be d1 mm. Next
he wants to determine the width of each slit by
investigating the interference pattern. He finds
that the 7th order interference maximum lines up
with the first diffraction minimum and thus
vanishes. What is the width of the slits?
7th order interference maximum dsin?7? so
sin?7?/d 1st diffraction minimum asin?1? so
sin??/a sin? must be equal for both, so ?/a7?/d
and ad/71/7 mm
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diffraction grating
consider a grating with many slits, each
separated by a distance d. Assume that for each
slit ?gtd. We saw that for 2 slits maxima appear
if d sin? m?, m0,1,2,3 This condition is not
changed for in the case of n slits.
d
Diffraction gratings can be made by scratching
lines on glass and are often used to analyze light
instead of giving d, one usually gives the number
of slits per unit distance e.g. 300
lines/mm d1/(300 lines/mm)0.0033 mm
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separating colors
d sin? m?, m0,1,2,3 for maxima (same as for
double slit) so ? sin-1(m?/d) depends on ?,
the wavelength.
cds can act as a diffraction grating (DVDs work
even better because their tracks are more
closely spaced.)
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question

• If the interference conditions are the same when
  using a double slit or a diffraction grating with
  thousands of slits, what is the advantage of
  using the grating to analyze light?
• a) the more slits, the larger the separation
  between maxima.
• b) the more slits, the narrower each of the
  bright spots and thus easier to see
• c) the more slits, the more light reaches each
  maximum and the maxima are brighter
• d) there is no advantage

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question
An diffraction grating has 5000 lines per cm. The
angle between the central maximum and the fourth
order maximum is 47.20. What is the wavelength of
the light?
d sin? m?, m 0,1,2,3 d 1/5000 2x10-4 cm
2x10-6 m m 4, sin(47.2)0.734 so ? d sin?/m
2x10-6x0.734/4 3.67x10-7 m 367 nm
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polarization

• We saw that light is really an electromagnetic
  wave with electric and magnetic field vectors
  oscillating perpendicular to each other. In
  general, light is unpolarized, which means that
  the E-field vector (and thus the B-field vector
  as long as it is perpendicular to the E-field)
  could point in any direction

E-vectors could point anywhere unpolarized
propagation into screen
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polarized light

• light can be linearly polarized, which means that
  the E-field only oscillates in one direction (and
  the B-field perpendicular to that)
• The intensity of light is proportional to the
  square of amplitude of the E-field. IEmax2

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How to polarize?

• absorption
• reflection
• scattering

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polarization by absorption

• certain material (such as polaroid used for
  sunglasses) only transmit light along a certain
  transmission axis.
• because only a fraction of the light is
  transmitted after passing through a polarizer the
  intensity is reduced.
• If unpolarized light passes through a polarizer,
  the intensity is reduced by a factor of 2

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polarizers and intensity
polarization axis
direction of E-vector
For unpolarized light, on average, the
E-field has an angle of 450 with the
polarizer. II0cos2?I0cos2(45)I0/2
?
If E-field is parallel to polarization axis, all
light passes
If E-field makes an angle ? pol. axis only the
component parallel to the pol. axis passes
E0cos? So II0cos2?
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question

• unpolarized light with intensity I0 passes
  through a linear polarizer. It then passes
  through a second polarizer (the second polarizer
  is usually called the analyzer) whose
  transmission axis makes and angle of 300 with the
  transmission axis of the first polarized. What is
  the intensity of the light after the second
  polarizer, in terms of the intensity of the
  initial light?

After passing through the first polarizer,
I1I0/2. After passing through the second
polarizer, I2I1cos2300.75I10.375I0
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polarization by reflection

• If unpolarized light is reflected, than the
  reflected light is partially polarized.
• if the angle between the reflected ray and the
  refracted ray is exactly 900 the reflected light
  is completely polarized
• the above condition is met if for the angle of
  incidence the equation tan?n2/n1
• the angle ?tan-1(n2/n1) is called the Brewster
  angle
• the polarization of the reflected light is
  (mostly) parallel to the surface of reflection

n1
n2
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question

• Because of reflection from sunlight of the glass
  window, the curtain behind the glass is hard to
  see. If I would wear polaroid sunglasses that
  allow polarized light through, I would be able
  to see the curtain much better.
• a) horizontally
• b) vertically

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sunglasses
wearing sunglasses will help reducing glare
(reflection) from flat surfaces (highway/water)
without with sunglasses
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polarization by scattering

• certain molecules tend to polarize light when
  struck by it since the electrons in the molecules
  act as little antennas that can only oscillate in
  a certain direction