interference, diffraction & polarization phy232 – spring 2007 jon pumplin pumplin/phy232 (ppt...
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Interference, Diffraction & Polarization
PHY232 – Spring 2007
Jon Pumplin
http://www.pa.msu.edu/~pumplin/PHY232
(Ppt courtesy of Remco Zegers)
PHY232 - Remco Zegers - interference, diffraction & polarization 2
light as waves
so far, light has been treated as if it travels in straight linesray diagramsrefraction, 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
PHY232 - Remco Zegers - interference, diffraction & polarization 3
interference
constructive interference destructive interference at any point in time one can construct the total amplitudeby adding the individual components
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Interference III
constructive interferencewaves in phase
demo: interference
+
=
destructive interferencewaves ½ out of phase
+
=
PHY232 - Remco Zegers - interference, diffraction & polarization 5
Interference in spherical wavesmaximum of wave minimum of wave
positive constructive interference
negative constructive interference
destructive interference
if r2-r1 = n then constructive interference occursif r2-r1 = (n+½) then destructive interference occurs
r1
r2
r1=r2
PHY232 - Remco Zegers - interference, diffraction & polarization 6
PHY232 - Remco Zegers - interference, diffraction & polarization 7
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 isno phase shift between the two sources
•the two sources of light must have identical wave lengths
PHY232 - Remco Zegers - interference, diffraction & polarization 9
Young’s interference experiment
there is a path difference: depending on its size the wavescoming from S1 or S2 are in or out of phase
PHY232 - Remco Zegers - interference, diffraction & polarization 10
Young’s 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, m=0,1,2,3…
m = 0: zeroth order, m=1: 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 = [m+1/2], m=0,1,2,3…)
path difference demo
PHY232 - Remco Zegers - interference, diffraction & polarization 11
distance between bright spots
if is small, then sin tan so: d sin = m, m=0,1,2,3… converts to dy/L = m difference between maximum m and maximum m+1:ym+1-ym= (m+1)L/d-mL/d= L/dym=mL/d
tan=y/L
L
demo
PHY232 - Remco Zegers - interference, diffraction & polarization 12
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
distance is equalso 1/2 difference:destructive int.
PHY232 - Remco Zegers - interference, diffraction & polarization 13
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?
a) use d sin = m with m=3 =sin-1(m/d)=sin-1(3x600x10-9/0.01)=0.01030
b) Ym = mL/d m=0: y0 =0 m=3: y3 = 3x600x10-9x5/0.01 = 9x10-4 m = 0.9 mm
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other ways of causing interference
remember
equivalent to:
1 2n1<n2
n1>n2
1 2
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phase changes at boundaries
If a light ray travels from medium 1 to medium 2 with n1<n2,the phase of the light ray will change by 1/2. This will not happen if n1>n2.
1 2
n1<n2
1/2 phase change
n1>n2
1 2
no phase changeIn a medium with index of refraction n, the wavelengthchanges (relative to vacuum) to /n
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thin film interference
n=1
n=1.5
n=1
The two reflected rays caninterfere. To analyze this system,4 steps are needed:
1. Is there phase inversion at the top surface?2. Is there phase inversion at the bottom surface3. What are the conditions for
constructive/destructive interference?4. what should the thickness d be for 3) to
happen?
PHY232 - Remco Zegers - interference, diffraction & polarization 17
n=1
n=1.5
n=1
thin film analysis1. top surface?2. bottom surface?3. conditions?4. d?
1. top surface: n1<n2 so phase inversion 1/2 2. bottom surface: n1>n2 so no phase inversion3. conditions:
1. constructive: ray 1 and 2 must be in phase2. destructive: ray 1 and 2 must be out of phase by 1/2
4. if phase inversion would not take place at any of the surfaces: constructive: 2d=m (difference in path length=integer number of wavelengths) due to phase inversion at top surface: 2d=(m+1/2) since the ray travels through film: 2d=(m+1/2)film =(m+1/2)/nfilm
destructive: 2d=mfilm =m/nfilm
1 2
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Note
The interference is different for light of differentwavelengths
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question
na=1
nb=1.5
nc=2
Phase inversion will occur ata) top surfaceb) bottom surfacec) top and bottom surfaced) neither surface
n1<n2 in both cases
constructive interference will occur if:a) 2d=(m+1/2)/nb
b) 2d=m/nb
c) 2d=(m+1/2)/nc
d) 2d=m/nc
note: if destructive 2d=(m+1/2)/nb
this is used e.g. on sunglasses to reduce reflections
PHY232 - Remco Zegers - interference, diffraction & polarization 20
another case
The air gap in between the plates has varying thickness.Ray 1 is not inverted (n1>n2)Ray 2 is inverted (n1<n2)where the two glasses touch: no path length difference:dark fringe.if 2t=(m+1/2) constructive interferenceif 2t=m destructive interference.
1 2
PHY232 - Remco Zegers - interference, diffraction & polarization 21
question
Given h=1x10-5 m30 bright fringes are seen,with a dark fringe at the leftand the right.What is the wavelength ofthe light?
2t=m destructive interference.m goes from 0 (left) to 30 (right).=2t/m=2h/m=2x1x10-5/30=6.67x10-7 m=667 nm
PHY232 - Remco Zegers - interference, diffraction & polarization 22
newton’s rings
spacing not equal
demo
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quiz (extra credit)
Two beams of coherent light travel different paths arrivingat point P. If constructive interference occurs at point P,the two beams must:a) travel paths that differ by a whole number of wavelengthsb)travel paths that differ by an odd number of halfwavelengths
PHY232 - Remco Zegers - interference, diffraction & polarization 24
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?a) the very narrow slits required would generate
different wavelength, thereby washing out the
interference patternb) the two slits would not emit coherent lightc) the fringes would be too close togetherd) in no direction could a path difference as large as
one wavelength be obtained
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diffraction
In Young’s experiment, two slits were used to producean 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/smallopenings.
In fact, every point in the slit openingacts as the source of a new wave front
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PHY232 - Remco Zegers - interference, diffraction & polarization 27
interference pattern from a single slit
pick two points, 1 and 2, one inthe top top half of the slit, one in the bottom half of the slit.Light from these two points interferesdestructively if:x=(a/2)sin=/2 so sin=/a
we could also have divided up the slitinto 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 m=1,2,3…
In between the minima, are maxima: sin = (m+1/2)/a m=1,2,3… AND sin=0 or =0
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slit width
if >a sin=/a > 1 Not possible, so nopatterns
if <<a sin=m/a is very smalldiffraction hardly seen
<a : interference pattern is seen
aa
PHY232 - Remco Zegers - interference, diffraction & polarization 29
the diffraction pattern
The intensity is not uniform:
I=I0sin2()/2 =a(sin)/
a a a a a a
PHY232 - Remco Zegers - interference, diffraction & polarization 30
question
light with a wavelength of 500 nm is used to illuminatea slit of 5m. 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 inverseimage, 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 witha diffraction pattern from each of the slit
due to diffraction
due to 2-slit interference
maxima dsin=m, m=0,1,2,3…minima dsin=(m+1/2), m=0,1,2,3…d: distance between two slits
minima asin=m, m=1,2,3…maxima asin=(m+1/2), m=1,2,3… and =0a: width of individual slit
PHY232 - Remco Zegers - interference, diffraction & polarization 33
double-slit experimenta
if >d, each slit acts as a singlesource of light and we geta more or less prefect double-slitinterference spectrum
if <d the interference spectrumis folded with the diffraction pattern.
d
PHY232 - Remco Zegers - interference, diffraction & polarization 34
7th7th
A person has a double slit plate. He measures the distance between the two slits to be d=1 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 andthus 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=/asin must be equal for both, so /a=7/d and a=d/7=1/7 mm
question
PHY232 - Remco Zegers - interference, diffraction & polarization 35
diffraction grating
d
consider a grating withmany slits, each separated bya distance d. Assume that foreach slit >d. We saw that for 2 slits maxima appear if:d sin = m, m=0,1,2,3…This condition is not changed for in the case of n slits.
Diffraction gratings can be madeby scratching lines on glass andare often used to analyze light
instead of giving d, one usuallygives the number of slits perunit distance: e.g. 300 lines/mmd=1/(300 lines/mm)=0.0033 mm
PHY232 - Remco Zegers - interference, diffraction & polarization 36
separating colors
d sin = m, m=0,1,2,3… for maxima (same as for double slit)so = sin-1(m/d) depends on , the wavelength.
cd’s can act as a diffraction grating(DVD’s work even better because their tracks are more closely spaced.)
PHY232 - Remco Zegers - interference, diffraction & polarization 37
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
PHY232 - Remco Zegers - interference, diffraction & polarization 38
question
An diffraction grating has 5000 lines per cm. The anglebetween the central maximum and the fourth ordermaximum 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 mm = 4, sin(47.2)=0.734so = d sin/m = 2x10-6x0.734/4 = 3.67x10-7 m = 367 nm
PHY232 - Remco Zegers - interference, diffraction & polarization 39
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
propagation into screen
E-vectors could pointanywhere: unpolarized
PHY232 - Remco Zegers - interference, diffraction & polarization 40
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. I~Emax
2
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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
PHY232 - Remco Zegers - interference, diffraction & polarization 43
polarizers and intensitypolarization
axisdirection of E-vector
If E-field is parallelto polarization axis,all light passes
If E-field makes an angle pol. axisonly the componentparallel to the pol. axispasses: E0cosSo I=I0cos2
For unpolarized light, on average, the E-fieldhas an angle of 450 withthe polarizer.I=I0cos2=I0cos2(45)=I0/2
PHY232 - Remco Zegers - interference, diffraction & polarization 44
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, I1=I0/2. After passing throughthe second polarizer, I2=I1cos230=0.75I1=0.375I0
PHY232 - Remco Zegers - interference, diffraction & polarization 45
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
n2
n1
PHY232 - Remco Zegers - interference, diffraction & polarization 46
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
horizontalverticaldirection of polarizationof reflected light
PHY232 - Remco Zegers - interference, diffraction & polarization 47
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