Transcript Interference, Diffraction & Polarization PHY232 – Spring 2007 Jon Pumplin http://www.pa.msu.edu/~pumplin/PHY232 (Ppt courtesy of Remco Zegers)

Interference, Diffraction & Polarization
PHY232 – Spring 2007
Jon Pumplin
http://www.pa.msu.edu/~pumplin/PHY232
(Ppt courtesy of Remco Zegers)
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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destructive interference
waves ½ out of phase
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Interference in spherical waves
maximum of wave r =r
1
2
minimum of wave
r1
r2
positive constructive interference
negative constructive interference
destructive interference
if r2-r1 = n
then constructive interference occurs
if r2-r1 = (n+½) then destructive interference occurs
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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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Young’s 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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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.
path difference
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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…)
demo
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distance between bright spots
tan=y/L
L
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-mL/d= L/d
demo
ym=mL/d
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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
distance is equal
 a) constructive interference
so 1/2 difference:
 b) destructive interference
destructive int.
+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?
a) use d sin = m with m=3
=sin-1(m/d)=sin-1(3x600x10-9/0.01)=0.01030
b) Ym = mL/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:
n1>n2
n1<n2
2
1
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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.
n1>n2
1
2
1
2
n1<n2
no phase change
1/2 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
n=1
The two reflected rays can
interfere. To analyze this system,
4 steps are needed:
n=1.5
n=1
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/destructive
interference?
4. what should the thickness d be for 3) to happen?
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thin film analysis
n=1
1
1.
2.
3.
4.
2
n=1.5
top surface?
bottom surface?
conditions?
d?
1. top
surface: n1<n2 so phase inversion 1/2
n=1
2. bottom surface: n1>n2 so no phase inversion
3. conditions:
1. constructive: ray 1 and 2 must be in phase
2. 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=m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
a) top surface
b) bottom surface
c) top and bottom surface
d) neither surface
na=1
nb=1.5
n1<n2 in both cases
nc=2
constructive interference will occur if:
a)
b)
c)
d)
2d=(m+1/2)/nb
2d=m/nb
2d=(m+1/2)/nc
2d=m/nc
PHY232 - Remco Zegers
note: if destructive 2d=(m+1/2)/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 (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 interference
if 2t=m destructive interference.
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question
Given h=1x10-5 m
30 bright fringes are seen,
with a dark fringe at the left
and the right.
What is the wavelength of
the 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
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newton’s 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:
a) 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?
a) the very narrow slits required would generate different
wavelength, thereby washing out the interference pattern
b) the two slits would not emit coherent light
c) the fringes would be too close together
d) 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 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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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 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
a
a
if >a sin=/a > 1
Not possible, so no
patterns
PHY232 - Remco Zegers
<a : interference
pattern is seen
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if <<a
sin=m/a is very small
diffraction hardly seen
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the diffraction pattern
The intensity is not uniform:
I=I0sin2()/2 =a(sin)/ 
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, m=1,2,3…
maxima asin=(m+1/2), m=1,2,3…
and =0
a: width of individual slit
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
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double-slit experiment
a
d
if >d, each slit acts as a single
source of light and we get
a more or less prefect double-slit
interference spectrum
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if <d the interference spectrum
is folded with the diffraction
pattern.
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question
7th
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 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 /a=7/d and a=d/7=1/7 mm
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diffraction grating
consider a grating with
many slits, each separated by
a distance d. Assume that for
each slit >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.
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
d=1/(300 lines/mm)=0.0033 mm
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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.)
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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 Efield) 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 Efield 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~Emax2
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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

If E-field is parallel
to polarization axis,
all light passes
PHY232 - Remco Zegers
For unpolarized light, on
average, the E-field
has an angle of 450 with
the polarizer.
I=I0cos2=I0cos2(45)=I0/2
If E-field makes an
angle  pol. axis
only the component
parallel to the pol. axis
passes: E0cos
So I=I0cos2
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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, I1=I0/2. After passing through
the second polarizer, I2=I1cos230=0.75I1=0.375I0
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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
vertical
direction of
polarization
of reflected
light
horizontal
 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
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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
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