Transcript 23.2 Physics 6C EM interference

Physics 6C
Interference of EM Waves
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Constructive Interference:
Waves add - larger amplitude.
These waves are “In Phase”
Destructive Interference:
Waves cancel - smaller amplitude.
These waves are “Out of Phase”
They are out of sync by ½λ
Interference in action:
http://phys23p.sl.psu.edu/phys_anim/waves/embeder1.203.html
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Thin Film Interference
Basic idea is that we will compare the two reflections.
FORMULAS
If they are in phase we have constructive interference (bright).
2 t  m   constructi ve
If they are out of phase we have destructive interference (dark).
2 t  (m  1 )  destructiv e
 no relative shift
2
Important details:
1) When light reflects from a higher-index medium it is phase-shifted by ½ of a wavelength.
If both reflected rays have this shift we can ignore it, but if only one of them is shifted,
we have to switch the formulas for constructive/destructive interference.
2) The wavelength in the formulas is the wavelength in the film, so we have to divide the
vacuum wavelength by the index of the film.
Thin film demo
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Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
Prepared by Vince Zaccone
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Assistance Services at UCSB
Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
1
2
Air n=1
Water n=1.33
Air n=1
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Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
1
2
Air n=1
Water n=1.33
Air n=1
The outgoing rays will interfere, and there
is a relative phase shift, since ray 1 reflects
from a higher index, but ray 2 does not.
This yields the following formulas:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
1
2
Air n=1
Water n=1.33
Air n=1
For part a) we use the constructive formula,
with m=0 (we want the thinnest film possible)
652 nm
1
(
)  2  t  t min  123 nm
2
1 . 33
The outgoing rays will interfere, and there
is a relative phase shift, since ray 1 reflects
from a higher index, but ray 2 does not.
This yields the following formulas:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
1
2
Air n=1
Water n=1.33
Air n=1
For part a) we use the constructive formula,
with m=0 (we want the thinnest film possible)
652 nm
1
(
)  2  t  t min  123 nm
2
1 . 33
The outgoing rays will interfere, and there For part b) we use the destructive formula, with
is a relative phase shift, since ray 1 reflects different values of m (we want visible wavelengths)
from a higher index, but ray 2 does not.
This yields the following formulas:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
1
2
Air n=1
Water n=1.33
Air n=1
For part a) we use the constructive formula,
with m=0 (we want the thinnest film possible)
652 nm
1
(
)  2  t  t min  123 nm
2
1 . 33
The outgoing rays will interfere, and there For part b) we use the destructive formula, with
is a relative phase shift, since ray 1 reflects different values of m (we want visible wavelengths)
from a higher index, but ray 2 does not.
0
2  t n
This yields the following formulas:
m
0
n
m 
n
 2  t  0 
m
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
a) What is the minimum soap-film thickness (n=1.33) in air that
will produce constructive interference in reflection for red
(λ=652nm) light? (assume normal incidence)
b) Which visible wavelength(s) will destructively interfere when
reflected from a soap film of thickness 613nm? Assume a range of
350nm to 750nm for visible light.
1
2
Air n=1
Water n=1.33
Air n=1
For part a) we use the constructive formula,
with m=0 (we want the thinnest film possible)
652 nm
1
(
)  2  t  t min  123 nm
2
1 . 33
The outgoing rays will interfere, and there For part b) we use the destructive formula, with
is a relative phase shift, since ray 1 reflects different values of m (we want visible wavelengths)
from a higher index, but ray 2 does not.
0
2  t n
This yields the following formulas:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
m 
n
 2  t  0 
m
m  3   0  544 nm
m  4   0  407 nm
Other values of m give
wavelengths that fall
outside of the visible range
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Here’s a sample problem:
A thin layer of magnesium fluoride (n=1.38) is used to coat a flint-glass lens (n=1.61).
What thickness should the MgF2 coating be to suppress the reflection of 595nm light?
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Here’s a sample problem:
A thin layer of magnesium fluoride (n=1.38) is used to coat a flint-glass lens (n=1.61).
What thickness should the MgF2 coating be to suppress the reflection of 595nm light?
1
2
Air n=1
MgF2 n=1.38
Glass n=1.61
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Here’s a sample problem:
A thin layer of magnesium fluoride (n=1.38) is used to coat a flint-glass lens (n=1.61).
What thickness should the MgF2 coating be to suppress the reflection of 595nm light?
1
2
Air n=1
MgF2 n=1.38
Glass n=1.61
We need destructive interference (no reflection).
In this case both outgoing rays reflect from a
higher index, so there is no relative phase shift.
Our formulas are:
m
0
n
 2  t  constructi ve

(m  1 )  0  2  t  destructiv e
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
A thin layer of magnesium fluoride (n=1.38) is used to coat a flint-glass lens (n=1.61).
What thickness should the MgF2 coating be to suppress the reflection of 595nm light?
1
2
Air n=1
MgF2 n=1.38
Glass n=1.61
We need destructive interference (no reflection).
In this case both outgoing rays reflect from a
higher index, so there is no relative phase shift.
Our formulas are:
m
0
n
 2  t  constructi ve

(m  1 )  0  2  t  destructiv e
2
n
We can use any integer m>0, so start
with m=0 and solve for t.
This will give the minimum thickness.
595 nm
(0  1 ) 
 2  t  t min  108 nm
2
1 . 38
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Here’s a sample problem:
A thin layer of magnesium fluoride (n=1.38) is used to coat a flint-glass lens (n=1.61).
What thickness should the MgF2 coating be to suppress the reflection of 595nm light?
1
2
Air n=1
MgF2 n=1.38
Glass n=1.61
We need destructive interference (no reflection).
In this case both outgoing rays reflect from a
higher index, so there is no relative phase shift.
Our formulas are:
m
0
n
 2  t  constructi ve

(m  1 )  0  2  t  destructiv e
2
n
To get other possible thicknesses that will
work, just use larger values for m:
We can use any integer m>0, so start
with m=0 and solve for t.
m  1  t  323 nm
This will give the minimum thickness.
etc ...
595 nm
(0  1 ) 
 2  t  t min  108 nm
2
1 . 38
m  2  t  539 nm
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Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
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Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
Water n=1.33
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
Water n=1.33
In this case ray 1 reflects from a higher index,
but ray 1 reflects from a lower index, so there
is a relative phase shift.
Our formulas are:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
We will use the destructive interference
formula for each given wavelength. Since
they are the only visible wavelengths that are
absent, we can deduce that they correspond
to consecutive values for m in the formula.
Water n=1.33
In this case ray 1 reflects from a higher index,
but ray 1 reflects from a lower index, so there
is a relative phase shift.
Our formulas are:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
Water n=1.33
In this case ray 1 reflects from a higher index,
but ray 1 reflects from a lower index, so there
is a relative phase shift.
We will use the destructive interference
formula for each given wavelength. Since
they are the only visible wavelengths that are
absent, we can deduce that they correspond
to consecutive values for m in the formula.
m
687 nm
1 . 40
(m  1 ) 
 2t
458 nm
1 . 40
 2t
Our formulas are:
m
0
n
 2  t  destructiv e

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
Water n=1.33
In this case ray 1 reflects from a higher index,
but ray 1 reflects from a lower index, so there
is a relative phase shift.
Our formulas are:
m
0
n
 2  t  destructiv e
We will use the destructive interference
formula for each given wavelength. Since
they are the only visible wavelengths that are
absent, we can deduce that they correspond
to consecutive values for m in the formula.
m
687 nm
1 . 40
(m  1 ) 
 2t
458 nm
1 . 40
 2t
At this point we have some algebra to do. My
preference is to find the integer value of m that
fits the formulas, then plug that in to find t.

(m  1 )  0  2  t  constructi ve
2
n
Prepared by Vince Zaccone
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Assistance Services at UCSB
Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
We will use the destructive interference
formula for each given wavelength. Since
they are the only visible wavelengths that are
absent, we can deduce that they correspond
to consecutive values for m in the formula.
Water n=1.33
m
In this case ray 1 reflects from a higher index,
but ray 1 reflects from a lower index, so there
is a relative phase shift.
Our formulas are:
m
0
n
 2  t  destructiv e
(m  1 ) 
2
0
n
 2  t  constructi ve
687 nm
1 . 40
(m  1 ) 
 2t
458 nm
1 . 40
 2t
At this point we have some algebra to do. My
preference is to find the integer value of m that
fits the formulas, then plug that in to find t.
m 
687 nm
1 . 40
 (m  1) 
458 nm
1 . 40
(m )( 687 )  (m  1)( 458 )
(m )( 687 )  (m )( 458 )  (1)( 458 )
229 m  458  m  2
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Here’s a sample problem:
A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only
colors that are absent from the reflected light are blue (458nm) and red (687nm).
Estimate the thickness of the film.
1
2
Air n=1
Oil n=1.40
We will use the destructive interference
formula for each given wavelength. Since
they are the only visible wavelengths that are
absent, we can deduce that they correspond
to consecutive values for m in the formula.
Water n=1.33
m
In this case ray 1 reflects from a higher index,
but ray 2 reflects from a lower index, so there
is a relative phase shift.
Our formulas are:
m
0
n
 2  t  destructiv e
(m  1 ) 
2
0
n
 2  t  constructi ve
687 nm
1 . 40
(m  1 ) 
 2t
458 nm
1 . 40
 2t
At this point we have some algebra to do. My
preference is to find the integer value of m that
fits the formulas, then plug that in to find t.
m 
687 nm
1 . 40
 (m  1) 
458 nm
1 . 40
(m )( 687 )  (m  1)( 458 )
(m )( 687 )  (m )( 458 )  (1)( 458 )
229 m  458  m  2
2
687 nm
1 . 40
 2  t  t  491 nm
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Young’s Double Slit Experiment
Monochromatic Light
In Young’s double-slit experiment light comes from the left and passes through the slits,
illuminating the screen some distance R away. The light rays from the 2 slits will travel
different distances to get to the screen (except in the center). Depending on the path length
difference the waves will be in phase or out of phase when they arrive at the screen. If they
are in phase, they combine to give constructive interference (a bright region). Out of phase
means destructive interference (dark region). Some geometry gives us a formula for this
difference in path length: dsin(θ). This yields the following formulas:
d  sin(  )  m   constructi
ve
d  sin(  )  (m  1 )   destructiv
2
tan(  ) 
y
R
e
m can be any integer
(m=0,±1,±2,±3,±4,…)
y = actual distance on screen (from center)
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Here’s a sample problem:
Light with wavelength 546nm passes through two slits and forms an interference pattern on a
screen 8.75m away. If the linear distance on the screen from the central fringe to the first bright
fringe above it is 5.16cm, what is the separation of the slits?
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Here’s a sample problem:
Light with wavelength 546nm passes through two slits and forms an interference pattern on a
screen 8.75m away. If the linear distance on the screen from the central fringe to the first bright
fringe above it is 5.16cm, what is the separation of the slits?
This should be a straightforward problem. We are given enough information to just
use our formulas. Start with the formula involving the distance on the screen:
tan(  ) 
y
R
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Here’s a sample problem:
Light with wavelength 546nm passes through two slits and forms an interference pattern on a
screen 8.75m away. If the linear distance on the screen from the central fringe to the first bright
fringe above it is 5.16cm, what is the separation of the slits?
This should be a straightforward problem. We are given enough information to just
use our formulas. Start with the formula involving the distance on the screen:
tan(  ) 
y
R
2
3
tan(  )  5 . 16 10 m  5 . 897  10
8 . 75 m
  0 . 338

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Here’s a sample problem:
Light with wavelength 546nm passes through two slits and forms an interference pattern on a
screen 8.75m away. If the linear distance on the screen from the central fringe to the first bright
fringe above it is 5.16cm, what is the separation of the slits?
This should be a straightforward problem. We are given enough information to just
use our formulas. Start with the formula involving the distance on the screen:
tan(  ) 
y
R
2
3
tan(  )  5 . 16 10 m  5 . 897  10
8 . 75 m
  0 . 338

Use this angle in the formula for bright fringes, with m=1
d  sin(  )  m 
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Here’s a sample problem:
Light with wavelength 546nm passes through two slits and forms an interference pattern on a
screen 8.75m away. If the linear distance on the screen from the central fringe to the first bright
fringe above it is 5.16cm, what is the separation of the slits?
This should be a straightforward problem. We are given enough information to just
use our formulas. Start with the formula involving the distance on the screen:
tan(  ) 
y
R
2
3
tan(  )  5 . 16 10 m  5 . 897  10
8 . 75 m
  0 . 338

Use this angle in the formula for bright fringes, with m=1
d  sin(  )  m 

d  sin( 0 . 338 )  546  10
d  9 . 26  10
5
9
m
m
Since the angle was small we could have used the approximate formula: y m  R
m
d
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Multiple Slits (diffraction gratings)
These work just like the double slit experiment (same formula), but the bright spots are
narrower, and the dark spots are wider. If the grating has more slits the result is a sharper
image, with narrower bright fringes.
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Diffraction
•When light encounters an obstacle it will exhibit diffraction effects as the
light bends around the object or passes through a narrow opening.
•Notice the alternating bright and dark bands around the edge of the razor
blade. This is due to constructive and destructive interference of the light
waves.
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Single Slit Diffraction
Formulas for Destructive
Interference (dark fringes)
a  sin(  )  (m )  
ym  R
(m )  
a
Formulas for Constructive
Interference (bright fringes)
a  sin(  )  (m 
These approximate
formulas work when
the angle is small
ym  R
(m 
1
2
1
2
)
)
a
•Similar to the double-slit experiment.
•The formulas are opposite (the geometry just comes out that way).
•Notice that the central maximum is double-width compared to the others.
•This is how you can tell a single-slit pattern from a multiple-slit pattern.
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Here’s a sample problem:
How many dark fringes will be produced on either side of the central maximum if green light
(λ=553nm) is incident on a slit that is 2µm wide?
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Here’s a sample problem:
How many dark fringes will be produced on either side of the central maximum if green light
(λ=553nm) is incident on a slit that is 2µm wide?
This is a single-slit problem, so the formula for the dark fringes is:
a  sin(  )  m 
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Here’s a sample problem:
How many dark fringes will be produced on either side of the central maximum if green light
(λ=553nm) is incident on a slit that is 2µm wide?
This is a single-slit problem, so the formula for the dark fringes is:
a  sin(  )  m 
Let’s find the angles to the first few dark fringes. We get a new angle for each value of m.
m
θ
1
16°
2
34°
3
56°
4
???
When we try to calculate with m=4 we get a
calculator error. Why doesn’t it work?
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Here’s a sample problem:
How many dark fringes will be produced on either side of the central maximum if green light
(λ=553nm) is incident on a slit that is 2µm wide?
This is a single-slit problem, so the formula for the dark fringes is:
a  sin(  )  m 
Let’s find the angles to the first few dark fringes. We get a new angle for each value of m.
When we try to calculate with m=4 we get a
calculator error. Why doesn’t it work?
m
θ
1
16°
2
34°
Recall the single-slit diffraction diagram.
3
56°
4
???
For the fringes to show up on the screen,
the angle must be less than 90°.
R
Of course the pattern gets very dim near
the edges, but mathematically the
formula will break down when sin(θ)>1.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Here’s a sample problem:
How many dark fringes will be produced on either side of the central maximum if green light
(λ=553nm) is incident on a slit that is 2µm wide?
This is a single-slit problem, so the formula for the dark fringes is:
a  sin(  )  m 
Let’s find the angles to the first few dark fringes. We get a new angle for each value of m.
When we try to calculate with m=4 we get a
calculator error. Why doesn’t it work?
m
θ
1
16°
2
34°
Recall the single-slit diffraction diagram.
3
56°
4
???
For the fringes to show up on the screen,
the angle must be less than 90°.
R
Of course the pattern gets very dim near
the edges, but mathematically the
formula will break down when sin(θ)>1.
So it looks like we will get 3 dark fringes.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Circular Aperture
Light passing through a circular opening gives a circular pattern.
A formula to find the first dark fringe is:
𝑠𝑖𝑛𝜃 = 1.22
𝜆
𝐷
This can be taken as the angular resolution of the aperture. When two light sources are close
together this angle limits our ability to “resolve” them as separate objects.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Circular Aperture
Example: You are driving at night on a long straight highway in the desert as another vehicle
approaches. What is the maximum distance at which you can tell that it is a car rather than a
motorcycle by seeing its headlights, which are separated by a distance of 1.5m?
a) Assume your visual acuity is limited only by diffraction. Use 550 nm for the wavelength, and
pupil diameter 6.0mm.
b) What answer do you get if you use a more realistic, typical visual acuity with θmin=5x10-4 rad?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB