Download Chapter 37 - UBC Physics

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Light
Section 20.5, Electromagnetic waves
Light is an electromagnetic which travels at 3.00×108m/s in a vacuum. It obeys the relationship:
λf = c just like other traveling waves.
For visible light, 700nm(red) > λ > 400 nm(indigo) See page 631 for wavelengths of colours
Index of refraction = n =
Speed of light in vacuum = c/v See table 23.1, page 722
Speed of light in medium
Recall that the velocity of light is different in different media. In Diamond it is about 40% as fast as in a vacuum. When light
moves from one medium to another, its velocity changes and its wavelength changes, but its frequency does not change. This
is like having a transverse wave travel from a thin rope to a thick rope.
When a light wave travels from one medium to another, its wavelength changes. Let λ be the
wavelength in vacuum. In a medium with index of refraction, n, the wavelength is λn.
λ = c/f; λn = vn/f ; then λn/λ = vn/c = 1/n and λn = λ/n
Section 20.7 Doppler effect for light
For a receding source, λ = √(1 + vs/c)/(1 - vs/c) λo
Red shift
For an approaching source, λ = √(1 - vs/c)/(1 + vs/c) λo
Blue shift
Light originating from other galaxies is
red shifted. This provides evidence for
the Big Bang- the time when the
universe started expanding from a hot
dense source. This happened about 14
billion years ago.
Section 21.6 Thin film optical coatings
An electromagnetic wave undergoes a phase change of 180o upon reflection from a medium that
has a higher index of refraction than the one in which the wave is travelling.
When light is reflected from a mirror it undergoes a phase change of 180o.
Phase change of 180o
Air
No phase change
medium(n>nair)
(Hard Reflection)
air
medium (n>nair)
(Soft Reflection)
air
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In Physics 101, we are not going to discuss the Physics of this phase change. The mechanical
wave analogy is the reflection of waves on a string which is rigidly attached to the wall as
compared to one which can move freely up and down.
Vs
Consider a thin film of uniform thickness t and index of refraction n with light rays incident from
nearly normal to the two surfaces of the film.
180 phase
change
No phase
change
This can be a model for soap
bubbles or peacock feathers.
air
n
t
air
Note that here we are
drawing the light waves at
an angle nearly normal to
the surface. Otherwise the
path length would not be 2t
and the path lengths of the
two rays outside the surface
would be different
When a light wave travels from one medium to another, its frequency does not change but its
wavelength changes:
λn = λ/n where λ is the wavelength in free space. (or air where n= 1.0003)
Applying the above rules, the condition for constructive interference of the two waves is that the
phase difference between the two outgoing waves is 2π
Phase difference between the two waves =
Phase change in film plus phase change upon reflection =
∆φ = 2π∆r/λ + ∆φo
For constructive interference, ∆φ = m2π; for destructive interference ∆φ = (m + ½ )2π
This is all you
need to know to
solve this kind
of problem.
For the above film: ∆φ = (2π) 2t/λn+ π = 2π(2t/λn + ½)
Then for constructive interference here: 2π(2t/λn + ½) = m2π, m =0, 1, 2, divide by 2π to get:
2t/λn = (m + ½), m =0, 1, 2, 3,..
These values make the phase change a multiple of 2π.
Note that m is always positive (or 0) since t > 0.
Note we could also have used 2nt = (m - ½)λ
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Or 2nt =(m + ½)λ
And for destructive interference: 2t/λn = m, m = 1, 2, 3..
These values make the phase change an odd multiple of π.
Or 2nt = mλ
For these problems, it is most important to understand the principles. You need not try to memorise the formulae for
each situation but rather you should work out each situation using λn = λ/n and the π phase change law.
Problem: A thin film of oil (n=1.25) is located on a smooth wet pavement. When viewed
perpendicular to the pavement, the film appears red (640)nm and has no blue (512nm) color.
How thick is the oil film?
1 2
t
Oil
Water
We assume that the light doesn’t
reflect from the dark surface of the
pavement
Note that the refractive index of water is 1.333 from table 23.1, hence the light wave undergoes a
180 phase change at both surfaces, hence the phase shift is entirely determined by the path length
difference.
∆φ = 2π∆r/λ + ∆φo
∆φo = 0
Then, the phase difference between path 1 and path 2 for constructive interference (red light) is
2π(2t/(λ(red)/noil)); for constructive interference, this phase difference is m2π so:
2t = mλ(red)/noil;
t = mλ(red)/2noil = m(640nm)/2.5 = m(256nm) = 256nm, 512nm, 768nm,
..
and for destructive interference (blue light)
2t = (m+ ½)λ(blue)/noil ; t = (m+½) (512nm)/2.50 = (m+½) (204.8nm) = 102.4nm, 307.2nm,
512nm
Then t = 512 nm
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Problem: An air wedge is formed between two glass plates separated at one edge by a very fine
wire. When the wedge is illuminated from above by 600nm light, 30 dark fringes are observed
across the plates. Calculate the radius of the wire.
1
y
2
t
Note that since the top and bottom plates have flat parallel surfaces, they result in uniform phase
shifts of the light waves. Only the air between the plates produces a phase shift which varies
along the horizontal direction.
nglass > nair, therefore there is a π phase shift when light reflects from the bottom surface of the air
cavity but no phase shift when light reflects from the top surface of the air cavity.
Recall ∆φ = 2π∆r/λ + ∆φo
Beam 1 undergoes no phase shift relative to the incident beam due to reflection, Beam 2 has a π
phase shift.
Phase shift between the two beams from the air gap:
φ = 2y/(λ/n)*2π + π = [2ny/λ](2π) + π = (m + ½)2π for destructive interference (dark fringe).
This simplifies to 2ny/λ = m
Note that y varies from 0 to t, the thickness of the wire, since the phase shift is π for y=0, the
pattern starts with a dark fringe at the left. As we move from left to right, 30 fringes, or 29 times
2π phase shift have occurred.
Then 2nt/λ = 29 and t = 29λ/(2n) = 29(600x10-9m)/(2×1.00) = 8.7x10-6m. Then the radius of the
wire is 4.35 µ A typical human hair is 50µ.
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Section 22-6: The Michelson Interferometer
The Michelson interferometer is used to measure light wavelengths or other lengths with great
precision.
M1
MS – Beam splitter
L1
M1 - fixed mirror
MS
L2
M2 - Adjustable mirror
M2
Light source
(coherent, monochromatic)
Observer
If M2 is not exactly perpendicular to M1, the observer observes a fringe pattern. If they were
exactly perpendicular, the image would be uniform.
At any point in the image at the observer, the intensity undergoes one complete cycle as M2 is
moved by λ/2. (since light travels a distance 2L2).
Problem: Light of wavelength 550.5 nm is used to calibrate a Michelson interferometer. How
many dark fringes are counted as the mirror is moved 0.180mm?
-3
-9
Answer: # fringes = ∆L/λ/2 = 2(0.180x10 m)/(550.2x10 m) = 654.31 = 654
Chapter 22: Wave Optics
Note that the Michelson
interferometer can be
used to measure index of
refraction (Mastering
Physics)
Huygens' Principle: Every point in a wavefront can be considered as a source of tiny
wavelets that spread out spherically at the speed of the wave itself. At a later time the shape
of the wave front is the line tangent to all the wavelets
Examples:
Diffraction: A
simple application of
Huygens' law.
At each point in the wave front, tiny wavelets move forward at the wave velocity with a spherical wavefront. This
is a wave property, particles. The property of light 'bending' around corners leads to interference and diffraction.
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Section 22-2 Interference of light
Young's Double Slit Experiment
Consider a system consisting of light source, a double slit and a viewing screen
Light
Source
r1
d
y
θ
θ
L
r2
If L >>d, the angles from each ray are equal
to θ.
dsinθ
Conditions for Interference (or light waves)
For observation of interference effects with light waves, the light source should be coherent (i.e.
constant phase relationship) and monochromatic (i.e. same wavelength).
Interference fringes are seen on viewing screen at a distance y from the centre line.
If we assume that r1 and r2 are parallel (note L>> d), then
when ∆r = r2 - r1 = dsin θ = mλ m = 0, ±1, ±2,.. there is constructive interference.
when ∆r = r1 - r2 = dsin θ = (m+½)λ m = 0, ±1, ±2, .. there is destructive interference.
tanθ = y/L
m is the order of the fringe. The intensity of the fringes is greatest for the central fringe (m=0)
and decreases for higher orders. This is due to diffraction, which we discuss later.
Problem: A Young's double slit experiment uses 589 nm light and a slit to screen distance of
2.00m. The 10th interference minimum is observed 7.26 mm from the central maximum. What is
the spacing of the slits?
y
θ
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Interference minima occur when dsinθ = (m+½)λ, m= 0,±1, ±2,.. The 10th interference
minimum occurs when m = +9; tanθ = y/L;
-1
-1
Note that the peaks are quite
close to each other- you
might need glasses
-3
Then θ = tan (y/L) = tan [(7.26x10 m)/(2.00m)] = 0.0036 rad
Then d = (9+½)λ/sinθ = 9.5(589x10-9m)/sin(0.0036) = 0.00155m = 1.55mm
Intensity in the Double Slit Interference Pattern
Consider two light beams passing through the double slit. They have electric fields given by
P
D1 = DMsin(ωt) and D2 = DMsin(ωt + ∆φ)
D1
We are covering the spatial phase dependence
by inserting the relative phase difference ∆φ.
θ
D2
dsinθ
The phase difference ∆φ is 2π times the difference in path lengths (d sinθ) divided by λ
i. e. ∆φ = 2πdsinθ/λ
The phase difference passes through 2π radians along each
wavelength. This is exactly analogous to the speaker problem
At the screen, we measure the superposition of the two beams
Recall the identity sin (a) + sin (b) = 2cos[(a-b)/2] sin[(a+b)/2]
Then at P, Dθ = D1 + D2 = 2DMcos(∆φ/2)sin(ωt + ∆φ/2)
The intensity of a light wave is proportional to the square of the wave amplitude (recall that I =
½ρω2DM2v for a single traveling wave)
So the light intensity, Iθ, at point P is proportional to Dθ2 = 4DM2cos2(∆φ/2)sin2(ωt + ∆φ/2)
The time-average of sin2(ωt + ∆φ/2) is ½. This variation of intensity with time at ~ 1014 Hz (=
c/λ= 3x108m/s/600x10-9m = 5x1014) is not visible.
Note that I is the intensity arising
1
Iθ = 4I1cos2(∆φ/2) or 4I1cos2(πdsinθ/λ) using ∆φ = 2πdsinθ/λ.
from a single slit. The factor ½ from
the sin2 average is included in I1.
Going back to the diagram for the Young's Double slit experiment, we see that for large L and
small y, we can write sin θ = y/L then ∆φ = 2πdy/λL and Iθ = 4I1cos2(πdy/λL).
This expression enables us to calculate the intensity anywhere in the double slit interference
pattern; before we could only find the maxima and the minima.
We will learn soon that due to diffraction that the central maximum is the most intense and the others fall off with
distance from the central maximum. This is due to interference effects between light rays going through the same slit.
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-4π
-3π
-2π
-π
0
π
2π
3π
4π
Intensity
∆φ
-2λL/d
-3/2λL/d
-λL/d
-λL/2d
-15.7 -12.56 -9.42
-6.28
0
-3.14
λL/2d
0
λL/d
3.14
3/2λL/d
2λL/d
9.42
12.56
6.28
15.7
PHI(rad)
Plot of theoretical intensity in the Young’s double slit measurement
Section 22-3 The diffraction grating
A diffraction grating contains a very large number of slits produced by a large number of parallel
lines on a glass plate. Gratings can contain 10,000 lines per cm.
Coherent
monochromatic
light source
d
θ
dsinθ
On the screen there are well defined peaks where dsinθ = mλ, m=0,1,2,3,4..
In contrast to the maxima observed for single slits and Young's two slit experiment, the maxima
observed for light passing through a diffraction grating are sharper and narrower.
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Recall that for the two slit experiment,
the maximum intensity is 4I1. For the
diffraction grating, the intensity of the
peaks is N2I1. N is a very large number!
Two slits:
Many slits:
Because the diffraction grating gives rise to much narrower lines, we can separately observe
maxima from light rays of different wavelengths.
Recall sinθ = mλ/d
All wavelengths contribute to the same m= 0 line which occurs at the centre of the viewing
screen. However, for the higher order peaks, light rays of different wavelength produce different
maxima.
Problem: Light of wavelength 440 nm produces a first order peak displaced 10 cm above the
central maximum on the viewing screen which is located 25 cm away from the diffraction
grating. (a) How many lines/cm are on the grating? (b) How many orders of peak can we observe
from this light ray?
10cm
25 cm
(a) Use dsinθ = mλ for a maximum. Here tanθ = 10cm/25cm = 0.4 θ = 21.8o.
Then d = mλ/sinθ = 440nm/sin(21.8o) = 440x10-9m/sin (21.8o) = 1.18×10-6m
This corresponds to the line spacing, then the number of lines/m is
1/d = 8.44×105 lines/m = 8,440 lines/cm.
(b) for the second order peak, sinθ = 2λ/d = 2(440×10-9m)/(1.18×10-6m) = 0.745,
for the third order peak, sinθ = 3λ/d = 1.1 Not possible, therefore, only two orders can be
observed from this light ray.
The diffraction grating can be used as a spectrometer to measure the wavelength of light from the
angle at which the light ray is diffracted.
Source
A diffraction spectrometer
θ
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Section 22.4
Single - Slit Diffraction
3λ/a
2λ/a
λ/a
θ
Single slit of thickness, a.
0
-λ/a
-2λ/a
-3λ/a
sinθ
The pattern of intensity on the viewing screen has a central bright fringe flanked by much weaker
maxima alternating with dark fringes.
Consider a single slit. Previously, we treated each slit as a point (or line) source of light. Each
portion of the slit acts as a source of light waves.
There is a viewing screen far off in
the distance.
a/2
θ
a
The path difference of a/2sinθ holds for the
comparison of any light ray from the bottom
half with its counterpart in the upper half.
a/2
a/2sinθ
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Divide the slit into two halves. The path difference between light rays in the upper half and the
lower half is a/2sinθ.
We increase θ from zero until we find the first minimum due to π phase shift between parallel
beams.
Destructive interference occurs when a/2sinθ = λ/2 or sinθ = λ/a
Divide the screen into four equal parts: Then the path difference is a/4sinθ and destructive
interference occurs when a/4sinθ = λ/2 which yields sinθ = 2λ/a
The general condition for destructive interference is sinθ = mλ/a. This determines the positions
of the fringes.
The complete intensity pattern for single slit diffraction follows the following relationship which
will not be derived here:
Intensity
I = Imax[sin(πasinθ/λ)]2
[πasinθ/λ]2
-40
-30
-20
-10
0
10
PHI (rad)
Theoretical plot of intensity from a single slit due to diffraction
20
30
40
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Problem A screen is placed 50.0 cm from a single slit, which is illuminated with 690 nm light.
If the distance between the first and third minima in the diffraction pattern is 3.00mm what is the
width of the slit?
Third minimum
First minimum
θ
L
Then the first minimum occurs at y = 1.5mm.
The first minimum occurs at sinθ = λ/a and the third occurs at sinθ = 3λ/a. So, 2λ/a = 3.00 mm.
using the small angle approximation since L>>a
The displacement along y is Ltanθ and the spacing between minima is 1.5mm
Then tanθ = 1.50x10-3m/50.0x10-2m = 0.00300 rad.
Then we can use the first minimum position to derive a = λ/sinθ = 690x10-9m/0.00300 =
2.30x10-4m or 0.230mm
Diffraction in the Double Slit Experiment
In the double slit experiment, we calculated the intensity of fringes as a function of θ.
I = Imaxcos2[πdsinθ/λ]
a
a
θ
d
Maxima occur at sinθ = mλ/d
Minima occur at sinθ = (m + ½ )λ/d
In fact, this was not rigorously correct. It turns out that due to the finite thickness of each slit this
intensity function must be multiplied by the diffusion correction.
[Sin(πasinθ/λ)]2
[πasinθ/λ]2
where a is the thickness of the slit.
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Intensity
Minima occur at sinθ = mλ/a, Hence the combined pattern looks like:
-40
-30
-20
-10
0
10
20
30
40
PHI (rad)
Theoretical intensity for light passing through two slits. The higher frequency oscillations are
determined by the distance between the two slits and peaks occur at sinθ = mλ/d where d is the
spacing between the two slits. The diffraction minima occur where sinθ = mλ/a where a is the
thickness of each slit.
Note that d is always larger than a.
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Problem from Exam Final 2006: A screen is placed 200 cm behind two narrow slits that are
separated by 0.480mm. The figure shows the two slit interference pattern not including the
effects of diffraction.
(a) What is the wavelength of the light?
(b) Consider point A on the screen. What is the path difference between the distance from A
to one slit and the distance to the other slit?
(c) A film of index of refraction n= 1.5 is now placed in front of one of the slits in the above
setup. What should be the thickness, t, of the film so that point A moves to the central
axis?
1 cm
A
y=0
(d) Light of wavelength 600nm is emerging from a single rectangular slit of width 0.1mm. A
screen of width .1m is placed at L=2.0m from the slit. How many diffraction minima will
appear on the screen?
(e)
0.1 mm
0.10m
(a) dsinθ = mλ; then λ = dsinθ/m; sin θ = tanθ = y/L;
then λ = dy/L/m = 0.480x10-3m*.0025m/2.00m = 600nm
(b) phase difference = 2π∆r/λ = 4π; ∆r= 2λ = 1200nm
(c) We need a phase change of 4π. The phase change with path length is
∆φ in air = 2π(t/λ); ∆φ in film = 2π(nt/λ) so the phase difference is:
2π(nt/λ- t/λ) = 4π; then t/λ(n-1) = 2; t = 2λ/(n-1) = 2400nm
(d) diffraction minima occur at asinθ = mλ or ay/L = mλ; then: y = mλL/a = m(.012) then we
have minima at ±0.012m, ±0.024m, ±0.036m, ±0.048m. or 8 minima.
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Section 22-5, 23-8 Circular Aperture diffraction and resolution
The ability of optical systems to distinguish between closely spaced objects is limited because of
the wave nature of light.
The opening represents the
lens of a camera or our eye.
θ
As the two light sources move closer together, the diffraction patterns overlap and we can no
longer separate the two images.
When the central maximum of one diffraction pattern falls on the first minimum of the other
diffraction pattern, the images are said to be just resolved. This limiting condition of resolution is
known as the Rayleigh criterion.
The first minimum occurs when sinθ = λ/a as discussed above. For λ<<a which holds in most
instances sinθ ~ θ and θmin = λ/a
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For a circular aperture, it ‘can be shown that’
θmin = 1.22λ/D where D is the diameter of the aperture.
Note also, from your text, page 700, the width of the central maximum of the diffraction pattern
from a circular aperture is: ∆θ = 2*1.22λ/D = 2.44λ/D. or width on screen, w = L∆θ = 2.44λL/D.
Resolution of the eye: At a distance of 25 cm, what is the minimum separation of light sources
that the eye can resolve?
The pupil diameter is approximately 2mm. The centre of the spectrum of visible light is 500 nm.
θmin = 1.22λ/D = 1.22(500x10-9m)/2x10-3m = 3. x10-4 rad
25 cm is about as close as we can focus our eyes.
s
θ
Pupil diameters vary from 1mm to 8mm. The eye
is most sensitive to 550nm light.
L
Minimum separation s = Lθ = (0.25m)(3x10-4) = 8x10-5m = 0.08 mm - the thickness of a human
hair.
Problem: When Mars is nearest the earth, the distance separating the two planets is 88.6x106m.
Mars is viewed by a telescope whose mirror has a diameter of 30.0cm.
(a) for light of wavelength 590nm, what is the angular resolution of the telescope?
(b) What is the smallest distance that can be resolved on Mars?
The mirror is equivalent to the aperture of the telescope. Then θmin = 1.22λ/D = 2.40x10-6 rad.
The smallest distance that can be resolved on Mars is then s = Lθmin = (88.6x106m)(2.40x10-6) =
213m.
Could you see life on Mars with this resolution?