The Michelson Interferometer
Invented by A.A. Michelson, also famous for measuring the
speed of light. The original purpose of an interferometer
was to measure lengths in terms of the wavelength of light,
but the interferometer is a very flexible arrangement for
setting up interference effects.
To create two beams of
light that are in phase, a
beam-splitter, or semireflecting mirror, is used:
To create a variable path
difference, one of the
mirrors can be moved.
The image of M1 in the
beam splitter is close to M2.
The reflections from the two
mirrors give an effect very
similar to thin film
interference.
The path difference between the two paths is 2(d2 – d1).
This has to increase by λ for a shift of one fringe
Alternatively, if the source is a
point or line source, the
images of the source seen in
the two mirrors are close
together and give similar
interference effects to the
double slit.
Problem 35-79. If mirror M2 is moved through 0.233 mm a
shift of 792 fringes occurs. What is the wavelength of the
light producing the fringe pattern?
2 x 0.233 mm = 792 λ
λ = 2 x 0.233 / 792 mm
As the moving mirror is
moved, on a micrometer
drive, the fringes move and
you can count the number of
fringes that appear or
disappear.
= 0.000588 mm
= 588 nm
If one of the paths contains a length L of transparent
material with refractive index n, the optical path length of
that part of the path is nL.
The Michelson-Morley experiment
Problem 35-81. An airtight chamber
5.0 cm long with glass windows is
placed in one arm of a Michelson
interferometer. Light of wavelength λ
= 500 nm is used. Evacuating the air
from the chamber causes a shift of 60
fringes. From these data, find the
index of refraction of air at
atmospheric pressure.
When Maxwell first developed his theory of light, there
seemed to be a difficulty. The equations predicted a single
value for the speed of light that didn’t depend on the speed
of the source or on the speed of the observer. This is not
how other waves, such as sound, behave.
An explanation proposed to understand this was that light
propagated in a medium called the ether. The speed
given by Maxwell’s equations was the speed of light
relative to the ether.
The change in optical path length of
the right hand beam is
2 x 5 cm x (nair – 1)
(nair – 1) = 60 x 500 nm / (2 x 5 x
The two beams are then
reflected from plane mirrors
and recombined in the beam
splitter.
107
nm) = 3 x
10-4
The ether had to have strange properties. It permeates
the whole of space, and the interiors of transparent
objects, but is has no detectable properties such as mass.
nair = 1.0003
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If the Earth is moving through the ether, the speed of light
should be (very slightly) different in different directions. This
should be detectable in the very sensitive interference fringes
shown in the Michelson interferometer.
Michelson and Morley did a
careful series of
experiments to look for
these effects, but failed to
see them.
Multiple slits – the Diffraction Grating
In a double slit
arrangement, the positions
of the light bands are given
by the equation
d sinθ = mλ
Conclusion: there is no
ether, light does behave
differently from other,
slower, waves. This paved
the way for the theory of
relativity.
This is the condition that the rays of light traveling from
the two slits to the point on the screen corresponding to
the angle θ have a path difference of a whole number of
wavelengths.
If there are more than two slits (“N slits”) a very similar
picture can be drawn.
If the condition d sinθ = mλ is
satisfied, the rays from all the
slits are in phase.
The difference in the pattern is that, as the number of slits
increases, the bright regions of the interference pattern
become narrower, until they are very small, limited only by
the size of the incident beam.
The condition for maxima in the
intensity of the interference
pattern of a large number of slits
is the same as for two slits:
d sinθ = mλ
“d” is the distance between
adjacent slits.
A commercial diffraction grating has thousands of lines.
For zero intensity, it is not necessary
that rays from adjacent slits should
be out of phase. All that is needed
is that rays from one half of the
grating should be out of phase with
rays from the other half.
λ/Nd is the half-width (in angle) of the central bright line.
In general, the angular half width of the line is
The first zero in the intensity occurs
when
The dispersion of a diffraction grating measures how far
apart the grating spreads similar wavelengths. This is
measured by the derivative dθ/dλ.
N
λ
d sin θ =
2
2
sin θ =
θ
λ
Nd
λ
Nd
because θ is small
Δθ hw =
λ
Nd cos θ
d sin θ = mλ
dθ
=m
dλ
dθ
m
D≡
=
d λ d cos θ
d cos θ
2
The resolving power of a grating measures how far apart
two wavelengths need to be for the grating to separate the
lines.
We can separate two lines if their angular separation is
greater than the angular half-width of each of the lines.
Combining the two previous results, the smallest
wavelength difference that can be resolved is
Δλ =
Δθ hw
λ
d cos θ
λ
=
×
=
dθ / d λ Nd cos θ
m
Nm
The resolving power is defined as
λ
= Nm
Δλ
Now use the difference in the two wavelengths, 0.6 nm, as
the angular half-width:
R=
N=
λ
Δλ
= Nm
λ
mΔλ
=
589.3 × 10−9
= 327
3 × 0.6 ×10−9
The total width of the grating is
Nd = 3.34 mm
(It might be more reasonable to understand this as the
width of the grating that is illuminated by the light.)
Problem 36-95. White light (400 nm < λ < 700 nm) is incident on a
grating. Show that, no matter what the value of the grating spacing d,
the second- and third- order spectra overlap.
In each order (i.e. for each value of m)
d sinθ = mλ
The first-order spectrum corresponds to m = 1. The range of θ values
this is spread over corresponds to
sinθ = 400 nm/d to
sinθ = 700 nm/d
The second-order spectrum corresponds to m = 2. It extends over a
range of θ values given by
sinθ = 2*400nm/d to sinθ = 2*700nm/d
= 800 nm/d
= 1400 nm/d
Problem 36-53. With a particular grating the sodium
doublet is viewed in the third order at 10o to the normal
and is barely resolved. Find (a) the grating spacing and
(b) the total width of the rulings.
Sodium has an almost monochromatic spectrum. (It has
been widely used as a street and parking lot light.) It’s
characteristic yellow color is formed by two spectral lines,
very close together, with wavelengths 589.0 and 589.6 nm.
Use the average value in
d sin θ = mλ
d=
mλ 3 × 589.3 × 10−9
=
sin θ
sin(100 )
= 1.02 ×10−5 m = 10.2 μ m
If the slits were illuminated by light containing a mixture of
wavelengths, such as white light, the maxima for the
different colors would come at different angles, and we
would see a series of spectra produced. This is the
principle of the diffraction grating. The name is a bad one.
The grating actually utilizes interference rather than
diffraction.
The pattern produced by a grating differs from the
spectrum produced by a prism, because the spectrum is
repeated for each value of m. An advantage of the
grating is that you can calculate exactly the angles for
each spectrum. A disadvantage is that the spectra for
different values of m (different orders) overlap.
Phasor addition of light rays
The total displacement in the two-slit interference
pattern is
y(x,t) = ymsin(kx- ωt+ φ) + ymsin(kx- ωt)
= [2ymcos(φ/2)]sin(kx-Tt+ φ/2)
The amplitude is 2ym cos(φ/2)
and the intensity is the square of this [2ym cos(φ/2)]²
We can get these results either from using trigonometric
identities or from phasor diagrams.
The third-order spectrum extends from
sinθ = 3*400 nm/d to sin = 3*700nm/d
= 1200 nm/d
= 2100 nm/d
The second and third order spectra must overlap.
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Phasor diagrams for different values of φ:
If I try to draw a phasor diagram for a large number of
slits, say 10, the problem is that unless the phase
difference between adjacent rays is very small, or very
close to a multiple of 2π, the phasor diagram closes up on
itself, and the net amplitude is small.
This accounts for the sharpness of the interference spots.
Single slit diffraction
When a light beam passes through a restricted aperture it
spreads out. This is called diffraction. The pattern of the
diffracted light shows a series of light and dark regions.
There is a simple argument to locate the minima in the
pattern.
Path difference between rays
drawn from the top and bottom
of the slit is
a sin θ
If a sin θ = mλ the rays from
the top and bottom of the slit
are in phase at the screen.
The net amplitude is the sum of all the rays of light that
pass through the opening, but there are now an infinite
number of rays of infinitesimal strength.
Now draw a ray from the
center of the slit. It will be
exactly out of phase with the
other two.
A sin θ = mλ is the formula for the zero’s in the pattern.
Problem 36-65. Sound waves with frequency 3000 Hz and
speed 343 m/s diffract through the rectangular opening of
a speaker cabinet and into a large auditorium. The
opening, which has a horizontal width of 30.0 cm faces a
wall 100 m away. Where along that wall will a listener be
at the first diffraction minimum and thus have the most
difficulty hearing the sound
At the first minimum
a sinθ = λ = v/f
sinθ = v/fa = 343/3000*0.3
= 0.38111
θ = 22.4025o
tanθ = 0.41222
At 100 m from the speaker this is
41.22 m from the center line.
(Large auditorium!)
Phasor diagram for single slit
Imagine dividing the width of the slit into a large number
N of narrow slices. If there is a phase difference φ
between the two sides, the phase difference between
adjacent slices is φ/N. The phasor diagram therefore
consists of a large number of very short segments with
small angles between them. If we take the limit of N
going to infinity, we get a smooth curved line:
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The intensity is the square of the amplitude, which in turn
is measured by the length of the resultant line, and in is
given by
I = Im [(sin α)/α)]²
α = φ/2
= π(a sinθ)/λ
Problem 36-11. Monochromatic light with wavelength 538
nm is incident on a slit with width 0.025 mm. The distance
from the slit to a screen in 3.5 m. Consider a point on the
screen 1.1 cm from the central maximum. (a) Calculate θ for
that point. (b) Calculate α. (c) Calculate the ratio of the
intensity at this point to the intensity at the central
maximum.
(a)
Double slit revisited
tanθ = .011m / 3.5 m
= .0031429
θ = .180072o
sin θ = .0031428
(b)
α = ½φ = πa sinθ / λ
= π * .025x10-3 * .0031428 / 538x10-9
= 0.45881
(c)
I = Im * (sin α / α)²
= Im * (.44288 / .45881)²
= 0.93177 * Im
The formula for the intensity of the two slit interference
pattern included the contributions from two point sources.
This was not really realistic and it did not properly describe
what you saw on the screen. Each of the two slits has a
finite width, and the light is diffracted through it in a single
slit diffraction pattern. It is these patterns that should be
superposed to give the final intensity. Usually, the width of
each of the slits is quite a bit less than the separation of
their centers. You can get the result by drawing a two-slit
interference pattern inside the envelope of a single slit
pattern.
Circular aperture
Similar ideas apply to two-dimensional openings. The
diffraction pattern of a circular opening looks a lot like the
diffraction pattern of a single slit, rotated about its center.
The first zero occurs at an
angle given by
d sin θ = 1.22 λ
Or
θ = sin-1 (1.22 λ/d)
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Does this matter?
All optical instruments, such as telescopes and
microscopes, that have circular lenses, are limited in
resolution by this behavior.
Star images seen in a good astronomical telescope are
each diffraction patterns of this form. If two stars are
close together (in angle) their diffraction patterns
overlap and you can’t tell that there are two star images
present.
Rayleigh criterion The images
of two stars are resolved if
their angular separation is
greater that 1.22 λ/d, where d
is now the aperture of the
telescope.
A telescope with a diameter of 9 cm (quite a small amateur
telescope), operating with light of wavelength 500 nm, has
a resolution of
1.22 × 500 × 10−9
= 6.8 × 10−6 radians
9 ×10−2
= .0004 degrees
Δθ =
= 1.4" (seconds of arc)
A radio telescope with a dish 50 m across (huge) operating
with a wavelength of 21 cm has a resolution of
1.22 × 21× 10−2
= .0051 radians
50
= 0.3 degrees
Δθ =
≈ 2 ' (minutes 0f arc)
BUT, it is possible to combine signals from an array of radio
telescopes. As long as the phase information is recorded
and combined, as far as resolution is concerned, the effect
is that of a much larger telescope.
By combining signals from radio telescopes in the USA and
in Australia, you can get the effect of a baseline roughly the
size of the Earth. The resolution is then much better than
optical telescopes can give. This method has given the
best information about what is going on at the center of our
galaxy, for example.
Assign problem 36-07, 589 nm -> (400 + 4n) nm
problem 36-43, 20o -> (21.0 + n/5)
o
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