Today’s Lecture
Interference
Diffraction Gratings
Electron Diffraction
Can we see any interference without a laser?
Some math: the slits are two coherent sources.
The distances to the observation points are r1 and r2. Their difference
r2 − r1 = d sin θ
θ = tan ( y / L) ≈ y / L
r2 − r1 ≈ dy / L
for small angles θ, and small y/L
−1
Approximation used:
d << L
θ
r2 − r1 = d sin θ
Constructive (a bright strip)
Destructive (a dark strip)
r2 − r1 ≈ dy / L
for small
d sin θ = mλ
d sin θ = (m + 1 / 2)λ
y/L
Approximation used:
d << L
θ
Constructive (a bright strip)
r2 − r1 = d sin θ
d sin θ = mλ
Consider two slits .075mm apart located
1.5m from screen. If the 3rd order bright fringe
is 3.8cm from the screen center what is λ?
Approximation used:
d << L
θ
Destructive (dark strip)
r2 − r1 = d sin θ
Consider two slits .075mm apart located
1.5m from screen. Given λ =633nm what is the
distance from screen center to the 3rd order dark
fringe?
To determine the intensity pattern on the screen we use
linear superposition of the electric fields at the screen:
Here δ is the difference in path length, δ = d sinθ.
Using the trigonometric identity
we can sum the two electric fields.
The result of summing the
electric fields is
The intensity pattern is proportional to the time average of Ep2 or
The intensity pattern for a single slit is proportional to S0 = E02/2,
In general, the distribution of intensity on the screen:
S0
− intensity of either wave alone
Bright fringes:
Dark fringes:
d sin θ = mλ
d sin θ = (m + 1 / 2)λ
Multiple-Slit Interference and Diffraction Gratings
From the diffraction limit with only two
slits the intensity pattern is
πd sin θ
S = 4 S 0 cos (
)
λ
2
What if you require better resolution?
Consider multiple slits.
For multiple slits
The location of the bright fringes still satisfies
d sin θ = mλ
The location of the dark fringes now satisfy
Here N is the number of slits and m is not an integer
multiple of N. This means there are N-1 dark fringes
between each bright fringe.
Multiple-Slit Interference and Diffraction Gratings
What if you require better resolution?
Consider multiple slits.
The location of
the bright fringes
still satisfies
d sin θ = mλ
The location of
the dark fringes
now satisfy
Here N is the number of slits and m is
not an integer multiple of N. This
means there are N-1 dark fringes
between each bright fringe.
Multiple-Slit Interference and Diffraction Gratings
Consider the first 4 slits below “screen”.
If the time dependence of electric field
from the first slit is eiωt, then the sum of
the electric fields from these 4 slits is:
The location of
the dark fringes
now satisfy
Multiple-Slit Interference and Diffraction Gratings
Consider the first 4 slits below “screen”.
For m=1 the sum of the electric fields
from these 4 slits becomes:
Similar results occur for m=2, 3. In
general it is straightforward to show
that the dark fringes satisfy:
Multiple-Slit Interference and Diffraction Gratings
More minima between bright fringes results
in better resolution of the bright fringes.
A set of very many closed spaced slits is called a
“Diffraction Grating”
White Light and a Diffraction Grating
This figure shows spectra that is observed when passing white
light through a diffraction grating (note m=0). The orders of the
diffraction, m, are separated vertically for presentation
purposes. Note the increased separation between violet and red
as the order of the dispersion increases.
http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/gratcal.html
Multiple-Slit Interference and Diffraction Gratings
More minima between bright fringes results
in better resolution of the bright fringes.
Using a diffraction grating with 6000 slits/cm what is the
angular separation between the hydrogen α spectral line
(λ=656.3nm) and the hydrogen β line (λ=486.1nm)?
The same criteria for the diffractions peaks applies, dsinθ =mλ.
For this grating the slit spacing is d=(1/600)mm. In general the
angular separation is larger for diffraction gratings since d is small.
Resolving Power and Diffraction Gratings
For a grating with N slits the criteria for the mth order
maximum is
The adjacent minimum occurs at
Maximum for λ’ = minimum for λ
Multiple-Slit Interference and Diffraction Gratings
A binary star system has one massive star at rest and
another smaller star rotating about it. The hydrogen α line
for the rotating star is Doppler shifted from λ=656.272nm
to λ=656.215nm. What order spectrum is required for
astronomers to resolve these lines when using a grating
that is 2.5cm wide with 2000 slits/cm?
Since the order must be an integer, the resolution of the
two wavelengths is third order, m = 3.
X-Ray Diffraction
When atomic spacing is
comparable to x-ray
wavelengths then the
reflected beam satisfies
Bragg condition:
Bragg scattering is used to determine the positions of atoms in a crystal
structure, or even the locations of atoms in organic molecules.
Watson and Crick won the Nobel prize for X-Ray diffraction analysis of the
“DNA molecule”
Interference in Thin Films
Constructive Interference:
Destructive Interference:
Is the wavelength
inside the film and
Newton’s Rings
Interference in thin films is a very sensitive test for optical systems.
Above you see the rings induced from the lens of a telescope.
Interference in Thin Films
Constructive Interference:
Destructive Interference:
(a) If the film were illuminated with 650nm light, how
many bright bands in the film with n=4/3 will appear?
There are 4 bright bands, why??
(b) What part of the film will be dark? Since the
smallest wavelength for visible light is λ=400nm,
No Interference with Bullets
Bullets come out randomly. If you block slit 2 then the bullets
have a probability distribution P1 with an analogous result when
you block slit 1.
If you open both slits then the bullets have the
probability distribution P12 = P1 + P2
Interference with Waves
With water waves (or monochromatic light), If you block slit
2 then the intensity, I1, is given by intensity distribution |h1|2
with an analogous result when you block slit 1.
If you open both slits then the interference results in an
intensity distribution I12 = |h1 + h2|2 NOT |h1|2+|h2|2.
Interference of Electron Waves?
Electrons of identical energy come out randomly. If, (b), you block slit 2
then the electrons have a probability distribution P1=|φ1|2. There is an
analogous result when you block slit 1.
If, (c), you open both slits then the electrons have the probability
distribution P12 = P1 + P2?? NO, P12=|φ1+φ2|2. The detector only
measures whole discrete electrons! but with the probability of a wave!
Interference of Electron Waves
with a Light Detector
If the light from the source interacts with the electrons then you should
observe a flash through one of the slits. But then the interference pattern
Disappears!
So are the electrons particles or waves??
Interference of Electron Waves
with a Light Detector
+
To answer this question, we turn down the intensity of the light source.
Now the flashes occur less often, but are just as bright when they do
occur. The resulting pattern is partially that of particles and waves.
Are the electrons somehow interfering with each other?
NO! The same interference pattern results even when
electrons arrive one at a time, but it takes longer to develop.
www.hqrd.hitachi.co.jp/em/doubleslit.cfm
Interference of Electron Waves
with a Light Detector
Remembering the diffraction limit for two slits, what happens when we
use light with ever increasing wavelengths?
When λ approximately equals d then a fuzzy flash appears to come
through both slits, and ??
The interference pattern returns!
It seems that if you measure a particle property of the electron,
it behaves as a particle, but it you measure a wave property it
behaves as a wave!
Interference of Electron Waves
with a Light Detector
DeBroglie wavelength is defined as p=h/λ or λ=h/p.
In this expression h is Planck’s constant, h=6.6x10-34Jsec.
This expression holds for both light particles (photons)
as well as any particle with mass!
If the momentum of the photon is large enough, when it
scatters off of an electron, it destroys the interference pattern.
Optical Interference for
Low Intensity Source
If light acts as particles, photons, carrying momentum and
scattering off of electrons then what is the effect of this
phenomena when viewing the interference pattern of light?
http://math.unipa.it/~grim/dott_HD_MphCh/GGiuliani_single_photon_interference_08.pdf
For light as well electrons it seems that if you measure a particle
property, a photon, it behaves as a particle, but it you measure a
wave property it behaves as a wave!
All of this is what physicists call “wave-particle duality”. Electrons,
photons, and all elementary particles interact with each other as
particles but in many other ways act as waves (e.g. traveling from one
position to another).
That’s All Folks!
This completes Physics 2C. I hope you enjoyed
at least some of it!
There will be a review session tomorrow that
should be helpful in preparing for the final! Good
Luck to Everyone!