Today’s Lecture
Interference
Diffraction
1D Interference
Destructive interference results when light paths differ by an
half odd integer multiple of a wavelength,
λ/2, 3λ/2, … (2n+1)λ/2.
Constructive interference results when light paths differ by
an integer multiple of a wavelength
λ, 2λ, … nλ.
2D Interference
If you go along the
green line (upward),
you will see a
pattern of nodes and
antinodes like in a
standing wave.
Two sources oscillating in phase.
Lines of nodes and antinodes?
Are antinodes the points, where disturbance is always high?
Are nodes the point, where disturbance is always low?
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r1
How do we know,
whether we are in a
node or an
antinode?
r2
If the two sources oscillate at the same frequency, it is all a function
of distances from the sources to the point of observation, r1 and r2.
If r1 = r2 the two waves are going to produce oscillations in phase
and constructive interference. If they are not equal, It is
straightforward to determine in general the phase difference.
How do we know,
whether we are in a
node or an
antinode?
r1
r2 r1
r2
Let’s translate it into the language of phases:
Add two waves of the same amplitude and frequency:
There is now a spatially varying amplitude, 2Acos(k(r1-r2)/2).
How do we know,
whether we are in a
node or an
antinode?
r1
r2 r1
r2
There is now a spatially varying amplitude, 2Acos(k(r1-r2)/2).
This new amplitude is a maximum (
) when k(r1-r2)/2 = mπ,
and zero when k(r1-r2)/2 = (m+1/2)π.
In terms of the wavelength these amplitude expressions become:
How do we know,
whether we are in a
node or an
antinode?
r1
r2 r1
r2
Hence the oscillations will be in phase and the interference will be
constructive if r1 - r2 = λ or, r1 - r2 = −λ , or in general:
r1 − r2 = mλ
where m is an integer number
With phase difference:
r1
How do we know,
whether we are in a
node or an
antinode?
r
r2 1
r2
The oscillations will be out of phase and the interference will be
destructive if r1 - r2 = λ/2 or, r1 - r2 = −λ/2 , or in general:
where m is an integer number
With phase difference:
What happens if the two sources are moved further apart?
Does the distance between two consecutive nodes/antinodes
decrease or increase?
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r1
r2
What if the two sources are out of phase with each other?
A cos( kr1 − ωt +ϕ 1) A cos( kr2 − ωt + ϕ 2 )
The phase difference is then:
k (r1 − r2 ) + ϕ1 − ϕ 2
2πm constructive
2πm + π destructive
Why do not we see any interference with two light bulbs?
What would be a characteristic distance between nodes, antinodes?
What is the phase,
a high rate?
ϕ1 , of a light source changes randomly in time at
k (r1 − r2 ) + ϕ1 − ϕ 2
2πm constructive
2πm + π destructive
Interference absolutely requires coherence of the two source:
their frequencies should be equal (beats otherwise);
the phase difference between their oscillations should be constant
(phase locking).
What does it takes in practical terms?
Say, our observation time is 0.001 sec and we want to see a steady
pattern of interference.
Light frequency is about 1015Hz. Frequency – identical down to 10-12.
Phase locked over 1012 cycles of oscillations… Wow!..
Coherence of light.
Coherent light sources…
Wishful thinking!
Realistic light sources, like a light
bulb… VERY (incoherent) bad guys…
A, B and C can be thought of as light waves
originating from different areas of the filament.
A+B+C is a light wave from a bulb, yak…
Where on the Earth do we find a good coherent light source to
observe interference?!
A laser?
Works better…
but two different lasers with frequency
– identical down to 10-12f and their
phases locked over 1012 cycles of
oscillations is far too much to ask for!
What do we do?
We make the light wave to interfere with itself!
Even an ugly one can do it! Coherence length is the issue here.
+
But laser light does it a bit
better…
Can we see any interference without a laser?
What is going on here?
We had a plane light
wave, so why do the fronts
get curved after passing
through the slits?
Diffraction – the wave front
gets curved if the wave
passes through a narrow
slit (or a small hole).
Diffraction
Diffraction
A simple equation for the
opening angle, θ :
θ
d
θ
wavelength
sin θ =
gap width
sin θ =
λ
d
Diffraction
For a single slit, how does diffraction occur?
What is the origin of the relation
θ
d
θ
sin θ =
λ
d
The answer is:
Huygen’s Priciple
All points on a wave front act as point sources
of spherically propagating “Wavelets”
??
Huygen’s Principle
For both the plane wave and spherical wave, the wavefronts act like a
set of infinitesimal point sources, “Huygen’s Wavelets”. A short time,
Δt, later the new wavefront is tangent to the wavefront resulting from
the sum (more precisely the integral) of the wavelets.
Huygen’s Principle
The diffraction occurs at the edges of the slit in the barrier. When the
width of the slit is much larger than the wavelength, the diffraction is
negligible. However when the slit size and wavelength are comparable,
the emerging wavefronts spead out in a broad beam. In the limit as the
width of the slit becomes much smaller than the wavelength the
emerging wavefronts become circular (spherical for a circular hole).
Huygen’s Principle
θ
d
θ
sin θ =
λ
d
First assume that the electric field of a wavelet
in the plane of the slit is of the form
Then integrate these wavelet sources over a small slit
opening from –d/2 to d/2 (see notes).
Huygen’s Principle
θ
d
θ
sin θ =
λ
d
From integrating the wavelets over the slit, we find that
the intensity on a screen a distance L behind the slit is
Huygen’s Principle
The intensity on a screen a distance L behind the slit is
This is a maximum at
Intensity is zero when
or
Diffraction Limit From a Single Slit
Note the position of the zeroes as a function of θ
for the single slit diffraction patterns below:
For d
For d
For d
= 10λ, the 1st zero occurs at ~ 6o and the 5th zero at 30o.
= 2λ the 1st zero occurs at 30o and the 2nd at 90o.
= λ the 1st zero occurs at 90o.
Diffraction Limit From a Single Slit
The intensity on a screen as a function of θ is:
Diffraction Limit From a Single Slit
Diffraction imposes a fundamental limit on the ability
of optical systems to resolve closely space objects
Rayleigh Criterion
Two peaks are can barely be
distinguished if the central
maximum of one coincides
with the first minimum of
the other.
If d is the slit width then
Electron microscopes versus optical microscopes
Diffraction Limit For Circular Apertures
Diffraction imposes a
fundamental limit on the ability
of optical systems to resolve
closely space objects.
Rayleigh Criterion
Two peaks are can barely be
distinguished if the central
maximum of one coincides
with the first minimum of
the other.
For a circular aperture
(a) Two point sources are far enough apart so that they are well resolved
(b) Two point sources are separated enough to barely satisfy Rayleigh Criterion
(c) Two point sources are so close that they are not resolved.
Resolution of the Eye
Assume light of wavelength of λ = 500nm
and a pupil diameter of d = 2mm.
(a) Estimate the limiting angle of resolution
for a human eye.
For a circular aperture
(b) Find the minimum distance of separation, d, for two objects that are
25cm from the observer.