1D Interference
Destructive interference results when light paths differ by an half odd integer multiple of a wavelength,
λ/
2
, 3λ/2
, … (2 n +1
)λ/2
.
Constructive interference results when light paths differ by an integer multiple of a wavelength
λ
, 2
λ
, … n
λ
.

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?
http://www.ngsir.netfirms.com/englishhtm/Interference.htm

r
1 r
2
How do we know, whether we are in a node or an antinode?
If the two sources oscillate at the same frequency, it is all a function of distances from the sources to the point of observation, r
1 and r
2
.
If r
1
= r
2 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?
r
1 r
2 r r
1
2
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, 2 A cos( k ( r
1
r
2
)/2) .

How do we know, whether we are in a node or an antinode?
r
1 r
2 r r
1
2
There is now a spatially varying amplitude,
This new amplitude is a maximum ( ) when k ( r and zero when k ( r
1
r
2
)/2 = ( m+ 1/2)
.
2 A cos( k ( r
1
r
2
)/2) .
1
r
2
)/2 = m
,
In terms of the wavelength these amplitude expressions become:

How do we know, whether we are in a node or an antinode?
r
1 r
2 r r
1
2
Hence the oscillations will be in phase and the interference will be constructive if r
1
- r
2
=
or, r
1
- r
2
=
, or in general: r
1
r
2
m
where m is an integer number
With phase difference:

How do we know, whether we are in a node or an antinode?
r
1 r
2 r r
1
2
The oscillations will be out of phase and the interference will be destructive if r
1
- r
2
=
λ
/2 or, r
1
- r
2
=
−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?
http://www.ngsir.netfirms.com/englishhtm/Interference.htm

r
1 r
2
What if the two sources are out of phase with each other?
A cos( kr
1
− ω t
+ ϕ
1
) A cos( kr
2
− ω t
+ ϕ
2
)
The phase difference is then: k ( r
1
− r
2
)
+ ϕ −
1 ϕ
2
2
π m
2
π m
+ constructive
π destructive

Why do not we see any interference with two light bulbs?
What would be a characteristic distance between nodes, antinodes?
ϕ a high rate?
k ( r
1
− r
2
)
+ ϕ −
1 ϕ
2
2
π m
2
π m
+ constructive
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 10 15 Hz . Frequency – identical down to 10 -12 .
Phase locked over 10 12 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 -12 f and their phases locked over 10 12 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).

θ d
θ
A simple equation for the opening angle,
θ
:
sin
= d

Diffraction
For a single slit, how does diffraction occur?
θ d
θ
What is the origin of the relation sin
θ =
λ
??
d
The answer is :
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,
Δ t , 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).

θ
Huygen’s Principle 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).

θ
Huygen’s Principle 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 = 10
For d = 2
For d =
λ
λ
, the 1 st zero occurs at ~ 6 o and the 5 th zero at 30 o
.
the 1 st zero occurs at 30 o
λ the 1 st zero occurs at 90 o
. and the 2 nd at 90 o
.

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.

Assume light of wavelength of
λ
= 500nm and a pupil diameter of d = 2 mm .
(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
25 cm from the observer.
At 25 cm objects can not be resolved if they are closer than 80
μ m !