Photonics

Unit - II
1.Wave Optics Describes the connection between waves and rays of light.
According to wave theory of light, light is a form of energy which travels
through a medium in the form of transverse wave.

2.Wave front The locus of all those particles which are vibrating in the same
phase at any instant is called wave front. Thus, wave front is a surface having
same phase of vibrating particles at any instant at every point on it.

3.Phase Speed Phase speed is the speed with which wave front moves and it
is equal to wave speed.

4.The shape of wavefront due to a

(i)point source is spherical
(ii)line source is cylindrical
(iii) source at infinity is a plane.

5.A line perpendicular to a wave front is called a ray. The direction of rays
are always perpendicular to the wavefront along the direction of propagation
of wave.
Huygens’ Principle

Huygens’
Principle says
that every
point along a
wave-front
emits a
spherical wave
that interferes
with all others.
Christiaan Huygens
1629 – 1695

Our solution for diffraction illustrates this idea, and it’s more rigorous.
 Huygens's Principle
Huygens’ principle, also called Huygens-Fresnel principle, a
statement that all points of a wavefront of sound in a transmitting
medium or of light in a vacuum or transparent medium may be
regarded as new sources of wavelets that expand in every direction at a
rate depending on their velocities.

Huygens’ Principle Huygens’ principle is essentially a geometrical

construction, which gives the shape of the wave front at any time, allows
us to determine the shape of the wavefront at a later time. According to
Huygens’ principle,
(i)Every point on a wave front behaves like a light source and emits
secondary wavelets.
(ii)The secondary wavelets spread in all directions in space (vacuum)
with the velocity of light.
(iii) The envelope of wave front of secondary wavelets, after a given time,
along forward direction gives the new position of wave front.
Huygens proposed that every point to which a luminous disturbance
reaches becomes a source of a spherical wave, and the sum of these
secondary waves determines the form of the wave at any subsequent
time.

when slit width is comparable in size to the wavelength w λ ,

notice more diffraction (spreading or bending of waves through
an aperture or around an obstacle) with the yellow barrier
Huygens-Fresnel principle Refraction on an aperture (slit) but
notice the secondary source space equally apart and start from
center ending 1/2 space from the edge of slit which is more
logical than end at the edge of the slit.
Infinitely many points (three shown) along the length d project phase
contributions from the wavefront, producing a continuously varying
intensity θ on the registering plate.
 Primary and Secondary Sources

• Huygens stated that light is a wave propagating through space like ripples in
water or sound in air. Hence, light spreads out like a wave in all directions from
a source.
• The locus of points that travelled some distance during a fixed time interval is
called a wavefront.
• Thus, from a point source of light, the locus of points that light has travelled
during a fixed time period is a sphere (a circle if you consider a 2D source).
Advantages and Disadvantages of Huygens Principle

Advantages:
•Huygens concept proved the reflection and refraction of light.
•The concepts like diffraction of light, as well as interference of light,
were proved by Huygens.

Disadvantage:
•Concepts like emission of light, absorption of light and polarisation of
light were not explained by Huygens principle.
•Huygens principle failed to explain the photoelectric effect.
•A serious drawback is that the theory proposes an all-pervading
medium required to propagate light called luminiferous ether. This was
proved to be false in the 20th century.
• How can we use coherent light to
visually see the difference in pit
density on CDs, DVDs and Blu-
Ray disks?
• Why does light from a point source
form light and dark fringes when it
shines on a razor blade?
• We will continue our exploration
of the wave nature of light with
diffraction.
• And we will see how to form
three-dimensional images using a
hologram.
 Diffraction
• According to geometric optics, a light source shining on an
object in front of a screen should cast a sharp shadow.
Surprisingly, this does not occur because of diffraction.
 Diffraction and Huygen’s Principle
• Huygens’s principle can be used to analyze diffraction.
• Fresnel diffraction: Source, screen, and obstacle are close
together.
• Fraunhofer diffraction: Source, screen, and obstacle are far apart.
• Figure below shows the diffraction pattern of a razor blade.
 Diffraction from a single slit
• In Figure below, the prediction of geometric optics in (a)
does not occur. Instead, a diffraction pattern is produced, as
in (b).

• The narrower the slit, the broader the diffraction pattern.

Diffraction and the Fourier Transform

Prof.
Rick
Trebino
Georgia
Tech

www.physics.gatech.edu/frog/lectures
Diffraction
Shadow of a
Light does not hand
always travel in a illuminated
straight line. by a
Helium-
Neon laser
It tends to bend
around objects.
This tendency is
called diffraction.

Any wave will do Shadow of

this, including a zinc oxide
matter waves and crystal
acoustic waves. illuminated
by a
electrons
Why it’s hard to see diffraction
Diffraction tends to cause ripples at edges. But a point source is
required to see this effect. A large source masks them.

Screen
with hole Rays from a
point source
yield a perfect
shadow of the
hole. Rays from
other regions
blur the shadow.

Example: a large source (like the sun) casts blurry shadows,

masking the diffraction ripples.
Diffraction of ocean water waves
Ocean waves passing through slits in Tel Aviv, Israel

Diffraction occurs for all waves, whatever the phenomenon.

Diffraction of a
wave by a slit

  slit size
Whether waves in water or
electromagnetic radiation in air,
passage through a slit yields a
diffraction pattern that will
appear more dramatic as the
size of the slit approaches the   slit size
wavelength of the wave.

  slit size
Diffraction

Transmission
by an Edge
x

Even without
a small slit,
diffraction
Light
can be passing
strong. by edge
Simple
propagation
past an edge
yields an
unintuitive Electrons
irradiance passing by
an edge
pattern.
(Mg0 crystal)
Radio waves diffract around mountains.

When the
wavelength is a
km long, a
mountain peak is
a very sharp
edge!

Another effect
that occurs is
scattering, so
diffraction’s role
is not obvious.
 Fresnel and Fraunhofer diffraction
by a single slit
• Figure below shows Fresnel (near-field) and
Frauenhofer (far-field) diffraction for a single slit.
 Locating the dark fringes
• Follow the single-slit diffraction discussion in the text.
• Figure below shows the geometry for Fraunhofer
diffraction.

( a / 2) sin  =  / 2
An example of single-slit diffraction
• Figure (bottom left) is a photograph of a Fraunhofer pattern of a single
horizontal slit.
• Example : You pass 633-nm light through a narrow slit and observe the
diffraction pattern on a screen 6.0 m away. The distance at the screen between
the center and the first minima on either side is 32 mm long. How wide is the
slit?
xm (6000 mm)(0.000633 mm)
ym =  a= = 0.24 mm
a 32 mm
 Intensity in the single-slit pattern
• Follow the text discussion of the intensity in the single-slit
pattern using the phasor diagrams in Figure below.
Diffraction Geometry
We wish to find the light electric field after a screen with a hole in it.
This is a very general problem with far-reaching applications.

Aperture
y y1
transmission
Observation
E(x,y) t(x,y) plane

r = ( x1 − x)2 + ( y1 − y)2 + z 2 P1
x x1
P

z
E(x1,y1)

Incident This region is assumed to be

wave much smaller than this one.

What is E(x1,y1) at a distance z from the plane of the aperture?

Diffraction Assumptions E (r , t ) = 0
E ( r , t ) = 0
The best assumptions were E (r , t )  E ( r , t )
determined by Kirchhoff:
E (r , t ) = 0
Incident E ( r , t ) = 0
1) Maxwell's equations wave

2) Inside the aperture, the field and its spatial derivative are the
same as if the screen were not present.

3) Outside the aperture (in the shadow of the screen), the field
and its spatial derivative are zero.

While these assumptions give the best results, they actually

over-determine the problem and can be shown to yield zero field
everywhere! Nevertheless, we still use them.
 Quantitative Intensity in the single-slit pattern
• Follow the text discussion of the intensity in the single-slit pattern using the
phasor diagrams in Figure 36.8 below.
• The angle b is the phase angle of the ray
from the top of the slit, while the phase
angle from the bottom of the slit is 0. The
vectors lie along a circle whose center is at
C, so Ep is a chord of the circle. The arc
length E0 is subtended by this same angle
b, so the radius of the circle is E0/b.
• From the diagram,
E0 b sin b / 2
Ep = 2 sin = E0
b 2 b /2
• Since
2
b= a sin 

• We have
 sin  a ( sin  ) /   
2

I = I0   (sinc function)
  a ( sin  ) /  
 Intensity maxima in a single-slit
pattern
• Figure 36.9 at the right shows the intensity versus
angle in a single-slit diffraction pattern.
• The minima occur when b is a multiple of 2, i.e.
at a sin  = m (m = 1,  2,  3, ...)

• The location of the maxima are found by taking 2
the derivative of  sin  a ( sin  ) /   
I = I0  
  a ( sin  ) /  
• and setting it to zero. Surprisingly, these are
not precisely where b = (2m + 1) (m = 0,1, 2, ...)
• In fact, there are no minima for m = 0 in this
expression. The central maximum is wider than
the others, and occurs at  = 0.
• Using these approximate values of b in the
intensity, we find I0
Im 
( m + 12 )  2
2
 Width of the single-slit pattern
• The single-slit diffraction pattern depends on the ratio of the slit width a to the
wavelength .

• Example 36.2: (a) The intensity at the center of a single-slit diffraction pattern is
I0. What is the intensity at a point in the pattern where there is a 66-radian phase
difference between wavelets from the two edges of the slit? (b) If this point is 7
degrees from the central maximum, how many wavelengths across is the slit?
 sin  a ( sin  ) /     sin 33 rad  
2 2
• (a) −4
I = I0   = I0   = 9.2 10 I 0
  a ( sin  ) /    33 rad 
• (b)  a ( sin  ) 33 rad
= 33 rad  a =  = 86
  ( sin 7)
 Two slits of finite width
• When we discussed two-slit interference in Chapter 35, we
ignored the width of each slit. When we demonstrated it,
however, we saw clearly the effect of the slit widths.
• The overall pattern of two finite-width slits is the product
of the two patterns, i.e.
b  sin x
I = I 0sinc 2 cos 2 sinc ( x ) =
2 2 x
where
2 d 2 a
= sin  b= sin 
 
 Several slits
• In Figure 36.13 below, a lens is used to give
a Fraunhofer pattern on a nearby screen.
It’s function is to allow the pattern to be
seen nearby, without having the screen
really distant.
• The phasor diagrams show the electric
vectors from each slit at different screen
locations.
Interference pattern of several slits
• The figure below shows the interference pattern for 2, 8, and 16 equally
spaced narrow slits.

• By making the slits really close together, the maxima become more
separated. If the light falling on the slits contains more than one wavelength
(color), there will be more than one pattern, separated more or less according
to wavelength, although all colors have a maximum at m = 0.
• This means that the different orders make rainbows—separating wavelengths
into a spectrum, with the separation being greater for greater order m.
 The diffraction grating
• A diffraction grating is an array of a large number of
slits having the same width and equal spacing. The
intensity maxima occur at
d sin  = m
• Example : The wavelengths of the visible spectrum
are approximately 380 nm (violet) to 750 nm (red).
(a) Find the angular limits of the first-order visible
spectrum produced by a plane grating with 600 slits
per millimeter when white light falls normally on the
grating. (b) Do the first order and second order
spectra overlap? What about the 2nd and 3rd orders?
1 mm
d= = 1.67 10−6 m
• (a) distance between slits is 600 slits
• Violet light for 1st order occurs at
• Red light for 1st order occurs at
• (b) recalculate for m = 2 and m = 3.
 = arcsin (  / d ) = arcsin ( 3.8 10−7 /1.67 10−6 ) = 13.2
 = arcsin (  / d ) = arcsin ( 7.5 10−7 /1.67 10−6 ) = 26.7
The 2nd-order spectrum extends from 27.1-63.9° while the 3rd order is from 43-90.
 Grating spectrographs
• A diffraction grating can be used to disperse light into a spectrum.
• The greater the number of slits, the better the resolution.
• Figure 36.18(a) below shows our sun in visible light, and in (b) dispersed into a
spectrum by a diffraction grating. See description of Eschelle spectrograph:
• http://www.vikdhillon.staff.shef.ac.uk/teaching/phy217/instruments/phy217_inst_echelle.html
Diagram of a grating spectrograph
• Figure below shows a diagram of a diffraction-grating
spectrograph for use in astronomy.
 X-ray diffraction
• When x rays pass through a crystal, the crystal behaves like
a diffraction grating, causing x-ray diffraction. Figure below
illustrates this phenomenon.
A simple model of x-ray diffraction
• Follow the text analysis using Figure 36.22 below.
• The Bragg condition for constructive interference is
2d sin = m.
• Follow the Example.
 Circular apertures
• An aperture of any shape forms a diffraction pattern.
• Figures below illustrate diffraction by a circular aperture. The airy disk
is the central bright spot.
• The first dark ring occurs at an angle given by sin1 = 1.22 /D.
 Diffraction and image formation
• Diffraction limits the
resolution of optical
equipment, such as
telescopes.
• The larger the aperture, the
better the resolution. Figure
(right) illustrates this effect.
Bigger telescope, better resolution
• Because of diffraction, large-diameter telescopes, such as
the VLA radio telescope below, give sharper images than
small ones.
• Follow Example
 What is holography?
• By using a beam splitter and mirrors, coherent laser light
illuminates an object from different perspectives.
Interference effects provide the depth that makes a three-
dimensional image from two-dimensional views. Figure
below illustrates this process.
 How does holography work?
• Follow the text analysis using Figure below.
 An example of holography
• Figure below shows photographs of a holographic image
from two different angles, showing the changing
perspective.
Fraunhofer
Diffraction from a
Square Aperture
The diffracted field is a
sinc function in both x1
and y1 because the
Fourier transform of a
rect function is sinc.

Diffracted
irradiance

Diffracted
field
Diffraction from a
Circular Aperture
A circular aperture yields a
diffracted "Airy Pattern,"
which looks a lot like a
sinc function, but actually
involves a Bessel function.

Diffracted field Diffracted Irradiance

Diffraction from
small and large
circular
Far-field
apertures intensity pattern
from a small
aperture

Recall the Scale Theorem!

This is the Uncertainty
Principle for diffraction.
Far-field
intensity pattern
from a large
aperture
 Fraunhofer diffraction
from two slits
t(x) w w

-a 0 a x

t(x) = rect[(x+a)/w] + rect[(x−a)/w]

E (k x )  F {t ( x)}

 sinc[w(kx1 / z ) / 2]exp[ +ia(kx1 / z )] +

sinc[w(kx1 / z ) / 2]exp[ −ia(kx1 / z )]

kx1/z
E ( x1 )  sinc( wkx1 / 2 z ) cos(akx1 / z)
Diffraction from one- and two-slit screens

Fraunhofer diffraction patterns

One slit

Two slits
Diffraction from Slit Diffraction
Pattern Pattern
multiple slits

Infinitely many equally

spaced slits (a Shah
function!) yields a far-field
pattern that’s the Fourier
transform, that is, the
Shah function.
Two Slits and Spatial Coherence
If the spatial coherence length is less
than the slit separation, then the
relative phase of the light transmitted
through each slit will vary randomly,
washing out the fine-scale fringes, and
a one-slit pattern will be observed.

Fraunhofer diffraction patterns

Good spatial
coherence

Poor spatial
coherence
Young’s Two Slit Experiment and
Quantum Mechanics
Imagine using a beam so weak that only one photon passes through
the screen at a time. In this case, the photon would seem to pass
through only one slit at a time, yielding a one-slit pattern.
Which pattern occurs?
Possible Fraunhofer diffraction patterns

Each photon
passes
through only
one slit

Each photon
passes
through
both slits
Dimming the light incident on two slits

Dimming the light in a two-slit experiment yields single photons

at the screen. Since photons are particles, it would seem that
each can only go through one slit, so then their pattern should
become the single-slit pattern.

Each
individual
photon goes
through both
slits!
Fresnel diffraction: example
Fresnel diffraction from a single slit:

Close Far
to the z from
slit the
slit

Slit

Incident
plane wave
Fresnel Diffraction from a Slit
This irradiance vs. position just after a slit illuminated by a laser.

Irradiance

x1
The Spot of Arago
If a beam encounters a stop, it develops a hole, which fills in as it
propagates and diffracts:

x x1 Interestingly,
the hole fills
in from the
center first!

Stop

Input beam Beam after

with hole some distance

This irradiance can be quite high and can do some damage!

Fresnel diffraction from an array of slits:
The Talbot Effect
One of the few Fresnel diffraction problems that can be solved
analytically is an array of slits.
The beam pattern alternates between two different fringe patterns.

Diffraction
Screen with patterns
array of slits

ZT = 2d2/
The Talbot
Carpet

What goes on in
between the solvable
planes?

The beam
propagates in this
direction.

The slits are here.

Problem:
When a monochromatic light source shines through a 0.2 mm wide slit onto a
screen 3.5 m away, the first dark band in the pattern appears 9.1 mm from
the center of the bright band. What is the wavelength of the light?

Solution:
•Reasoning:
Dark fringes in the diffraction pattern of a single slit are found at angles θ
for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... . For the first
dark fringe we have w sinθ = λ. Here we are asked to solve this equation for
λ.

•Details of the calculation:

z = 9.1 mm = 9.1*10-3 m.
L = 3.5 m.
w = 0.2 mm = 2*10-4 m.
L >> z, therefore sinθ ~ z/L and λ = zw/(mL).
λ = (9.1*10-3 m)(2*10-4 m)/(3.5 m).
λ = 5.2*10-7 m = 520 nm.
Problem:
Consider a single slit diffraction pattern for a slit width w. It is
observed that for light of wavelength 400 nm the angle between the
first minimum and the central maximum is 4*10-3 radians. What is
the value of w?

Solution:
•Reasoning:
Dark fringes in the diffraction pattern of a single slit are found at
angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ...

•For the first dark fringe we have w sinθ = λ. Here we are asked to
solve this equation for w.

•Details of the calculation:

First minimum: w sinθ = λ,
•w = (400 nm)/sin(4*10-3 radians) = 1*10-4 m.
 Fraunhofer Diffraction from a slit
Fraunhofer Diffraction from a slit is simply the Fourier Transform of a
rect function, which is a sinc function. The irradiance is then sinc2.

t(x) = rect[x/w]

E (k x )  F {t ( x)}

E (k x )  sinc( wk x / 2)
E ( x1 )  sinc( wkx1 / 2 z ) wkx1
2z

I (k x )  sinc2 ( wk x / 2)
I ( x1 )  sinc2 ( wkx1 / 2 z )