# Part 2 - Light as EM Wave 1

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```							Light as an Electromagnetic Wave

Light as EM Wave 1
Maxwell (late 19th century) unified electricity
and magnetism with his famous equations and
showed that light is an electromagnetic wave.

                     B
E  0          E  
t

                1 E
B  0          B  2                      James Clerk Maxwell
c t                      (1831-1879)

Where the E and B vectors are the electric and magnetic fields
and the L vector is:
Light as an Electromagnetic Wave

Light as EM Wave 2
Combining his equations,Maxwell found that there are
electric (and magnetic) fields which are solutions to the
wave equation, eg:

These particular waves propagate at the speed c (in
vacuum) and Maxwell made the connection between
these wave solutions and light. Hertz showed that
propagating electromagnetic waves could be
generated by oscillating charges (electrons).
Light as an Electromagnetic Wave

Light as EM Wave 3
A charge undergoing
SHM at angular
frequency  gives rise to
electric and magnetic
fields which propagate
through space as
harmonic waves with
angular frequency .

Far from the source, the
waves propagate as
plane waves.
Light as an Electromagnetic Wave
Animation
Light as EM Wave 4

EM_wave_anim.gif
Light as an Electromagnetic Wave

Light as EM Wave 5
The electric and magnetic fields propagate as “vector”
waves eg the wave

has three vector components:
Light as an Electromagnetic Wave

Light as EM Wave 6s points in
Electromagnetic waves  Unit vector
are “transverse” waves :                   the propagation
direction.

In optics, we generally describe light as an oscillating
electric field and forget about the oscillating magnetic field
component for 2 reasons:

1. Light interacts with the charges (electrons) in a (generally
non-magnetic) medium more strongly through E than B.
The (Lorentz) force of B on an electron is proportional to
ve/c where ve is the electron’s speed. Usually ve/c <<1.

2. Once we know the electric field it’s trivial to find the
magnetic field:
Light as an Electromagnetic Wave

Light as EM Wave 7
Energy transported

Consider a light
beam striking a
surface:

We define the “Irradiance” , I , as the average energy per unit
time arriving at a unit area of surface (or emitted by a unit area
of surface). Units: J s-1 m-2 = W m-2

Strictly speaking the “Intensity” of a light beam is defined
somewhat differently, but, in our lectures, we will use the words
“I” for both of them.
Light as an Electromagnetic Wave

Light as EM Wave 8
The relation between irradiance and electric field in a light wave:
Consider the field component:

This field oscillates extremely rapidly (1015 Hz = 1 Terahertz). This
means that, even over very short measurement times, many
oscillations take place and the field averages to zero! Our eyes
and measuring instruments do not respond to the field directly.
Light as an Electromagnetic Wave

Light as EM Wave 9
Detection of light involves the measurement of energy deposited
per unit time on the detector: Irradiance I is measured.

A general property of waves of any type is that they deliver
energy at a rate proportional to the square of the wave amplitude.

For our field:

The average
value of Ey2
over many
oscillations is
E0y2 /2

(not zero!).
Light as an Electromagnetic Wave

Light as EM Wave 10
Electromagnetic theory shows that the irradiance is given by:

Usually we consider near-normal incidence:

Here:  is the electric permittivity of free space, c is the speed of
light in vacuum, n is the refractive index and the angle brackets < >
denote an average value over one harmonic cycle..

The value of a function
averaged over one
harmonic cycle of
period T is:
Light as an Electromagnetic Wave
Example:
Light as EM Wave 11
For:

For the complex
representation:

Make use of:

for  constant.
Light as an Electromagnetic Wave
Thus for:   Light as EM Wave 12
Light as an Electromagnetic Wave
Thus for:   Light as EM Wave 13

Same result whether we express the propagating field in the
real or the complex representation! Good!
Light as an Electromagnetic Wave
Note: for the field expressed in complex notation:
Light as EM Wave 14
Take the dot product of the field with its own complex conjugate:

Our previous result for irradiance I (taking cycle averages):

We can simply write (no cycle average!):

We will use this relation from now on with
the field written in complex representation.
Interference

Interference 1
“Principle of Superposition”

The disturbance at some point in the medium due to a
number of waves propagating through that point is the
sum of disturbances due to the individual waves.

Rewrite this somewhat for light waves:

“Principle of Superposition”

If a number of propagating electromagnetic fields (light
waves) fall on a surface, then the overall field at the
surface is the sum of the individual fields:
“superposition field”.
Interference

Interference 2
The irradiance (intensity) at the surface is due to the
superposition field there.

Because some fields can cancel others (crests meet
troughs), the overall intensity at the surface can be
smaller than the sum of the intensities due to the
individual fields.
“Destructive Interference”.

It also turns out that the individual fields can reinforce
each other (crests meet crests) so that the intensity at the
surface can be greater than the sum of intensities of the
individual fields.
“Constructive Interference”.
Two Beam Interference

Two Beam Interference 0
Two Source Interference Pattern.

Dark regions correspond to        Bright regions correspond to
cancellation by destructive       reinforcement by constructive
interference: “Nodes”             interference: “Antinodes”
Two Beam Interference

Two Beam Interference 1
What is the irradiance falling on a detector (surface)
when two optical fields are superposed?

Let the fields be:
Two Beam Interference

Two Beam Interference 2

These fields have identical:

These fields have different:

They would be produced by the two charges q1 and q2
oscillating (SHM) “out of phase” by:
Two Beam Interference

Two Beam Interference 3
The superposition field at the detector is:

The irradiance at the detector is:
Two Beam Interference

Two Beam Interference 4
Some mathematics - simplifying the irradiance formula:

The dot products:

(similarly)
Two Beam Interference

Two Beam Interference 5
Some mathematics - simplifying the irradiance formula:

The dot products:

This result assumes that E1
and E2 vectors point in the
same direction
 same “polarization”.

(similarly)
Two Beam Interference

Two Beam Interference 6
Some mathematics - simplifying the irradiance formula:

Combining dot products:

where  = 02 – 01 is the phase difference between
the two fields superposing at the detector.
Two Beam Interference

Two Beam Interference 7

becomes:

Now we make use of the definitions:

intensity due to field E1 only at the detector.

intensity due to field E2 only at the detector.
Two Beam Interference

Two Beam Interference 8

The General Two Beam Interference Formula.

For equal amplitude sources: I1 = I2  I0 .
I0 is the irradiance due to one source alone.

The Two Beam Interference Formula.
for equal amplitude sources.

 = 02 – 01 is the phase difference between
the two fields superposing at the detector.
Two Beam Interference
The General Two Beam Interference Formula.
Two Beam Interference 9
 is the phase
difference between
the two fields.

For  = even multiple of 
 = 2m (with integer m)  cos  = 1

Constructive interference
Imax > I1 + I2

For  = odd multiple of 
 = (2m+1) (with integer m)  cos  = -1

Destructive interference
Imin < I1 + I2
Two Beam Interference
The Two Beam Interference Formula (equal amplitude sources).
Two Beam Interferencephase
 is the
10
difference between the
two fields.

For  = even multiple of 
 = 2m (with integer m)  cos  = 1

Constructive interference
Imax = 4I0 > I1 + I2 = 2I0

For  = odd multiple of 
 = (2m+1) (with integer m)  cos  = -1

Total Destructive
interference
Imin = 0
Young’s Double Slit
An application of the two beam interference formula.
Young’s Double Slit 1
Observation
Point
Incident
Plane
Wave.

Screen

Huygens’ Principle: illumination of the slits by a plane
wave is equivalent to setting up two, identical, in-phase
sources; S1 and S2.

Call the field at S1 or S2:
Young’s Double Slit
Superposing wavefronts at P will have different phases
Young’s Double Slit 2
due to the different lengths (strictly, OPLs) of paths
S1 P and S2  P

To calculate the phase difference between the two fields
superposing at P, we recall the difference in phase between two
wavefronts belonging to a single field:
Young’s Double Slit
Superposing wavefronts at P will have different phases
Young’s Double Slit 3
due to the different lengths (strictly, OPLs) of paths
S1 P and S2  P

To calculate the phase difference between the two fields
superposing at P, we recall the difference in phase between two
wavefronts belonging to a single field:

B - A is the phase difference
between wavefronts at A and B
Young’s Double Slit
Making use of this result:
Double Slit
Young’sThe field at observation 4
point P due to S1 is:

The field at observation
point P due to S2 is:

Field E2 can be written as field E1 times a phase factor:
Young’s Double Slit

Young’s Double Slit 5
The superposition field
at observation point P is:

Where the two fields have the same amplitude ES and
a phase difference of:

To find the irradiance at P we use the two beam
interference formula:

where I0 is the irradiance at P due to one slit alone.
Young’s Double Slit

Young’s Double Slit 6

where I0 is the irradiance at P due to one slit alone.

with

The phase difference is determined by the path length
difference of paths S1 P and S2  P , namely r2 – r1 .

 = 2m constructive interference at P
 = (2m+1) destructive interference at P
Young’s Double Slit
Note: more strictly the phase difference is determined by the
Young’s Double Slit 6a
optical path difference (OPD) for paths S1 P and S2  P .

More strictly:

where k0 is the vacuum
wavenumber and OPL=nr
Young’s Double Slit
Calculating r2-r1 : The “Fraunhofer Condition”
Young’s Double Slit 7
If point P is “far enough” away from the slits, then the
paths S1 P and S2  P are nearly parallel.

for parallel paths

Path difference:

Phase difference:
Young’s Double Slit
Calculating r2-r1 : The “Fraunhofer Condition”
Young’s Double Slit 8
If point P is “far enough” away from the slits, then the
paths S1 P and S2  P are nearly parallel.

for parallel paths

In terms of angle  :
Path difference:

Phase difference:
Young’s Double Slit

Young’s Double Slit 9
Thus, for a point P far away, the irradiance at P is given by:

For observation in the “near forward” direction, the angle  is
small and the small angle approximation sinx  x is used:

Constructive Interference condition (Bright Fringe):
ka = 2m  m = a (m integer)

Destructive Interference condition (Dark Fringe):
ka = (2q+1)  (q+½) = a (q integer)
Young’s Double Slit

Young’s Double Slit 10

Distant observation screen

I = 2I0(1+cos(ka))

ka = 2m      Bright fringe (m integer)
ka = (2q+1)  Dark fringe (q integer)
Young’s Double Slit
In practice,a lens is often used to satisfy the Fraunhofer Condition.
Young’s Double Slit 11

The lens selects fields propagating along parallel paths to
superpose at the observation point P in the focal plane.
Young’s Double Slit
How do we calculate (optical) path difference with a lens?
Young’s Double Slit 12
Recall Fermat’s Principle applied to image formation. All paths
from object point P0 to image point P have the same OPL!
Young’s Double Slit

Young’s Double Slit 13
Let the object point move to infinity.

Consider a plane  lying perpendicular to the incident parallel
rays from distant point P0 . Paths from any point (S , S’ , ….)
lying on  to image point P in the lens focal plane all have the
same OPL.
Young’s Double Slit

Young’s Double Slit 14

Since OPL(S1 P) = OPL(S2’ P) , the phase
change undergone by fields from S1 P and S2’ P
is the same. A phase difference at P in the fields from
S1 and S2 arises from the path difference r .
Young’s Double Slit

Young’s Double Slit 15

Lens Focal Plane

I = 2I0(1+cos(ka))

ka = 2m      Bright fringe (m integer)
ka = (2q+1)  Dark fringe (q integer)
Young’s Double Slit
Note: Image formation as an interference (diffraction)
Young’s Double Slit 16
phenomenon.

Fermat’sPrinciple: OPL (PP’) is the same along all paths.
Thus, the phase change of the fields from along all paths
PP’ is the same  Constructive Interference!

P’ (the image point) is an irradiance maximum due to
constructive interference.

```
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