Physical Optics
The wave nature of light
Interference
Diffraction
Polarization
Huygens’ Principle
• Every point on a propagating
wavefront serves as the
source of spherical wavelets,
such that the wavelets at
sometime later is the
envelope of these wavelets.
• If a propagating wave has a
particular frequency and
speed, the secondary
wavelets have that same
frequency and speed.
“Isotropic”
Diffraction
a
• Diffraction – Bending of light into the shadow region
• Grimaldi - 17th Century observation of diffraction
• Diffraction vs. Refraction?
a
Explanation of Snell’s Law
sin 1 
BD v1t

AD AD
AC v 2 t
sin 2 

AD AD
sin 1 v1t c / n1


sin 2 v 2 t c / n 2
n1 sin 1  n 2 sin 2
Superposition of waves
Constructive
Interference
Destructive
Interference
Conditions for Interference
• To observe interference in light waves, the
following two conditions must be met:
1) The sources must be coherent
• They must maintain a constant phase with respect to
each other
2) The sources should be monochromatic
• Monochromatic means they have a single
wavelength
Young’s Experiment
Young’s Experiment
Young’s Experiment
Maxima occur when:
  r1  r2  m
m  2
S2
m  1
sin  
m0
d
m  1
  dsin   m
m  2
S1

d
maxima
Minima occur when:
1

  r1  r2   m   
2

s
r2
d
S2
S1
  r1  r2

r1
s  a
y
1

  d sin    m   
2

What the pattern looks like
Intensity Distribution,
Electric Fields
• The magnitude of each
wave at point P can be
found
– E1 = Eo sin ωt
– E2 = Eo sin (ωt + φ)
– Both waves have the
same amplitude, Eo
φ
2π
2π
δ
d sin θ
λ
λ
Intensity Distribution, Resultant
Field
• The magnitude of the resultant electric field comes
from the superposition principle
– EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]
• This can also be expressed as
AB
 AB
sin A  sin B  2sin 
 cos 

 2 
 2 
φ
φ 
EP  2Eo cos   sin  ωt  
2
2 
– EP has the same frequency as the light at the slits
– The magnitude of the field is multiplied by the factor 2
cos (φ / 2)
Intensity Distribution, Equation
• The expression for the intensity comes from the
fact that the intensity of a wave is proportional to
the square of the resultant electric field magnitude
at that point
• The intensity therefore is
 πd sin θ 

2  πd
I  I max cos 
y
  I max cos 
λ


 λL 
2
Resulting Interference Pattern
• The light from the two slits
forms a visible pattern on a
screen
• The pattern consists of a series
of bright and dark parallel
bands called fringes
• Constructive interference
occurs where a bright fringe
occurs
• Destructive interference results
in a dark fringe
Example
A He-Ne Laser has a wavelength of 633 nm. Two slits are
placed immediately in front of the laser and an interference
pattern is observed on a screen 10 m away. If the first
bright band is observed 1 cm from the central bright fringe,
how far apart are the two slits?
At what angle Q is the fourth dark band found?
Thin Films
Constructive Interference (maxima)
d ABC  m n
o
n 
n
Destructive Interference (minima)
d ABC
1

  m   n
2

Chapter 35 - Problem 45
Stealth aircraft are designed to not reflect radar
whose wavelength is 2 cm, by using an antireflecting coating. Ignoring any change in
wavelength in the coating, estimate its
thickness.
Phase shift on reflection
External Reflection
Internal Reflection
Now, If one reflection is internal and one reflection is external half
wavelength path differences will result in constructive interference
Phase Changes Due To
Reflection
• An electromagnetic wave
undergoes a phase change
of 180° upon reflection
from a medium of higher
index of refraction than
the one in which it was
traveling
– Analogous to a pulse on
a string reflected from a
rigid support
Lloyd’s Mirror
• An arrangement for
producing an interference
pattern with a single light
source
• Waves reach point P either
by a direct path or by
reflection
• The reflected ray can be
treated as a ray from the
source S’ behind the mirror
Lloyd’s Mirror
S
y

a
2


r1  r2
S1
s
 a sin  m     m2
Phase shift on reflection  
s  a

ym
1

a sin  m   2m 1    m
  a m  a
2
2
s

Interference in Thin Films Again
•
•
•
•
Assume the light rays are traveling
in air nearly normal to the two
surfaces of the film
Ray 1 undergoes a phase change of
180° with respect to the incident ray
Ray 2, which is reflected from the
lower surface, undergoes no phase
change with respect to the incident
wave
For constructive interference
=2t = (m + ½)λn (m = 0, 1, 2 …)
• This takes into account both the
difference in optical path length for
the two rays and the 180° phase
change
•
For destructive interference
=2t = mλn (m = 0, 1, 2 …)
Newton’s Rings
o

n
Maxima - bright
1

2d m    m  
2

Minima - dark
2dm  m
Chapter 37 – Problem 62
Show that the radius of the mth dark Newton’s ring
as viewed from directly above is given by:
 o Rm
xm 
n
Where R is the radius of curvature of the curved
glass surface and  is the wavelength of the light
used. Assume that the thickness of the air gap is
much less than R at all points and that x<<R
Newton’s Rings
R  x  R  d 
2
R
Rd
2
x 2  R 2   R  d   R 2  R 2  d 2  2 Rd
2
d

x
x 2  d 2  2 Rd  2 Rd
Minima
2d  m n
o
x 2m
2
m
2R
n
 o Rm
xm 
n
Newtons rings are a
special case of Fizeau fringes.
They are useful for testing
surface accuracy of a lens.
o

n
2
Maxima
o R 
1
xm 
m 
n 
2
Wedges – Fringes of Equal Thickness
Minima (destructive)
d
2t  m  m

x
t  x
Fringes of this type are also
known as Fizeau fringes
o
n
Maxima (constructive)
1

2t    m  
2

x m 
 
1
m



2 
2
x m 

2
Example: thin film of air
Michelson Interferometer
d
Compensator
Plate
• A lens can be used to form
fringes of equal inclination
(rings)
• Tilting the mirrors can cause
fringes of equal thickness.
• Accurate length
measurements are
accomplished by fringe
counting as one of the
mirrors is moved.
Detector
  2d  m
d 
m
2
42.
Monochromatic light is beamed into a Michelson
interferometer. The movable mirror is displaced 0.382
mm, causing the interferometer pattern to reproduce itself
1 700 times. Determine the wavelength of the light. What
color is it?
  4.49 107 m  449 nm
5.
Young’s double-slit experiment is performed with 589-nm
light and a distance of 2.00 m between the slits and the screen.
The tenth interference minimum is observed 7.26 mm from the
central maximum. Determine the spacing of the slits.
7.
Two narrow, parallel slits separated by 0.250 mm are
illuminated by green light (λ = 546.1 nm). The interference
pattern is observed on a screen 1.20 m away from the plane of
the slits. Calculate the distance (a) from the central maximum to
the first bright region on either side of the central maximum and
(b) between the first and second dark bands.
17.
In Figure 37.5, let L = 120 cm and d = 0.250 cm. The
slits are illuminated with coherent 600-nm light. Calculate the
distance y above the central maximum for which the average
intensity on the screen is 75.0% of the maximum.
32.
A thin film of oil (n = 1.25) is located on a smooth wet
pavement. When viewed perpendicular to the pavement, the film
reflects most strongly red light at 640 nm and reflects no blue
light at 512 nm. How thick is the oil film?
33.
A possible means for making an airplane invisible to
radar is to coat the plane with an antireflective polymer. If radar
waves have a wavelength of 3.00 cm and the index of refraction
of the polymer is n = 1.50, how thick would you make the
coating?
37.
A beam of 580-nm light passes
through two closely spaced glass plates,
as shown in Figure P37.37. For what
minimum nonzero value of the plate
separation d is the transmitted light
bright?
Mach Zehnder Interferometer
Detector
•
•
•
•
No compensating plate needed
Test cell easily inserted in one leg.
No factor of two as in the Michelson interferometer.
Difficult to align
Sagnac Interferometer
s  r
v  r
v
2
R
2
R
2
R
v
4R 2
tcw 
v
c
2
tccw
t  tcw  tccw

t 
t 
Detector
4R 2

v
c
2
8R
8R

2c  v
2c  v
8R
8R

2c  R
2c  R



8R
1
1  8R  R 
R  
t 

1


1






2c 1  R 1  R 
2c 
2c 
2c  

2c
2c 
Ring laser gyro.
The rotation effectively shortens the path
taken by one direction over the other.
8R  2 R  8R 2
t 

  c2
2c  2c 
8R 2
  ct 
c
Phase Angle
i 1 t 
E  E1  E2  E01e
i  2 t 
 E02e
i   t 
E2
E
2
 1 E1
E01 cos 1
E  E0 e 
E02 sin  2
E01 sin 1
tan   
E02 cos  2
E 
E01 sin 1  E02 sin  2
E01 cos 1  E02 cos  2
E01  E02  2 E01 E02 cos  2  1 
2
2