Chapter 30: Reflection and Refraction
The nature of light
 Speed
of light (in vacuum)
c = 2.99792458 x 108 m/s measured but it is now the definition
Michelson’s 1878 Rotating Mirror Experiment
• German American physicist A.A. Michelson realized, on putting together
Foucault’s apparatus, that he could redesign it for much greater accuracy.
• Instead of Foucault's 60 feet to the far mirror, Michelson used 2,000 feet..
• Using this method, Michelson was able to calculate c = 299,792 km/s
• 20 times more accurate than Foucault
• Accepted as the most accurate measurement of c for the next 40 years.
Picture credit
The nature of light
 Waves,
wavefronts, and rays
• Wavefront: The locus of all adjacent points at which the phase of
vibration of a physical quantity associated with the wave is the same.
rays
source
wavefronts
spherical wave
plane wave
Reflection and refraction
 Reflection and refraction
• When a light wave strikes a smooth interface of two transparent
media (such as air, glass, water etc.), the wave is in general partly
reflected and partly refracted (transmitted).
reflected rays
incident rays
r
a
a
b
b
refracted rays
b
a
Reflection and refraction
 Reflection
• The incident, reflected, and refracted rays, and the normal to the
surface all lie in the same plane.
• The angle of reflection r is equal to
the angle of incidence  a for all
wavelengths and for any pair of
material.
reflected rays
incident rays
r
a
a
r  a
b
b
refracted rays
Reflection and refraction
 Refraction
• The index of refraction of an optical material (refractive index), denoted
by n, is the ratio of the speed of light c in vacuum to the speed v in the
material.
wavelength in vacuum. Freq. stays the same.
reflected rays
incident rays
n  c / v ;   0 / n
• The ratio of the sines of the angles
 a and b , where both angles are
measured from the normal to the
surface, is equal to the inverse ratio
of the two indices of refraction:
sin  a nb

sin  b na
r
a
a
b
b
Snell’s law
refracted rays
Total internal reflection
 Total internal reflection
n2
sin  2 , sin 2  1 when n2 / n1  1& n2  n1 sin 1.
Since sin 1 
n1
When this happens, 2 is 90o and 1 is called critical angle. Furthermore
when 1   crit , all the light is reflected (total internal reflection).
Total internal reflection
 Optical fibers
Dispersion
 Dispersion
• The index of refraction of a
material depends on wavelength
as shown on the right. This is
called dispersion. It is also true that,
although the speed of light in vacuum
does not depends on wavelength,
in a material wave speed depends
on wavelength.
Diversion
 Examples
Huygens’ principle
 Huygens’ principle
Every point of a wave front may be considered the source of secondary
wavelets that spread out in all directions with a speed equal to the speed
of propagation of the wave.
Plane waves
Huygens’ principle (cont’d)
 Huygens’ principle for plane wave
• At t = 0, the wave front is
indicated by the plane AA’
• The points are
representative sources
for the wavelets
• After the wavelets have
moved a distance cDt, a
new plane BB’ can be
drawn tangent to the
wavefronts
Huygens’ principle (cont’d)
 Huygens’ principle for spherical wave
Huygens’ principle (cont’d)
 Huygens’ principle for spherical wave (cont’d)
• The inner arc represents
part of the spherical wave
• The points are
representative points
where wavelets are
propagated
• The new wavefront is
tangent at each point to
the wavelet
Huygens’ principle (cont’d)
 Huygens’ principle for law of reflection
• The Law of Reflection
can be derived from
Huygen’s Principle
• AA’ is a wavefront of
incident light
• The reflected wave
front is CD
Huygens’ principle (cont’d)
 Huygens’ principle for law of reflection (cont’d)
• Triangle ADC is congruent to triangle AA’C
• Angles 1 = 1’
• This is the law of reflection
Huygens’ principle (cont’d)
 Huygens’ principle for law of refraction
• In time Dt, ray 1 moves
from A to B and ray 2
moves from A’ to C
• From triangles AA’C and
ACB, all the ratios in the
law of refraction can be
found:
n1 sin 1 = n2 sin 2
 sin 1  v1Dt;  sin  2  v2 Dt

v1Dt
v Dt
c
c
 2 , v1  , v2 
sin 1 sin  2
n1
n2
  AC
Polarization
 EM wave
y

E ( z, t )  iˆEmax cos( kz  t )

B( z, t )  ˆjBmax cos( kz  t )
 
EB
In the text:
^
x
E(x,t)=jE
maxcos(kx-t)
^
B(x,t)=kB
maxcos(kx-t)
z

 Polarization (defined by the direction of E )
Linear polarization
Circular polarization
Polarization (cont’d)
 Polarization (defined by the direction of
Circular polarization

E
)
Polarization (cont’d)
 Polarizing filters
Polarization (cont’d)
 Polarization by reflection
plane of incidence
p
p
na
b
nb
When the angle of incident coincides with
the polarizing angle or Brewster’s angle,  p
the reflected light is 100% polarized.
na sin  p  nb sin b  nb sin( 90   p )  nb cos  p
n
tan  p  b Brewsters’s law of the polarizing angle
na
Example: depth of a swimming pool
Pool depth s = 2m
person looks straight
down.
2
1
L
the depth is judged by
the apparent size of
some object of length
L at the bottom of the
pool (tiles etc.)
na sin 1  sin  2
2
tan 1 
L
s
tan  2 
L
L

s  Ds s '
 s tan 1  ( s  Ds ) tan  2
for small angles: tan ->sin
1
s sin 1  ( s  Ds ) sin  2
s sin 1  ( s  Ds ) na sin 1
L
Ds  s
na  1
1
 ( 2 m )  50 cm.
na
4
Example: Flat refracting surface
• The image formed by a flat
refracting surface is on the
same side of the surface as
the object
– The image is virtual
– The image forms between
the object and the surface
– The rays bend away from
the normal since n1 > n2
n1 n2
n2
 q
p
p q
n1
| q | tan  2  L, | p | tan 1  L  q tan  2  p tan 1
tan   sin    for   1
 q sin  2  p sin 1
 n1 q  n2 p ( n1 sin 1  n2 sin  2 )
1
L
2
2
Prism example
• Light is refracted twice – once entering and once leaving.
• Since n decreases for increasing , a spectrum emerges...
Analysis: (60 glass prism in air)
sin 1 = n2 sin 2
n2 sin 3 = sin 4
n1 = 1
60
Example: 1 = 30
1

2

3
 sin(30 ) 
o
  19 .5
 1.5 
 2  sin 1 
4
 3  (60 o   2 )  40 .5o
 4  sin 1 1.5 sin  3   76 .9 o
n2 = 1.5
++60o = 180o
3 = 90 - 
 = 90 - 2

3 = 60 - 2
Atmospheric Refraction and Sunsets
•
•
•
Light rays from the sun are
bent as they pass into the
atmosphere
It is a gradual bend because
the light passes through layers
of the atmosphere
– Each layer has a slightly
different index of refraction
The Sun is seen to be above
the horizon even after it has
fallen below it
Mirages
• A mirage can be observed
when the air above the
ground is warmer than the
air at higher elevations
• The rays in path B are
directed toward the ground
and then bent by refraction
• The observer sees both an
upright and an inverted
image
Problem 1
Exercises
The prism shown in the figure has a refractive
index of 1.66, and the angles A are 25.00 . Two
light rays m and n are parallel as they enter
m
the prism. What is the angle between them
they emerge?
n
A
A
Solution
na sin  a
1 1.66 sin 25.0
na sin  a  nb sin b  b  sin (
)  sin (
)  44.6.
nb
1.00
Therefore the angle below the horizon is b  25.0  44.6  25.0  19.6,
and thus the angle between the two emerging beams is 39.2.
1
Exercises
Problem 2
Light is incident in air at an angle
on the upper surface of a transparent
plate, the surfaces of the plate being
plane and parallel to each other. (a)
t
Prove that  a   a' . (b) Show that this
is true for any number of different parallel
plates. (c) Prove that the lateral displacement
d of the emergent beam is given by the
sin(  a  b' )
relation:
d t
,
a
n
Q
n’
n
b'
b P
 a'
d
cos b'
where t is the thickness of the plate. (d) A ray of light is incident at an angle
of 66.00 on one surface of a glass plate 2.40 cm thick with an index of
refraction 1.80. The medium on either side of the plate is air. Find the lateral
Displacement between the incident and emergent rays.
Problem 2
Exercises
Solution
a
(a) For light in air incident on a parallel-faced
plate, Snell’s law yields:
n sin  a  n' sin  b  n' sin  b'  n sin  a'  sin  a  sin  a'   a   a' .
n
t
n’
b'
Q
(b) Adding more plates just adds extra steps
P
n b
in the middle of the above equation that
L
d
 a'
always cancel out. The requirement of  n  n'
parallel faces ensures that the angle
and the chain of equations can continue.
(c) The lateral displacement of the beam can be calculated using geometry:
d  L sin(  a   b ), L 
(d)
t sin(  a   b )
t
d 
.
cos  b
cos  b
n sin  a
sin 66.0
)  sin 1 (
)  30.5
n'
1.80
(2.40cm) sin( 66.0  30.5)
d 
 1.62 cm.
cos 30.5
 b  sin 1 (