Download Lecture 22: Reflection and Refraction of Light

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
 Speed
of light (in vacuum)
Foucault’s experiment
 Speed
of light (in vacuum)
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.
Nature of light
 Waves,
wave fronts, and rays
• Wave front: 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
wave fronts
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
 Reflection (cont’d)
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.
n  c / v ;   0 / n
f=v
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
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  s s '
 s tan 1  ( s  s ) tan  2
for small angles: tan ->sin
1
s sin 1  ( s  s ) sin  2
s sin 1  ( s  s ) na sin 1
L
s  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
L
n1
n2
n2
   s'   s
s
s'
n1
| s ' | tan 1  L, | s | tan  2  L  s' tan 1  s tan  2
tan   sin    for   1
 s' sin 1  s sin  2
 n1 s'  n2 s ( n1 sin 1  n2 sin  2 )
s’
s
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
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
Prism
 Applications of prism
• A prism and the total reflection
can alter the direction of travel
of a light beam.
• All hot low-pressure gases emit
their own characteristic spectra.
A prism spectrometer is used
to identify gases.
Diversion
 Diversion
• The index of refraction of a material
depends on wavelength as shown
on the right. This is called diversion.
• 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 st, 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 wave front of incident
light
• The reflected wave front is CD
• 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 t, 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  v1t;  sin  2  v2 t

v1t
v t
c
c
 2 , v1  , v2 
sin 1 sin  2
n1
n2
  AC
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
Example
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
Example
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.
Exercises
Problem
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
in the middle of the above equation that
P
n b
always cancel out. The requirement of
L
d
 a'
parallel faces ensures that the angle  n   n'
and the chain of equations can continue.
(c) The lateral displacement of the beam can be calculated using geometry:
t
t sin( a  b' )
d  L sin( a   ), L 
d 
.
cos b'
cos b'
'
b
(d)
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 (