Transcript Ch. 35: Reflection and Refraction of Light

Geometrical Optics
Geometrical Optics
•
Optics is usually considered as the study of the behavior of
visible light (although all electromagnetic radiation has the
same behavior, and follows the same rules).
•
The propagation of light can be described in two alternative
views:
a) As electromagnetic waves
b) As rays of light
•
When the objects with which light interacts are larger than
its wavelength, the light travels in straight lines called rays,
and its wave nature can be ignored.
•
This is the realm of geometrical optics.
Geometrical Optics
Light can be described using geometrical optics,
as long as the objects with which it interacts,
are much larger than the wavelength of the light.
This can be described using
geometrical optics
This requires the use of full
wave optics (Maxwell’s equations)
Propagation of Light
Light propagates in straight lines (rays).
This is valid as long as the light does
not change the medium through which
it propagates (air, water, glass, plastic),
or finds an obstacle (interface).
The velocity of light in air is c
c = 3x108 m/s
The velocity of light in other media
may be different from c (less than c).
Reflection and Transmission
Most materials reflect light (partially or totally). For example,
metals reflect light (almost totally) because an incident oscillating
light beam causes the metal’s nearly free electrons to oscillate,
setting up another (reflected) electromagnetic wave.
Opaque materials absorb light (by, say, moving electrons into
higher atomic orbitals).
Transparent materials transmit light. These are usually insulators
whose electrons are bound to atoms, and which would require
more energy to move to higher orbitals than in materials which are
opaque.
Geometrical Optics
q1 = angle of incidence
q1
Normal to surface
Incident ray
Surface
Angles are measured with respect to the normal to the surface
Reflection
q1
q’1
The Law of Reflection
q1 = q’1
Light reflected from a
surface stays in the plane
formed by the incident ray
and the surface normal
and
the angle of reflection equals
the angle of incidence
(measured to the normal)
Reflection
q1
q’1
Specular and Diffuse Reflection
q1 = q’1
Smooth  specular  shiny
Rough  diffuse  dull
Two mirrors are placed at right angles as shown.
An incident ray of light makes an angle of 30° with the x axis.
Find the angle the outgoing ray makes with the x axis.
Refraction
q1
q’1
Medium 1
Medium 2
q2
More generally, when light passes
from one transparent medium to
another, part is reflected and part
is transmitted. The reflected ray
obeys q1 = q’1.
Refraction
q1
q’1
q2
Medium 1
More generally, when light passes
from one transparent medium to
another, part is reflected and part
is transmitted. The reflected ray
obeys q1 = q’1.
Medium 2
The transmitted ray obeys
Snell’s Law of Refraction:
It stays in the plane, and the
angles are related by
n1sinq1 = n2sinq2
Here n is the “index of refraction” of a medium.
Refraction
q1
q’1
Medium 1
Reflected ray
Incident ray
Medium 2
Refracted ray
q2
q1 = angle of incidence
q’1= angle of reflection
q2 = angle of refraction
Law of Reflection
q1 = q’1
Law of Refraction
n1 sinq1= n2 sinq2
n  index of refraction
ni = c / vi
vi = velocity of light in
medium i
Index of Refraction
The speed of light depends on the medium trough which it travels.
The speed of light in a given medium is determined by the medium’s
index of refraction n.
c
n
v
n  index of refraction
v  velocity of light in medium
c  3 108 m
s
Air, n = 1.000293
Glass, 1.45  n  1.66
Water, n = 1.33
Refraction
l1=v1T
1
q1
q1
2
q2
q2
The period T doesn’t change, but
the speed of light can be different.
in different materials. Then the
wavelengths l1 and l2 are
unequal. This also gives rise to
refraction.
l2=v2T
The little shaded triangles have the same
hypoteneuse: so l1/sinq1= l2/sinq2, or
v1/sinq1=v2/sinq2
Define the index of refraction: n=c/v.
Then Snell’s law is: n1sinq1 = n2sinq2
Example: air-water interface
If you shine a light at an incident angle of 40o onto the
surface of a pool 2m deep, where does the beam hit the
bottom?
Air: n=1.00
Water: n=1.33
40
air
water
2m
q
d
(1.00)sin40 = (1.33)sinq
sinq=sin40/1.33 so q=28.9o
Then d/2=tan28.9o which gives
d=1.1 m.
Example: air-water interface
If you shine a light at an incident angle of 40o onto the
surface of a pool 2m deep, where does the beam hit the
bottom?
Air: n=1.00
Water: n=1.33
40
air
water
2m
q
d
(1.00)sin40 = (1.33)sinq
sinq=sin40/1.33 so q=28.9o
Then d/2=tan28.9o which gives
d=1.1 m.
Example: air-water interface
If you shine a light at an incident angle of 40o onto the
surface of a pool 2m deep, where does the beam hit the
bottom?
Air: n=1.00
40
air
water
2m
q
d
Water: n=1.33
(1.00) sin(40) = (1.33) sinq
Sinq = sin(40)/1.33 so q = 28.9o
Then d/2 = tan(28.9o)
which gives  d=1.1 m.
Turn this around: if you shine a light from the bottom at
this position it will look like it’s coming from further right.
Some common
refraction effects
Air-water interface
Air: n1 = 1.00
Water: n2 = 1.33
n1 sinq1 = n2 sinq2
n1/n2 = sinq2 / sinq1
q1
air
water
q2
When the light travels from air to
water (n1 < n2) the ray is bent
towards the normal.
When the light travels from water
to air (n2 > n1) the ray is bent
away from the normal.
This is valid for any pair of materials with n1 < n2
Total Internal Reflection
n1 > n2
q2
q1
q2
q1
n2
qc
q1
n1
n2sin p/2 = n1sin q1
... sin q1 = sin qc = n2 / n1
Some light is refracted
and some is reflected
Total internal reflection:
no light is refracted
Total Internal Reflection
• The critical angle is when q2 = p / 2,
which gives qc = sin-1(n2/n1).
• At angles bigger than this “critical angle”,
the beam is totally reflected.
Find the critical angle for light traveling from glass (n = 1.5) to:
a) Air (n = 1.00)
b) Water (n = 1.33)
Example: Fiber Optics
An optical fiber consists of a core with index n1
surrounded by a cladding with index n2, with n1 > n2.
Light can be confined by total internal reflection,
even if the fiber is bent and twisted.
Example: Fiber Optics
Find the minimum angle of incidence for guiding in the fiber,
for n1 = 1.7 and n2 = 1.6
Example: Fiber Optics
Find the minimum angle of incidence for guiding in the fiber,
for n1 = 1.7 and n2 = 1.6
sin qC = n2 / n1
 qC = sin-1(n2 / n1)
 sin-1(1.6/1.7)  70o.
(Need to graze at < 20o)
Reflection and Transmission
at Normal Incidence
Geometrical optics can’t tell how much is reflected and how
much transmitted at an interface. This can be derived from
Maxwell’s equations. These are described in terms of the
reflection and transmission coefficients R and T, which are,
respectively, the fraction of incident intensity reflected and
transmitted. For the case of normal incidence, one finds:
TI
I
RI
2
n2  n1 
4n1 n2

R
, T  1 R 

( n2  n1 )2
 n2  n1 
Notice that when n1=n2 (so that there is not really any
interface), R = 0 and T = 1.