Announcements
 Exam 3 is one week from tomorrow. I will need to know by
Wednesday of any students who have special needs different
than for exam 2.
Today’s material WILL NOT be covered on Exam 3!
Material for Exam 3:
Chapters 28, 29, 32
 Friday of this week is the last day to withdraw.
Today’s agenda:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s Law).
You must be able to determine the path of light rays using the laws of reflection and
refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs,
and apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
Light
Normally, “light” refers to the type of electromagnetic wave
that stimulates the retina of our eyes.
Light acts like a wave except when it acts like particles.
How many physicists does it take to change a light bulb?
Light—Waves or Particles?*
http://www.nearingzero.net (quantum007.jpg)
*Light is both! Take Physics 2305 for enlightenment!
The Speed of Light
Light is a type of electromagnetic wave and travels with the
speed c = 2.9979x108 m/s in a vacuum. (Just use 3x108!)
How many physicists does it take to change a light bulb?
Eleven. One to do it and ten to co-author the paper.
Visible light is a small part of the electromagnetic spectrum.
News Flash!
Light slows down when it passes from
a vacuum into matter!
Details later!
Geometric Optics
Although light is actually an electromagnetic wave, it generally
travels in straight lines (like particles do!).
We can describe many properties of light by assuming that it
travels in straight-line paths in the form of rays.
A ray is a straight line along which light is propagated. In
other contexts, the definition of ray might be extended to
include bent or curved lines.
http://www.lab-initio.com (nz341.jpg)
A light ray is an infinitely thin beam of light. Of course, there
really isn’t such a thing, but the concept helps us visualize
properties of light.
thankfully, there really isn’t such a thing
http://www.lab-initio.com (nz342.jpg)
Light rays from some
external source strike an
object and reflect off it in all
directions.
We only see those light rays
that reflect in the direction
of our eyes.
If you can see something, it
must be a source of light!
Zillions* of rays are simultaneously reflected in all directions
from any point of an object. Later, when we study mirrors and
lenses, we won’t try do draw them all! Just enough
representative ones to understand what the light is doing.
*one zillion = 10a big number
Today’s agenda:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s Law).
You must be able to determine the path of light rays using the laws of reflection and
refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs,
and apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
Reflection
Light striking a surface may be reflected, transmitted, or
absorbed. Reflected light leaves the surface at the same angle
it was incident on the surface:
i = r
i r
Real Important Note: the angles are measured relative to the surface normal.
Possible homework hint: the sum of the angles in a 4-sided polygon is 360.
Reflection from a
smooth surface is
specular (mirrorlike). Reflection
from a rough
surface is diffuse
(not mirror-like).
http://acept.la.asu.edu/PiN/rdg/reflection/reflection.shtml
http://www.mic-d.com/java/specular/
Refraction
Light travels in a straight line
except when it is reflected or
when it moves from one
medium to another.
http://id.mind.net/~zona/mstm/physics/light/rayOptics/refraction/refraction1.html
Refraction—the “bending,” or change of direction of light rays
when light moves from one medium to a different one—takes
place because light travels with different speeds in different
media.
The speed of light in a vacuum is c = 3x108 m/s. The index of
refraction of a material is defined by
c
n =
,
v
If you study light in advanced classes,
you’ll find it is more complex than this.
where c is the speed of light in a vacuum and v is the speed of
light in the material.
The speed and wavelength of light change when it passes
from one medium to another, but not the frequency, so
c

v=
and n = .
n
n
Because light never travels faster than c, n  1.* For water, n =
1.33 and for glass, n 1.5. Indices of refraction for several
materials are listed in your text.
Example: calculate the speed of light in diamond (n = 2.42).
c
v =
n
3×108 m/s
v =
2.42
v = 1.24×108 m/s
*Actually, not true but don’t worry about it unless you take advanced courses in optics.
Because n  1, we
see this:
If n < 1, we would
see this:
Thanks to Dr. Xiaodong Yang for the images.
Snell’s Law
When light moves from one medium into another, some is
reflected at the boundary, and some is transmitted. “Toy”
The transmitted light is refracted (“bent”).
a is the angle of incidence, and b is the angle of refraction.
a
b
air (na)
water (nb)
water (na)
b
nb>na
air (nb)
a
na>nb
na sin  θa  = nb sin  θb 
Light passing from air (n  1) into water (n  1.33).
Light “bends” towards the normal to the surface as it slows
down in water.
a
air (na)
water (nb)
b
(1)  sin  θa   = (1.33)  sin θb  
θa > θb
nb>na
na sin  θa  = nb sin  θb 
Light passing from water (n  1.33) into air (n  1).
Light “bends” away from the normal to the surface as it
speeds up in air.
(1.33)  sin  θa   = (1)  sin  θb  
b
air (nb)
water (na)
θa < θb
a
na>nb
na sin  θa  = nb sin  θb 
Snell’s law, also called the law of refraction, gives the
relationship between angles and indices of refraction:
na sin  θa  = nb sin  θb  .
air (na)
air (na)
a
a
b
water (nb)
b
water (nb)
You are free to choose which is “a” and which is “b.”
 is the angle the ray makes with the normal!
Example: a 45-45-90 glass (n=1.50) prism is surrounded by
water (n=1.33). Light is incident at a 23 angle, as shown in
the diagram. What angle does the light make when it exits the
prism?
45
ng=1.50
45
i
nw=1.33
Example: a 45-45-90 glass (n=1.50) prism is surrounded by
water (n=1.33). Light is incident at a 23 angle, as shown in
the diagram. What angle does the light make when it exits the
prism?
45
1.33 sin θi = 1.50 sin θg
ng=1.50
1.33
sin θg =
sin 23
1.50
g
45
i
nw=1.33
θg = 20.27
keep an extra digit to
reduce roundoff error
Example: a 45-45-90 glass (n=1.50) prism is surrounded by
water (n=1.33). Light is incident at a 23 angle, as shown in
the diagram. What angle does the light make when it exits the
prism?
Trig…
45
20.27
65.27
90 - 20.27 = 69.73
180 - 69.73 - 45 = 65.27
69.73
45
nw=1.33
23
90 - 65.27 = 24.73
There are a variety of
ways to get this.
Example: a 45-45-90 glass (n=1.50) prism is surrounded by
water (n=1.33). Light is incident at a 23 angle, as shown in
the diagram. What angle does the light make when it exits the
prism?
45

24.73
45
nw=1.33
Example: a 45-45-90 glass (n=1.50) prism is surrounded by
water (n=1.33). Light is incident at a 23 angle, as shown in
the diagram. What angle does the light make when it exits the
prism?
1.50 sin 24.73 = 1.33 sin 
45

1.50
sin  =
sin 24.73
1.33
24.73
45
nw=1.33
 = 28.2
Interactive applet: Fun with Snell’s Law.
Same applet as slide 20, try glass into water.
Total Internal Reflection; Fiber Optics
n1 sin  θ1  = n2 sin  θ2 
sin  θ1  =
n2
sin  θ2 
n1
Suppose n2<n1. The largest possible value of sin(2) is 1
(when 2 = 90). The largest possible value of sin(1) is
sin  θ1,max 
n2
= .
n1
For 1 larger than this, Snell’s
Law cannot be satisfied!
Example:
=1.5
n2=1
sin angle,
2=0.667
sin of
2.
This
valuen1of
 isand
called
thethen
critical
C.sinFor
any angle
1=1/1.5
If
you set larger
thenCn,1 all
sinof
1=n
sin 2incident
has no solution.
A ray
incidence
than
the2 light
at an interface
1=45,
coming
in at and
a 45none
angle
pass through the interface and
is reflected,
is cannot
transmitted.
be refracted.
1 < C
1 close to C
1
1
1 > C
1
n2

n1>n2
Ray incident normal to surface is not “bent.” Some is reflected,
some is transmitted.
n2

Increasing angle of incidence…
n1>n2
n2

Increasing angle of incidence…more…
n1>n2
n2

n1>n2
Increasing angle of incidence…more…critical angle reached…
some of incident energy is reflected, some is “transmitted
along the boundary layer.
n2

n1>n2
Light incident at any angle beyond C is totally internally
reflected.
application: fiber optics
http://laser.physics.sunysb.edu/~wise/wise187/janfeb2001/reports/andr
ea/report.html
Example: determine the incident angle i for which light strikes
the inner surface of a fiber optic cable at the critical angle.
Light is incident at
an angle i on a
transparent fiber.
i
f
nf>1
ni=1 (air)
The light refracts at an angle f.
ni sin  θi  = nf sin  θf 
sin  θi  = nf sin  θf 
Light strikes the fiber
wall an an angle of
90-f normal to the
surface.
90
i
f
90-f
nf>1
ni=1 (air)
At the critical angle, instead of exiting the fiber, the refracted
light travels along the fiber-air boundary. In this case, 90-f is
the critical angle.
nf sin  90 - θf  =nf sin  θc  =na sin  90   1
1
sin  90 - θ f  =
nf
Solve the above for f and use sin  θi  = nf sin  θf  to solve for
i.
Numerical example:
what i will result in the
critical angle if nf=1.4?
90
i
90-f
f
nf>1
ni=1 (air)
Begin the analysis at the fiber-into-air interface:
1.4  sin  90 - θf  = 1 sin  90 
1
sin  90 - θ f  =
1.4
90 - θf = 45.58
θf = 44.41
keep an extra digit to
reduce roundoff error
Numerical example:
what i will result in the
critical angle if nf=1.4?
90
i
θf = 44.41
90-f
f
nf>1
ni=1 (air)
Next consider the air-into-fiber interface.
1 sin θi  = 1.4  sin  44.41
sin  θi  = 0.980
θi = 78.5
This is a very large angle of incidence! If you want the incident light to be
nearly parallel to the fiber axis, you must surround the fiber with a coating
with noutside<ncoating<nfiber.
application: swimming underwater
If you are looking up from underwater, if your angle of sight
(relative to the normal to the surface) is too large, you see an
underwater reflection instead of what’s above the water.
application: perfect mirrors
(used in binoculars)
application: diamonds
Today’s agenda:
Introduction to Light.
You must develop a general understanding of what light is and how it behaves.
Reflection and Refraction (Snell’s Law).
You must be able to determine the path of light rays using the laws of reflection and
refraction.
Total Internal Reflection and Fiber Optics.
You must be able to determine the conditions under which total internal reflection occurs,
and apply total internal reflection to fiber optic and similar materials.
Dispersion.
You must understand that the index of refraction of a material is wavelength-dependent.
Picture from the Exploratorium (http://www.exploratorium.edu/).
Dispersion
We’ve treated the index of refraction of a material as if it had a
single value for all wavelengths of light.
In fact, the speed of light in a material substance depends on
the wavelength, so the index of refraction is generally
wavelength- (or color-) dependent.
http://www.physicsclassroom.com/class/refrn/Lesson-4/Dispersion-of-Light-by-Prisms