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AP PHYSICS 2
 What path did the light follow to
reach the wall?
UNIT 6
Geometric and
physical optics
 Represent the path from the
laser to the wall with an arrow.
 Why can’t you see the beam of
light itself but you can see the
spot on the wall?
CHAPTER 21
Reflection and
refraction
 Sprinkle chalk dust along the
line of propagation. What are
your observations?
RAY MODELS
MODEL 1
MODEL 2
 We can see objects (even tiny ones such as
dust) illuminated by light. The path of light is a
straight line from the source of light to the object
and then (assuming that the behavior does not
change) another straight line of reflected light
from the object to our eyes)
WHITEBOARD
How would the shadow on the wall would look like with
each ray model? (One sketch of each ray model, for each
experiment)
EXPERIMENT 1
(object close to screen)
EXPERIMENT 2
(object NOT so close to screen)
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WHITEBOARD
 What How would you see on the wall. Explain using each
ray model.
EXPERIMENT 3
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Ray model of light
Ray diagrams
 Diagram that represents the travel of light
from one location to another, drawn as a
straight line and an arrow.
 Testing experiments show that model 1 is inconsistent
with experimental evidence.
 Model 2 is supported:
 Each point on an extended light source emits light in
many different directions.
 This light can be represented by multiple rays
diverging from that point.
Shadows and semi-shadows
 A sharp shadow is called an umbra.
 A shadow is a region behind the object where no light
reaches.
 A semi-shadow is called a penumbra.
 A semi-shadow is a region where some light reaches
and some does not. It appears as a fuzzy shadow.
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WHITEBOARD
 On a sunny day, a
streetlight pole casts a
9.6-m-long shadow on
the ground. You have a
meter stick that, when
held vertical, casts a
0.70-m shadow. Use this
information to determine
the height of the pole.
WHITEBOARD
 You place a lit candle
several meters from the wall
in an otherwise dark room.
Between the candle and the
wall (and close to the
candle), you place a piece
of stiff paper (or cardboard)
with a small hole in it. Use
the ray model of light
propagation to predict what
you will see on the wall.
Pinhole camera
 Cardboard with a small hole in it is the foundation of the
pinhole camera, also called a pin hole camera.
 It consists of a lightproof box with a very small hole in
one wall and a photographic plate or film inside the box
on the opposite wall.
 Before the invention of modern cameras that use lenses,
pinhole cameras were used to make photographs.
Ray Model of Light
• Light is a particle that
propagates in straight
lines, unless it is
reflected or enters a new
medium.
• Pioneered by Isaac
Newton in his book
Opticks
Theories of Light
• Ray Model, Wave Model, Photon Model
• The Ray Model, although not perfect,
explains a great deal of phenomena
– Reflection, refraction, mirrors, and
lenses
• The Ray Model was the first attempt for
scientists to model the behavior of light.
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Light is fast! Really fast!
• 300,000,000 m/s
• 1/10th of a second to go from NY to LA
• 1 light year = the distance that light travels in
one year
• The speed of light in a vacuum is the speed limit
of the universe
c = 3x108 m/s
(speed of light in a vacuum)
Ray Model of Light
• One of the most important components of the Ray Model is
the way that objects emit light.
• In a well-lit room, all objects are visible from every angle.
• Not only this, but every point on the object is visible from all
directions!
This is because when objects reflect
light, the light rays are so numerous that
each point on the surface of the object
constantly emits light in all directions.
The room is completely filled with light,
constantly reflecting off of objects!
Reflection of light
Many, many rays…
However, we can only detect the
ones that reach our eye!
• Light from a laser pointer shines on a flat mirror.
In fact, there are so many rays
constantly reflecting off of
ordinary objects that the Ray
Model assumes an essentially
infinite number of light rays
coming out in all directions.
The Law of Reflection
incident ray
normal
reflected ray
When the mirror is curved, the normal line is
basically just the radius of the mirror.
incident ray
θi θr
mirror
normal
mirror
The ray reflects symmetrically across the normal line.
The normal line is perpendicular to the surface of the mirror,
and touches the point where the ray hits the mirror.
θi = θr
θi
θr
Think of the
mirror as a part
of a circle!
reflected ray
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Reflection of light
 Incident light: light striking the mirror
 Normal line: a line perpendicular to
surface where the incident light hits
mirror
 Angle of incidence: the angle between
incident beam and the normal line
 Angle of reflection: the angle between
reflected beam and the normal line
Law of reflection
the
the
the
the
• When a narrow beam of light, represented by one ray,
shines on a smooth surface such as a mirror, the angle
between the incident ray and the normal line
perpendicular to the surface equals the angle between
the reflected ray and the normal line.
• The incident beam, reflected beam, and the normal line
are in the same plane.
reflection = incidence
WHITEBOARD
• Two mirrors stand on a table, with their faces forming an
angle greater than 90o. Place an phone on the table in
front of mirror 2.
• Use the rule of reflection to predict how to aim a laser
beam so that it hits first mirror 1 and then mirror 2, and
finally hits the center of the target.
• Draw a top view
(bird eyes view)
Specular and diffuse reflection
WHITEBOARD
(discussion)
• On a sunny day, if you look at a house with its
lights off, the uncovered windows look almost
black but the outside walls do not. How can we
explain this difference?
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Red eye effect
REFRACTION
• When a camera flash illuminates the open iris,
light reflects from the red blood vessels in the
retina on the back of the eye.
• Some of this reflected light passes back out of
the pupil and makes the pupil appear red.
Refraction of light
Ever wonder why this happens?
• At the shore of a lake, you see sunlight reflecting off the
water's surface.
– You also see rocks and sea plants under the surface.
– To see them, light must have entered the water,
reflected off the rocks and plants, returned to the
water surface, and then traveled from the surface to
your eyes.
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Or this?
These and many more phenomena can be
understood by learning how light refracts
Refract – to change direction upon entering
a new type of material
This is responsible for all sorts of image
distortions, and we can use it to our
advantage!
n=
c
v
Index of refraction of a material actually the
ratio of how fast light travels in a vacuum
divided by how fast light travels in the material.
Every material has an index of
refraction
Symbol: n
The index of refraction of a material is a measure of
how slowly light travels in that material.
The slower light goes in it,
the higher its index of refraction will be!
Index of
refraction
of material
c
n=
v
Speed of light
in a vacuum
Speed of light in
the material
Some common indexes of refraction
Whiteboard Quick Question!
What range of values will n have?
What values are impossible for n?
Answer: Since c (3 × 108 m/s) is the fastest possible speed
for light to have (unhindered in a vacuum), v will always
be less than c.
This means that n for any material other than a vacuum
must be a number greater than 1!
More dense materials tend to have a higher index
of refraction (slower for light).
Whiteboard Question!
How fast does light travel in water?
n=
c
v
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Why does light travel slower in matter?
When light travels
through matter, it
is constantly
being absorbed
and re-emitted by
atoms.
In general, more
dense materials
will hinder the
speed of light.
What really happens when
light hits a boundary
What do we know so far?
When light travels from one medium to
another, its frequency does not change
(frequency depends on the source)
Video 1: https://www.youtube.com/watch?v=wlELYZJ5JF4
Video 2: Here
What really happens when
light hits a boundary
What really happens when
light hits an interface
v = λf
However, its velocity and wavelength will
change proportionally.
As a result of this change in speed, it is easily
observed that light will also change direction.
When light travels from one medium to another,
it refracts (changes direction).
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When light travels from one medium to another,
it refracts (changes direction)
air
To show how light refracts, we first need to know how to
draw a normal line at the point where the ray strikes the
new medium.
A normal line is perpendicular to the surface, and crosses
through the point where the ray hits the new material.
glass
Example: Normal lines on various surfaces
air
v
v
v
v
v
v
The ray above refracts twice:–
once entering the glass and once leaving the glass
Concepts of Refraction!
If a ray goes from a fast
medium into a slow medium,
it bends toward the normal.
θi
If a ray goes from a slow
medium into a fast medium, it
bends away from the normal.
Concepts of Refraction!
Lower n to higher n. Light
bends toward the normal.
θi
n1
θr
θr
θrefracted < θincident
θrefracted > θincident
A useful analogy!
When a car travels from the road
(fast medium) to mud (slow
medium), the tire that hits the
mud first will slow down first.
This will cause the car to
turn toward the normal
(just like light!)
When a car travels from mud
(slow medium) to the road (fast
medium), the tire that hits the
road first will speed up first.
n2
θr
n1 < n2
Higher n to lower n. Light
bends away from the normal.
n1
n2
θi
θr
n1 > n2
Whiteboard Showdown
Using the concepts of refraction and drawing the normal lines,
estimate the complete path of the incoming ray as it enters and
as it leaves the given object.
This will cause the car to
turn away from the normal
(just like light!)
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A WORD OF CAUTION
Light rays will never bend past the normal (see below).
The ray will always end up on the other side of the normal line.
(Or along the normal line if it came in along the normal)
Refraction of light
SNELL’S LAW!
• We can develop a mathematical relationship between
the angle of incidence and the angle of refraction.
n1 sinq1 = n2 sinq2
n1
θ1
θ2
The math! It’s just so beautiful!
Rewriting it like
n1 sinq 2
=
n2 sinq1
If n1 < n2, θ2 < θ1
θ1
θ2
Gee, thanks Snell!
Refractive indexes
shows the concept
If n1 > n2, θ2 > θ1
n1
n1
n2
n2
Fast to slow, bends towards
n2
This works for slowfast and fast-slow
transitions BOTH!
θ1
θ2
Slow to fast, bends away
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n1 sinq1 = n2 sinq2
n1
θ1
The angles will always be between
0° and 90°
θ2
This makes the sine function behave
nicely, ranging between 0 and 1.
n2
Give it a go!
air
n ≈ 1.00
Draw and label the
complete path of the
light ray through the
block of diamond
60°
Ignore this
madness
diamond
n ≈ 2.417
You will need to use
some geometry here
=)
n1 sinq1 = n2 sinq2
n ≈ 1.00
30
°
This is why you can see your reflection in a
window, but also see light coming through the glass
1.00sin30 = 2.417sinq2
q2 = sin-1 (sin30 / 2.417)
60
°
12
12 °
°
n ≈ 2.417
30°
θ2 ≈ 12°
Alternating interior angles show
that 12° is also the incident
angle for the second refraction
This makes the second refraction
just the reverse of the first!
Light traveling from a more optically
dense medium to a less optically dense
medium
purdy cool, huh
Light traveling from a more optically
dense medium to a less optically dense
medium
(increasing 1)
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Increasing 1 to the point that light
refracts at a 90°
(1 = C)
Total internal reflection
• At the critical angle of incidence, the refraction
angle is 90o. The refracted ray travels along the
water-air interface.
• If 1 > C, light is totally reflected.
But wait, there’s more!
Total Internal Reflection
As θ1 approaches
the critical angle,
θ2 approaches 90°.
n1
θ1
n2
It doesn’t all add up yet! Think about it – if light
goes from a slow medium to a fast medium
n1
θ2
Substituting into Snell’s Law, we get
n1 sinq1 = n2 sinq2
n1 sinqc = n2 sin90
n1 sinqc = n2
q c = sin -1
n2
n1
The critical angle!
n2
θ1
n1
θ2
n2
n1
θ1
θ2
n2
θ1
θ2
There is a fundamental limit on how far we can go
before the refracted ray can’t bend any further!
This is called the critical angle.
This gives rise
to some truly
beautiful results
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The index of refraction for diamonds is very high
compared to ordinary glass (2.4 vs 1.5)
FIBER OPTICS
APPLICATION OF TOTAL INTERNAL
REFLECTION
As a result, the critical angle for light to be reflected
totally as it travels from diamond to air is small (24 vs
42°for regular glass). Therefore most light is rereflected back from a diamond. This gives the
characteristic brilliance to a diamond
Thanks Russell
Fiber optics
• We can understand fiber
optics by using total internal
reflection.
– Fiber optic filaments are used in
telecommunications
to transmit high-speed
light-based data and in medicine
to see inside the human body
during surgery.
Pro Tip
If you ever get “DOMAIN ERROR” when you try to
calculate the critical angle, it means you have flipped
the indices of refraction.
You can never take the inverse sine of
a number that is not between -1 and 1!
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WHITEBOARD
WHITEBOARD
Draw the reflected and refracted rays
below. Indicate angles.
air
n = 1.00
You shine a laser light into the water at an
incident angle of 42°relative to the horizontal.
Determine the angle of the light in the water
relative to the normal line.
air
n = 1.00
50
°
water
n = 1.33
40
°
air
n=1
water
n = 1.33
water
n = 1.33
WHITEBOARD
WHITEBOARD
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟 = 𝑛𝑤𝑎𝑡𝑒𝑟 𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟
air
n=1
water
n = 1.33
48˚
42˚
𝜃𝑤𝑎𝑡𝑒𝑟
𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟 =
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
sin 48°
1.33
𝜃𝑤𝑎𝑡𝑒𝑟 = 33.97°
The equation below describes a physical process. Make
up a problem for which the equation would provide a
solution (Sketch it).
1.60 sin 30° = 1.33 sin𝜽𝟐
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
1.60 ∙ sin 30°
1.33
water
n = 1.33
glass
n = 1.60
𝜃𝑤𝑎𝑡𝑒𝑟 = 36.98°
30˚
WHITEBOARD
What is the critical
angle for total internal
reflection of light going
from water (n=1.33)
into glass of refractive
index 1.56?
?
𝜃𝑔𝑙𝑎𝑠𝑠 = 90°
water
n = 1.33
glass
n = 1.56
𝑛𝑤𝑎𝑡𝑒𝑟 𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑛𝑔𝑙𝑎𝑠𝑠 𝑠𝑖𝑛𝜃𝑔𝑙𝑎𝑠𝑠
𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟 =
𝑛𝑔𝑙𝑎𝑠𝑠 𝑠𝑖𝑛𝜃𝑔𝑙𝑎𝑠𝑠
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
𝑛𝑔𝑙𝑎𝑠𝑠 𝑠𝑖𝑛𝜃𝑔𝑙𝑎𝑠𝑠
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
1.56 ∙ sin 90°
1.33
Answer: Not possible
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WHITEBOARD
Light from a coin at the bottom of a
fountain reaches your eye at an
angle of 27.0° below the
horizontal.
WHITEBOARD
Answer: 42.1°
Sketch the actual path of the light.
Determine the angle between the
incident coin and the normal line.
WHITEBOARD
WHITEBOARD
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟 = 𝑛𝑤𝑎𝑡𝑒𝑟 𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟
𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟 =
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
sin 63°
1.33
𝜃𝑤𝑎𝑡𝑒𝑟 = 42.06°
WHITEBOARD
Answer: 45°
WHITEBOARD
A mosquito fish hides from a kingfisher bird at the
bottom of a shallow lake, 0.40 m below the surface. A
leaf has blown onto the lake and floats above the
mosquito fish. How big should the leaf be so the
kingfisher cannot see its prey from any location above
the water?
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WHITEBOARD
𝑅 =?
𝑛𝑤𝑎𝑡𝑒𝑟 𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑠𝑖𝑛𝜃𝑤𝑎𝑡𝑒𝑟 =
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑛𝑤𝑎𝑡𝑒𝑟
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛
−1
𝜃𝑤𝑎𝑡𝑒𝑟 = 𝑠𝑖𝑛−1
Answer:
C = 48.75°
R = 0.46 m
𝑛𝑎𝑖𝑟 𝑠𝑖𝑛𝜃𝑎𝑖𝑟
𝑛𝑤𝑎𝑡𝑒𝑟
1 ∙ sin 90°
1.33
ℎ = 0.4 𝑚
𝜃𝑤𝑎𝑡𝑒𝑟 = 48.75°
𝑅 = ℎ ∙ tan 𝜃𝑤𝑎𝑡𝑒𝑟
𝑅 = 0.456 𝑚
𝐿𝑒𝑎𝑓 (2𝑅) = 0.912 𝑚
𝜃𝑤𝑎𝑡𝑒𝑟 = 48.75°
WHITEBOARD: Fiber optics
• Imagine that you have a long glass block of
refractive index 1.56 surrounded by air. Light
traveling inside the block hits the top horizontal
surface at a 41o angle. What happens next?
Prisms
• The refractive index of prism glass is greater for
violet light and smaller for red light.
• The light is totally internally reflected during the
first incidence on the upper surface. From there
in moves down and to the right and hits the
bottom surface at 41°
Prisms for reflection
• Prisms reflect almost
100% of the light incident
on them, whereas mirrors
reflect somewhat less
than 100%.
• Prisms do not tarnish like
mirrors.
• Prisms can invert an
image—that is, make it
appear upside down.
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Mirages
Mirages
• On a hot day, hot air may hover just above the pavement.
This hot air is less dense and has a lower index of
refraction than the cooler air above it.
– When light from the sky passes through air with a
gradually changing index of refraction, its path
gradually bends, leading us to perceive that the source
of light is at a different location than it actually is.
Color of the sky
Particle model of light
• Due to their sizes, atmospheric particles reflect
blue light more efficiently than other colors.
Wave model of light
Wave model and refraction
• Imagine a light wave moving in a less optically
dense medium 1 and reaching an interface with a
denser medium 2 at a nonzero angle of incidence.
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