Chapter 22
Reflection and Refraction of Light
Waves and Optics
Geometric Optics (1.5 weeks)
Learning Objectives:
1. Reflection and refraction (pgs. 487-493)
Students should understand the principles of reflection and refraction, so they can:
a) Determine how the speed and wavelength of light change when light passes from
one medium into another.
b) Show on a diagram the directions of reflected and refracted rays.
c) Use Snell’s Law to relate the directions of the incident ray and the refracted ray,
and the indices of refraction of the media.
d) Identify conditions under which total internal reflection will occur.
2. Mirrors(pgs. 446-474)
Students should understand image formation by plane or spherical mirrors, so they
can:
a) Locate by ray tracing the image of an object formed by a plane mirror, and
determine whether the image is real or virtual, upright or inverted, enlarged or
reduced in size.
b) Relate the focal point of a spherical mirror to its center of curvature.
c) Locate by ray tracing the image of a real object, given a diagram of a mirror with
the focal point shown, and determine whether the image is real or virtual, upright or
inverted, enlarged or reduced in size.
d) Use the mirror equation to relate the object distance, image distance and focal
length for a lens, and determine the image size in terms of the object size.
3. Lenses (pgs. 494-511)
Students should understand image formation by converging or diverging lenses, so
they can:
a) Determine whether the focal length of a lens is increased or decreased as a result of
a change in the curvature of its surfaces, or in the index of refraction of the material
of which the lens is made, or the medium in which it is immersed.
b) Determine by ray tracing the location of the image of a real object located inside or
outside the focal point of the lens, and state whether the resulting image is upright
or inverted, real or virtual.
c) Use the thin lens equation to relate the object distance, image distance and focal
length for a lens, and determine the image size in terms of the object size.
d) Analyze simple situations in which the image formed by one lens serves as the
object for another lens.
Read and take notes on pages 421-426 and
432-436 in
Conceptual Physics Text
If you are not doing the readings from the Conceptual Physics Text
then read and take notes on the following throughout this unit
Reflection: Lesson 1 C and D
Refraction: Lesson 1 C-D
Lesson 2 A and B
Lesson 3 A-C
Read and take notes on
pages 732-735 in
College Physics Text
Nature of Light
• Light has a dual nature.
– Particle
– Wave
• Wave characteristics will be discussed in this
chapter.
– Reflection
– Refraction
– These characteristics can be used to understand
mirrors and lenses.
Introduction
A Brief History of Light
• Early models of light
– It was proposed that light consisted of tiny particles.
• Newton
– Used this particle model to explain reflection and
refraction
• Huygens
– 1678
– Explained many properties of light by proposing light was
wave-like
Section 22.1
A Brief History of Light, Cont.
• Young
– 1801
– Strong support for wave theory by showing
interference
• Maxwell
– 1865
– Electromagnetic waves travel at the speed of light.
Section 22.1
A Brief History of Light, Final
• Einstein
– Particle nature of light
– Explained the photoelectric effect
– Used Planck’s ideas
Section 22.1
The Particle Nature of Light
• “Particles” of light are called photons.
• Each photon has a particular energy.
– E=hƒ
– h is Planck’s constant
• h = 6.63 x 10-34 J s
– Encompasses both natures of light
• Interacts like a particle
• Has a given frequency like a wave
Section 22.1
Dual Nature of Light
• In some experiments light acts as a wave and in
others it acts as a particle.
– Classical electromagnetic wave theory provides
explanations of light propagation and interference.
– Experiments involving the interaction of light with matter
are best explained by assuming light is a particle.
• Light has a number of physical properties, some
associated with waves and others with particles.
Section 22.1
Either
Link to Bright Storm on Reflection and Refraction
(Start to minute 3:12)
Or
Khan Academy on Reflection and Refraction
(Start to minute 6:08)
Reflection and Refraction
• The processes of reflection and refraction can
occur when light traveling in one medium
encounters a boundary leading to a second
medium.
• In reflection, part of the light bounces off the
second medium.
• In refraction, the light passing into the second
medium bends.
• Often, both processes occur at the same time.
Section 22.2
Geometric Optics – Using a Ray Approximation
• Light travels in a straight-line path in a
homogeneous medium until it encounters a
boundary between two different media.
• The ray approximation is used to represent
beams of light.
• A ray of light is an imaginary line drawn along
the direction of travel of the light beams.
Section 22.2
Ray Approximation
• A wave front is a surface
passing through points of a
wave that have the same
phase and amplitude.
• The rays, corresponding to
the direction of the wave
motion, are perpendicular
to the wave fronts.
Section 22.2
Read and take notes on pages 442-459 in
Conceptual Physics Text
Khan Academy on
Specular and Diffuse
Reflection
More from Khan Academy on
Specular and Diffuse
Reflection
(optional)
Reflection of Light
• A ray of light, the incident ray, travels in a
medium.
• When it encounters a boundary with a second
medium, part of the incident ray is reflected
back into the first medium.
– This means it is directed backward into the first
medium.
Section 22.2
Specular Reflection
• Specular reflection is
reflection from a smooth
surface.
• The reflected rays are
parallel to each other.
• All reflection in this text is
assumed to be specular.
Diffuse Reflection
• Diffuse reflection is
reflection from a rough
surface.
• The reflected rays travel
in a variety of
directions.
• Diffuse reflection makes
the dry road easy to see
at night.
Section 22.2
Law of Reflection
• The normal is a line
perpendicular to the
surface.
– It is at the point where the
incident ray strikes the
surface.
• The incident ray makes an
angle of θ1 with the normal.
• The reflected ray makes an
angle of θ1’ with the normal.
Section 22.2
Law of Reflection, Cont.
• The angle of reflection is equal to the angle of
incidence.
• θ1= θ1’
Section 22.2
EXAMPLE 22.1 The Double Reflecting Light Ray
Mirrors M1 and M2 make an angle of 120° with each other.
Goal Calculate a resultant angle from two reflections.
Problem Two mirrors make an angle of 120° with each
other, as in the figure. A ray is incident on mirror M1 at
an angle of 65° to the normal. Find the angle the ray
makes with the normal to M2 after it is reflected from both mirrors.
Strategy Apply the law of reflection twice. Given the incident ray at angle θinc, find
the final resultant angle, βref.
SOLUTION
Apply the law of reflection to M1 to find the angle of reflection, θref.
θref = θinc = 65°
Find the angle that is the complement of the angle θref.
= 90° - θref = 90° - 65° = 25°
Find the unknown angle α in the triangle of M1, M2, and the ray traveling from M1
to M2, using the fact that the three angles sum to 180°.
180° = 25° + 120° + α → α = 35°
The angle α is complementary to the angle of incidence, βinc, for M2.
α + βinc = 90° → βinc = 90° - 35° = 55°
Apply the law of reflection a second time, obtaining βref.
βref = βinc = 55°
LEARN MORE
Remarks Notice the heavy reliance on elementary geometry and trigonometry in
these reflection problems.
Question In general, what is the relationship between the incident angle θinc and the
final reflected angle βref when the angle between the mirrors is 90.0°?
θinc - βref = 90.0°
θinc + βref = 180.0°
θinc + βref = 90.0°
Read and take notes on
pages 736-739 in
College Physics Text
Refraction of Light
• When a ray of light traveling through a transparent
medium encounters a boundary leading into another
transparent medium, part of the ray is reflected and
part of the ray enters the second medium.
• The ray that enters the second medium is bent at the
boundary.
– This bending of the ray is called refraction.
Section 22.2
Refraction of Light, Cont.
• The incident ray, the
reflected ray, the refracted
ray, and the normal all lie on
the same plane.
• The angle of refraction, θ2,
depends on the properties
of the medium.
Section 22.2
Following the Reflected and Refracted
Rays
• Ray  is the incident ray.
• Ray  is the reflected ray.
• Ray  is refracted into the
Lucite.
• Ray  is internally reflected
in the Lucite.
• Ray  is refracted as it
enters the air from the
Lucite.
Section 22.2
More About Refraction
• The angle of refraction depends upon the
material and the angle of incidence.
• The path of the light through the refracting
surface is reversible.
Section 22.2
Refraction Details, 1
• Light may move from a
material where its
speed is high to a
material where is speed
is lower.
• The angle of refraction
is less than the angle of
incidence.
– The ray bends toward
the normal.
Section 22.2
Refraction Details, 2
• Light may move from a
material where its speed is
low to a material where is
speed is higher.
• The angle of refraction is
greater than the angle of
incidence.
– The ray bends away from the
normal.
Link to Bright Storm on
Index of Refraction
(Minute 3:12- End)
The Index of Refraction
• When light passes from one medium to
another, it is refracted because the speed of
light is different in the two media.
• The index of refraction, n, of a medium can be
defined
Section 22.3
Index of Refraction, Cont.
• Some values of n
– For a vacuum, n = 1
– For other media, n > 1
• n is a unitless ratio
• As the value of n increases, the speed of the
wave decreases.
Section 22.3
EXAMPLE 22.2 Angle of Refraction for Glass
Goal Apply Snell's law to a slab of glass.
Problem A light ray of wavelength 589 nm (produced by a
sodium lamp) traveling through air is incident on a smooth,
flat slab of crown glass at an angle of 30.0° to the normal,
as sketched in the figure. Find the angle of refraction, θ2.
Strategy Substitute quantities into Snell's law and solve for
the unknown angle refraction, θ2.
SOLUTION
Solve Snell's law for sin θ2.
(1)
sin θ2 = (n1/n2) sin θ1
Using a refraction index, find n1 = 1.00 for air and n2 = 1.52 for crown glass.
Substitute these values into Equation (1) and take the inverse sine of both sides.
sin θ2 = (1.00 /1.52)(sin 30.0°) = 0.329
θ2 = sin-1 (0.329) = 19.2°
LEARN MORE
Remarks Notice that the light ray bends toward the normal when it enters a
material of a higher index of refraction. If the ray left the material following the
same path in reverse, it would bend away from the normal.
Question If the glass is replaced by a transparent material with smaller index of
refraction than that of glass, the refraction angle θ2 will be:
smaller
larger
unchanged
Frequency Between Media
• As light travels from one
medium to another, its
frequency does not change.
– Both the wave speed and the
wavelength do change.
– The wavefronts do not pile
up, nor are created or
destroyed at the boundary, so
ƒ must stay the same.
Section 22.3
Index of Refraction Extended
• The frequency stays the same as the wave travels
from one medium to the other.
• v=ƒλ
• The ratio of the indices of refraction of the two
media can be expressed as various ratios.
Section 22.3
Some Indices of Refraction
Section 22.3
EXAMPLE 22.3 Light in Fused Quartz
Goal Use the index of refraction to determine the effect of a medium on light's
speed and wavelength.
Problem Light of wavelength 589 nm in vacuum passes through a piece of fused
quartz of index of refraction n = 1.458. (a) Find the speed of light in fused quartz.
(b) What is the wavelength of this light in fused quartz? (c) What is the frequency
of the light in fused quartz?
Strategy Substitute values into the speed of light in media equation and the
wavelength in media equations.
SOLUTION
(a) Find the speed of light in fused quartz.
Obtain the speed from the speed of light in medium equation.
c 3.00 108 m/s
= 2.06 108 m/s
v= =
1.458
n
(b) What is the wavelength of this light in fused quartz?
Use the relevant equation to calculate the wavelength:
λn =
λ0 589 nm
=
= 404 nm
n 1.458
(c) What is the frequency of the light in fused quartz?
The frequency in quartz is the same as in vacuum. Solve c = fλ for the frequency.
c 3.00 108 m/s
= 5.09 1014 Hz
f= =
-7
λ 5.89 10 m
LEARN MORE
Remarks It's interesting to note that the speed of light in vacuum, 3.00 108 m/s, is
an upper limit for the speed of material objects. In studying relativity, we will find
that this upper limit is consistent with experimental observations. However, it's
possible for a particle moving in a medium to have a speed that exceeds the speed
of light in that medium. For example, it's theoretically possible for a particle to
travel through fused quartz at a speed greater than 2.06 108 m/s, but it must still
have a speed less than 3.00 108 m/s.
Question If light with wavelength λ in glass passes into water with index nw, the
new wavelength of the light is:
λ / nw
λ
λng/nw
λnw/ng
Either
Link to Bright Storm on Snell’s Law
(Start to minute 3:25)
Or
Khan Academy on Snell’s Law
(Minute 6:08 to End)
Khan Academy on Refraction in
water
(Optional but short and cool)
Snell’s Law of Refraction
• n1 sin θ1 = n2 sin θ2
– θ1 is the angle of incidence
– θ2 is the angle of refraction
• The experimental discovery of this
relationship is usually credited to Willebrørd
Snell (1591 – 1626).
Section 22.3
Khan Academy on example
problems of Snell’s law
(must watch)
More from Khan Academy on
example problems of Snell’s law
(This is a sample AP test Question, MUST WATCH!!!)
EXAMPLE 22.4 Light Passing Through a Slab
When light passes through a flat slab of material, the emerging beam is parallel to the
incident beam; therefore, θ1 = θ3.
Goal Apply Snell's law when a ray passes into and out of
another medium
Problem A light beam traveling through a transparent
medium of index of refraction n1 passes through a thick
transparent slab with parallel faces and index of refraction n2.
Show that the emerging beam is parallel to the incident beam.
Strategy Apply Snell's law twice, once at the upper surface and once at the lower
surface. The two equations will be related because the angle of refraction at the
upper surface equals the angle of incidence at the lower surface. The ray passing
through the slab makes equal angles with the normal at the entry and exit points.
This procedure will enable us to compare angles θ1 and θ3.
SOLUTION
Apply Snell's law to the upper surface.
(1)
n1
sin θ2 = sin θ1
n2
Apply Snell's law to the lower surface.
(2)
n2
sin θ3 = sin θ2
n1
Substitute Equation (1) into Equation (2).
n2
sin θ3 =
n1
(
n1
n2 sin θ1
) = sin θ
1
Take the inverse sine of both sides, noting that the angles are positive and less than
90°.
θ3 = θ1
LEARN MORE
Remarks The preceding result proves that the slab doesn't alter the direction of the
beam. It does, however, produce a lateral displacement of the beam, as shown in the
figure.
Question Suppose an additional slab with index n3 were placed below the slab of
glass. Would the exit angle at the bottom surface of this second slab still equal the
incident angle at the upper surface of the first slab?
No, never.
Sometimes yes and sometimes no, depending on the refractive
indices of the media.
to the first surface.
They are equal only if the incident light is perpendicular
Yes, always.
EXAMPLE 22.5 Refraction of Laser Light in a Digital Videodisc (DVD)
(a) A micrograph of a DVD surface showing tracks
and pits along each track. (b) Cross section of a
cone-shaped laser beam used to read a DVD.
Goal Apply Snell's law together with
geometric constraints.
Problem A DVD is a video
recording consisting of a spiral track
about 1.0 µm wide with digital information. (See Fig. a.) The digital information
consists of a series of pits that are "read" by a laser beam sharply focused on a track
in the information layer. If the width a of the beam at the information layer must
equal 1.0 µm to distinguish individual tracks and the width w of the beam as it
enters the plastic is 0.7000 mm, find the angle θ1 at which the conical beam should
enter the plastic. (See Fig. b.) Assume the plastic has a thickness t = 1.20 mm and
an index of refraction n = 1.55. Note that this system is relatively immune to small
dust particles degrading the video quality because particles would have to be as
large as 0.700 mm to obscure the beam at the point where it enters the plastic.
Strategy Use right-triangle trigonometry to determine the angle θ2 and then apply
Snell's law to obtain the angle θ1.
SOLUTION
From the top and bottom of figure b, obtain an equation relating w, b, and a.
w = 2b + a
Solve this equation for b and substitute given values.
w − a 7.000 10-4 m − 1.0 10-6 m
=
= 349.5 µm
b=
2
2
Now use the tangent function to find θ2.
349.5 µm
b
tan θ2 = =
→ θ2 = 16.2°
3
t 1.20 10 µm
Finally, use Snell's law to find θ1.
n1 sin θ1 = n2 sin θ2
n2 sin θ2 1.55 sin 16.2°
sin θ1 =
=
= 0.433
1.00
n1
θ1 = sin-1 (0.433) = 25.7°
LEARN MORE
Remarks Despite its apparent complexity, the problem isn't that different from
calculating the refraction of light by glass.
Question Suppose the plastic were replaced by a material with a higher index of
refraction. How would the width of the beam at the information layer be affected?
It would decrease.
It would increase.
It would remain the same.
Read and take notes on
pages 742-746 in
College Physics Text
Dispersion
• The index of refraction in anything except a vacuum
depends on the wavelength of the light.
• This dependence of n on λ is called dispersion.
• Snell’s Law indicates that the angle of refraction
made when light enters a material depends on the
wavelength of the light.
Section 22.4
Variation of Index of Refraction with
Wavelength
• The index of refraction
for a material usually
decreases with
increasing wavelength.
• Violet light refracts
more than red light
when passing from air
into a material.
Section 22.4
Refraction in a Prism
• The amount the ray is bent
away from its original
direction is called the angle
of deviation, δ
• Since all the colors have
different angles of
deviation, they will spread
out into a spectrum.
– Violet deviates the most.
– Red deviates the least.
Section 22.4
Prism Spectrometer
• A prism spectrometer uses a prism to cause the wavelengths
to separate.
• The instrument is commonly used to study wavelengths
emitted by a light source.
Section 22.4
Using Spectra to Identify Gases
• All hot, low pressure gases emit their own
characteristic spectra
• The particular wavelengths emitted by a gas serve as
“fingerprints” of that gas.
• Some uses of spectral analysis
– Identification of molecules
– Identification of elements in distant stars
– Identification of minerals
Section 22.4
The Rainbow
• A ray of light strikes a drop of water in the
atmosphere.
• It undergoes both reflection and refraction.
– First refraction at the front of the drop
• Violet light will deviate the most.
• Red light will deviate the least.
Section 22.5
The Rainbow, 2
• At the back surface the light
is reflected.
• It is refracted again as it
returns to the front surface
and moves into the air.
• The rays leave the drop at
various angles.
– The angle between the white
light and the violet ray is 40°
– The angle between the white
light and the red ray is 42°
Section 22.5
Observing the Rainbow
• If a raindrop high in the sky
is observed, the red ray is
seen.
• A drop lower in the sky
would direct violet light to
the observer.
• The other colors of the
spectra lie in between the
red and the violet.
Section 22.5
Read and take notes on
pages 746-749 in
College Physics Text
Christian Huygens
• 1629 – 1695
• Best known for
contributions to fields of
optics and dynamics
• Deduced the laws of
reflection and refraction
• Explained double refraction
Section 22.6
Huygen’s Principle
• Huygen assumed that light is a form of wave
motion rather than a stream of particles.
• Huygen’s Principle is a geometric construction
for determining the position of a new wave at
some point based on the knowledge of the
wave front that preceded it.
Section 22.6
Huygen’s Principle, Cont.
• All points on a given wave front are taken as
point sources for the production of spherical
secondary waves, called wavelets, which
propagate in the forward direction with
speeds characteristic of waves in that
medium.
– After some time has elapsed, the new position of
the wave front is the surface tangent to the
wavelets.
Section 22.6
Huygen’s Construction for a 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 cΔt, a
new plane BB’ can be
drawn tangent to the
wavefronts.
Section 22.6
Huygen’s Construction for a Spherical Wave
• 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.
Section 22.6
Huygen’s Principle and the 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.
Section 22.6
Huygen’s Principle and the Law of Reflection,
Cont.
• Triangle ADC is congruent to triangle AA’C.
• θ1 = θ1’
• This is the Law of Reflection.
Section 22.6
Huygen’s Principle and the Law of Refraction
• In time Δt, 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 θ1 = n2 sin θ2
Section 22.6
Khan Academy on Total Internal
Reflection
Total Internal Reflection
• Total internal reflection can
occur when light attempts
to move from a medium
with a higher index of
refraction to one with a
lower index of refraction.
– Ray 5 shows internal
reflection
Section 22.7
Critical Angle
• A particular angle of
incidence will result in
an angle of refraction of
90°
– This angle of incidence is
called the critical angle.
Section 22.7
Critical Angle, Cont.
• For angles of incidence greater than the critical
angle, the beam is entirely reflected at the boundary.
– This ray obeys the Law of Reflection at the boundary.
• Total internal reflection occurs only when light is
incident on the boundary of a medium having a
lower index of refraction than the medium in which
it is traveling.
Section 22.7
Fiber Optics
• An application of internal
reflection
• Plastic or glass rods are
used to “pipe” light from
one place to another.
• Applications include
– Medical use of fiber optic
cables for diagnosis and
correction of medical
problems
– Telecommunications
EXAMPLE 22.7
Goal Apply the concept of total internal reflection.
Problem (a) Find the critical angle for a water-air
boundary. (b) Use the result of part (a) to predict what a
fish will see if it looks up toward the water surface at
angles of 40.0°, 48.6°, and 60.0°.
Strategy After finding the critical angle by substitution, use that the path of a light
ray is reversible: at a given angle, wherever a light beam can go is also where a
light beam can come from, along the same path
SOLUTION
(a) Find the critical angle for a water-air boundary.
Substitute to find the critical angle.
n2 1.00
sin θc = =
= 0.750
n1 1.333
θc = sin-1 (0.750) = 48.6°
(b) Predict what a fish will see if it looks up toward the water surface at angles of
40.0°, 48.6°, and 60.0°.
At an angle of 40.0°, a beam of light from underwater will be refracted at the
surface and enter the air above. Because the path of a light ray is reversible (Snell's
law works both going and coming), light from above can follow the same path and
be perceived by the fish. At an angle of 48.6°, the critical angle for water, light from
underwater is bent so that it travels along the surface. So light following the same
path in reverse can reach the fish only by skimming along the water surface before
being refracted toward the fish's eye. At angles greater than the critical angle of
48.6°, a beam of light shot toward the surface will be completely reflected down
toward the bottom of the pool. Reversing the path, the fish sees a reflection of some
object on the bottom.
LEARN MORE
Question If the water is replaced by a transparent fluid with a higher index of
refraction, the critical angle of the fluid-air boundary is:
larger.
the same as for water.
smaller.
Link to Webassign
Unit 4 B Reflection and Refraction
Discussion Questions
Link to Webassign
Unit 4 B Reflection and Refraction
Homework 1-7
Grading Rubric for Unit 4B Reflection and Refraction
Name: ______________________
Conceptual Physics Text
Pgs 421-426 ---------------------------------------------------_____
Pgs 432-336 ---------------------------------------------------_____
Pgs 442-459 ---------------------------------------------------_____
Advanced notes from text book:
Pgs 732-735 --------------------------------------------------_____
Pgs 736-739 ---------------------------------------------------_____
Pgs 742-746 ---------------------------------------------------_____
Pgs 746-749 --------------------------------------------------_____
Example Problems:
22.1 (a) ----------------------------------------------------------_____
22.2 (a) ----------------------------------------------------------_____
22.3 (a-c) --------------------------------------------------------_____
22.4 (a) ----------------------------------------------------------_____
22.5 (a) ----------------------------------------------------------_____
22.7 (a-b) -------------------------------------------------------_____
Web Assign 1 (a-d), 2 (a-c), 3 (a-d), 4, 5, 6, 7 (a-f) -----------------------____