Light and Reflection
Electromagnetic Radiation

The difference among waves is the
nature of the vibrating source.

An electromagnetic wave is the coupled
vibration of electric and magnetic fields
free of matter.

An electromagnetic wave is produced
by a vibrating (moving) electric field or
magnetic field.

All electromagnetic waves travel at the
same speed.
• Speed of light (c):
2.997 924 58 x 108 m/s in a vacuum
186000 mi/s
2.997 09 x 108 m/s in air
c = fl
• Dual Nature of Light
• Light waves
• Light particles = photons
E = hf
• h = 6.626 x 10-34 J.s
• Electromagnetic waves are oscillating electric and magnetic fields.
Visible light is electromagnetic radiation with a frequency range of
4.3 x 1014 to 7 x 1014 Hz.
Optics: the study of the behavior and properties of light

The study and description of light in terms of straight line
propagation is called optics . In optics, lines, called rays,
are drawn perpendicular to the wave front.

Ray optics are used to analyze reflection and refraction
of light .

Ray optics do not account for the wave properties of
diffraction and interference.


When light waves encounter a barrier (change
media) they are absorbed, transmitted, or
reflected.
The texture of the reflecting surface affects
how it reflects light.
 Diffuse reflection is reflection from a rough, texture
surface such as paper or unpolished wood.
 Specular reflection is reflection from a smooth,
shiny surface such as a mirror or a water surface.
A good mirror can reflect
90% of the incident light.
The Law of Reflection

When light is reflected it obeys the Law of
Reflection
The angle of incidence and the angle of
reflection are always equal.

The angle of incidence is the angle between a ray
that strikes a surface and the line perpendicular (the
normal) to that surface at the point of contact.

The angle of reflection is the angle formed by the
normal and the direction in which a reflected ray
moves.
The normal is an imaginary line drawn perpendicular
to the reflecting surface used to measure the
incident and reflected angles.


Mirror and Lens Terminology
 Object distance is p.
 Image distance is q.
 Object height is h.
 Image height is h’.
 Angle of incidence is q.
 Angle of reflection is q’.
A dental mirror

Mirror and Lens Terminology
 Virtual Image: The image
formed by rays that appear to
come from the image point
behind the mirror—but never
really do .
 A virtual image is not real.
 A virtual image can never be
displayed on a physical surface.
A dental mirror

Mirror and Lens Terminology
 Real Image: The image
formed by rays of light that
actually pass through a
point on the image.
 A real image can be
displayed on a physical
surface (a screen).
A dental mirror

Flat Mirrors
 Flat mirrors form virtual images
that are the same distance from
the mirror’s surface as the
object is.

The Law of Reflection
 The angle of incidence and the angle of reflection are
always equal.
 qi = qr
qi
qr
Q’i
Q’r

Curved Mirror and Lens Terminology
 R is the radius of curvature,
which is the same as the
radius of the spherical shell of
which the mirror is a small
part.
 C is the center of curvature.
 The principal axis (optic axis)
is the line that extends
infinitely from the center of
the mirror’s surface through
the center of curvature.

Curved Mirror and Lens Terminology
 The focal point is the location at which
a mirror or a lens focuses rays parallel
to the principal axis or from which
such rays appear to diverge.
 f is the focal length
 f = 1/2R.

The Mirror Equation 
 Image location can be predicted by using the mirror equation.
1 1 1
 
p q f
1
1
1


object distance image distance focal length


The Mirror Equation 

1 1 1
 
p q f
1
1
1


Signs and
objectMeanings
distance image distance focal length
+ p: object is in front of mirror
+ q: image is real and in front of mirror
- q: image is virtual and is located behind mirror
+ R and +f: concave mirror
+h and +h’: object and image are above the principal axis
–h’: image is inverted and below the principal axis

The Equation for Magnification 
h' q
M –
h
p
image
distance
What does
thisheight
mean inimage
words?
magnification 
=–
object height object distance
 Signs and Meanings
+M: upright and virtual image
-M: inverted and real image

Ray Diagrams are drawings that use simple geometry to locate
an image formed by a mirror or lens. The image will be located at
the place where the rays intersect. Only two rays are necessary to
locate the image on a ray diagram, but it's useful to add the third
as a check.
1.
2.
Sketch the object and mirror and relative distance between them.
Draw rays
1. The parallel ray is drawn from the tip of the object parallel to the
principal axis. It then reflects off the mirror and either passes through
the focal point, or can be extended back to pass through the focal point.
2. The second ray (the chief ray) is drawn from the tip of the object to the
mirror through the center of curvature. This ray will hit the mirror at a
90° angle, reflecting back the way it came through the center. The
chief and parallel rays meet at the tip of the image.
3. The third ray, the focal ray, is a mirror image of the parallel ray. The focal
ray is drawn from the tip of the object through (or towards) the focal
point, reflecting off the mirror parallel to the principal axis. All three
rays should meet at the same point.

Ray Diagrams

Concave Mirror Image
 If the object is outside the focal length, a concave mirror
will form a real, inverted, and diminished (smaller)
image.
o=p
i=q
positive
in front of the mirror.

Concave Mirror Image
 If an object is placed inside the focal length of a concave
mirror, and enlarged virtual and erect image will be
formed behind the mirror.
o=p
i=q
positive
in front of the mirror.
1. A 1.50 m tall child is in a mirror gallery at the amusement park. She is
standing in front of a concave mirror with a radius of 4.00 m. She starts
walking toward the mirror from a distance of 9.00 m, and she stops every
meter to observe her image.
a. Find the focal point of this mirror and label it F.
b. Mark the child’s locations 9.00 m, 5.00 m, and 1.00 m in front of the
mirror, and label them A, B, C.
c. Sketch ray diagrams to locate the image formed when the child is at A.
Measure the distance from the image to the mirror and record it below.
Distance of A’s image =
d. Repeat question c for the object at positions B and C.
Distance of B’s image =
Distance of C’s image =
2. Calculate the image location for the object at A, B, and C in item 1, using
the mirror equation. Compare your results with your diagrams.
Distance of A’s image =
Distance of B’s image =
Distance of C’s image =
1. A 1.50 m tall child is in a mirror gallery at the amusement park. She is
standing in front of a concave mirror with a radius of 4.00 m. She starts walking
toward the mirror from a distance of 9.00 m, and she stops every meter to
observe her image.
a. Find the focal point of this mirror and label it F.
b. Mark the child’s locations 9.00 m, 5.00 m, and 1.00 m in front of the
mirror, and label them A, B, C.
A
B
F
C
1. A 1.50 m tall child is in a mirror gallery at the amusement park. She is
standing in front of a concave mirror with a radius of 4.00 m. She starts walking
toward the mirror from a distance of 9.00 m, and she stops every meter to
observe her image.
c. Sketch ray diagrams to locate the image formed when the child is at A.
Measure the distance from the image to the mirror and record it below.
Distance of A’s image = 2.5 m
Ray 1
Ray 3
Ray 2
A
F
1.
c. Sketch ray diagrams to locate the image formed.
d. Repeat question c for the object at positions B (5.0 0 m) .
Distance of B’s image = 3.3 m
Ray 1
Ray 3
B
F
Ray
2
1.
c. Sketch ray diagrams to locate the image formed.
d. Repeat question c for the object at positions C (1.0 0 m) .
Distance of C’s image = -1.8 m
Ray 1
Ray 2
Ray 3
F
C
o
i
2. Calculate the image location for the object at A (9.00 m), B (5.00 m), and
C (1.00 m) in item 1, using the mirror equation. Compare your results with your
diagrams.
Distance of A’s image = 2.57 m
1/f = 1/p + 1/q
f = focal point, p = object’s distance to mirror, q = image’s distance to mirror
f = 2.00 m ,
p = 9.00 m,
q=?
1/f – 1/p = 1/q
1/q = 1/f – 1/p
1/q = 1/2.00 m – 1/9.00 m
1/q = 0.500 m-1 – 0.111 m-1
1/q = 0.389 m-1
q = 1/0.389 m
= 2.57 m
2. Calculate the image location for the object at A (9.00 m), B (5.00 m), and
C (1.00 m) in item 1, using the mirror equation. Compare your results with your
diagrams.
Distance of B’s image = 3.33 m
1/f = 1/p + 1/q
f = focal point, p = object’s distance to mirror, q = image’s distance to mirror
f = 2.00 m ,
p = 5.00 m,
q=?
1/f – 1/p = 1/q
1/q = 1/f – 1/p
1/q = 1/2.00 m – 1/5.00 m
1/q = 0.500 m-1 – 0.200 m-1
1/q = 0.300 m-1
q = 1/0.300 m
= 3.33 m
2. Calculate the image location for the object at A (9.00 m), B (5.00 m), and
C (1.00 m) in item 1, using the mirror equation. Compare your results with your
diagrams.
Distance of C’s image = -2.00 m
1/f = 1/p + 1/q
f = focal point, p = object’s distance to mirror, q = image’s distance to mirror
f = 2.00 m ,
p = 1.00 m,
q=?
1/f – 1/p = 1/q
1/q = 1/f – 1/p
1/q = 1/2.00 m – 1/1.00 m
1/q = 0.500 m-1 – 1.00 m-1
1/q = - 0.500 m-1
q = - 1/0.500 m
= - 2.00 m

A convex spherical mirror is a mirror
whose reflecting surface is the
outward-curved segment of a
sphere.

Diverging mirror

Light rays diverge upon reflection
from a convex mirror, forming a
virtual image that is always smaller
than the object.

Image: virtual, upright, diminished

(-) f, (-) q, (+) p, (+)M < 1
A convex mirror forms a virtual image.

Convex or Diverging Mirrors
 A convex mirror forms a virtual image.

The Mirror Equation 
1 1 1
 
p q f
1
1
1


object distance image distance focal length
 Signs and Meanings for Convex Mirrors
+ p: object is in front of mirror
- q: image is virtual and is located behind mirror
- R and -f: convex mirror
+h and +h’: object and image are above the principal axis
–h’: image is inverted or below the principal axis
Chapter 13
Section 3 Curved Mirrors
Sample Problem
Convex Mirrors
An upright pencil is placed in front of a convex
spherical mirror with a focal length of 8.00 cm. An
erect image 2.50 cm tall is formed 4.44 cm behind
the mirror. Find the position of the object, the
magnification of the image, and the height of the
pencil.
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 13
Section 3 Curved Mirrors
Sample Problem, continued
Convex Mirrors
Given:
Because the mirror is convex, the focal length is
negative. The image is behind the mirror, so q is
also negative.
f = –8.00 cm q = –4.44 cm h’ = 2.50 cm
Unknown:
p=? h=?
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 13
Section 3 Curved Mirrors
Sample Problem, continued
Convex Mirrors
Diagram:
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 13
Section 3 Curved Mirrors
Sample Problem, continued
Convex Mirrors
2. Plan
Choose an equation or situation: Use the mirror
equation and the magnification formula.
1 1 1
h'
q
 
and
M –
p q f
h
p
Rearrange the equation to isolate the unknown:
1 1 1
 –
p f q
and
p
h  – h'
q
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 13
Section 3 Curved Mirrors
Sample Problem, continued
Convex Mirrors
3. Calculate
Substitute the values into the equation and solve:
1
1
1

–
p –8.00 cm –4.44 cm
1 –0.125 –0.225 0.100

–

p
cm
cm
cm
p  10.0 cm
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 13
Section 3 Curved Mirrors
Sample Problem, continued
Convex Mirrors
3. Calculate, continued
Substitute the values for p and q to find the magnification of the image.
q
–4.44 cm
M– –
M  0.444
p
10.0 cm
Substitute the values for p, q, and h’ to find the height
of the object.
p
10.0 cm
h  – h'  –
(2.50 cm)
q
–4.44 cm
h  5.63 cm
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Copyright © by Holt, Rinehart and Winston. All rights reserved.

Spherical Aberration and Parabolic Mirrors
 The blurring of an image that occurs when light from the margin of a
mirror or lens with a spherical surface comes to a shorter focus than
light from the central portion.

Color

Color
 The retina is lined with a variety of
light sensing cells known as rods and
cones. The rods on the retina are
sensitive to the intensity of light, but
cannot distinguish wavelengths
(colors). Three types of cone cells in
the eye allow humans to see in color.
 Light of different wavelengths
stimulate a combination of cone cells
so that a wide range of colors can be
perceived.
cyclohexane.tumblr.com

Color
 The color of an object depends on which wavelengths of light shine
on the object and which wavelengths are reflected.
Primary Additive Colors of light
produce white light when added.
Red
Orange
Yellow
Green
Cyan
Blue
violet
Red
Green
Blue
White

Color
 R +G
 G +B
 B+R
Y
C
M
• Additive primary colors produce white light when
combined.
•R +G + B
W
• Light of different colors can be produced by adding light
consisting of the primary additive colors (red, green, and blue).
• Adding different colors of light is used in color
television, color computer monitors, and on-stage
theater lighting.
•Complementary colors are two colors that add together to
produce white light. Two primary colors combine to produce
the complement of the third primary color.
C + R
W Cyan is the complementary of red.
Y +B
W Yellow is the complementary of blue.
M +G
W Magenta is the complementary of green.

Color
Filters are transparent materials which selectively absorb (or block) one
or more primary colors of light and allow the remaining colors of light to
pass through (or be transmitted). The color of the filter describes which
color of light is transmitted by the filter.
Opaque objects contain pigments that selectively absorb light and
reflect whatever light colors are not absorbed. The color of pigments
describes which color(s) of light are reflected.
The color of an object is the color of light the object reflects.

Color
 A pure pigment absorbs a single frequency of
light.
▪ Yellow absorbs blue.
▪ Magenta absorbs green.
▪ Cyan absorbs red.
 Three primary colors of paint (primary pigments):
yellow, cyan, and magenta
 An artist can create any color by using varying
amounts of the three primary colors of paint.
 Each primary color of paint absorbs one primary
color of light, its complementary color.

Color
 What pigment(s) (yellow, cyan, or magenta) would
you use to produce the following colors on white
canvas that is going to be illuminated with white
light?
 Yellow
yellow
 Red
 Green
yellow and magenta
yellow and cyan
 Blue
magenta and cyan
 Cyan
cyan
 Magenta
magenta
Chapter 13
Section 4 Color and Polarization

Blue Sky and Red Sunset
 The sky is blue because the atmosphere (N2 and O2) scatters blue light.
 Sunsets and sunrises are red because the sunlight passes through more
atmosphere and the longer wavelengths are not scattered as much by the
small atmospheric particles.
 White clouds?
 Brown smog?
When a light wave vibrates in a variety of
directions, the light is said to be unpolarized.
When a light wave's are isolated to a single
plane, the light is said to be polarized.
A Polaroid filter polarizes light by blocking part
of the vibrations while letting through those that
are in a specific plane
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