rty
nt
me
ern
v
Go
pe
Pro
R
NO
O
TF
L
SA
E
Senior High School
NOT
General Physics 2
Quarter 4 - Module 3
Geometric Optics
Department of Education ● Republic of the Philippines
General Physics 1 – Grade 12
Alternative Delivery Mode
Quarter 4 – Module 3: Geometric Optics
First Edition, 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any
work of the Government of the Philippines. However, prior approval of the
government agency or office wherein the work is created shall be necessary for
exploitation of such work for profit. Such agency or office may, among other things,
impose as a condition the payment of royalty.
Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand
names, trademarks, etc.) included in this book are owned by their respective
copyright holders. Every effort has been exerted to locate and seek permission to
use these materials from their respective copyright owners. The publisher and
authors do not represent nor claim ownership over them.
Published by the Department of Education – Division of Cagayan de Oro
Schools Division Superintendent: Dr. Cherry Mae L. Limbaco, CESO V
Development Team of the Module
Author/s:
Gina Lamparas
Reviewers:
Krista Marie Llenas
Illustrator and Layout Artist:
Arian M. Edullantes
Management Team
Chairperson:
Cherry Mae L. Limbaco, PhD, CESO V
Schools Division Superintendent
Co-Chairpersons:
Rowena H. Para-on, Ph.D.
Asst. Schools Division Superintendent
Members
Lorebina C. Carrasco, OIC-CID Chief
Jean S. Macasero, EPS - Science
Joel D. Potane, LRMS Manager
Lanie O. Signo, Librarian II
Gemma Pajayon, PDO II
Printed in the Philippines by
Department of Education – Bureau of Learning Resources (DepEd-BLR)
Office Address :
Fr. William F. Masterson Ave Upper Balulang, Cagayan de Oro
Telefax
:
(08822)855-0048
E-mail Address :
cagayandeoro.city@deped.gov.ph
Senior
High
School
Senior
High
School
General Physics 2
Quarter 4 - Module 3:
Geometric Optics
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities.
We encourage teachers and other education stakeholders to email their
feedback, comments, and recommendations to the Department of Education
at cagayandeoro.city@ deped.gov.ph.
We value your feedback and recommendations.
FAIR USE AND CONTENT DISCLAIMER: This SLM (Self Learning Module) is for
educational purposes only. Borrowed materials (i.e. songs, stories, poems, pictures, photos,
brand names, trademarks, etc.) included in these modules are owned by their respective
copyright holders. The publisher and authors do not represent nor claim ownership over
them.
Department of Education ● Republic of the Philippines
Table of Contents
What This Module is About
What I Need to Know
How to Learn from this Module
Icons of this Module
What I Know
i
i
i
ii
.iii
Lesson 1: Geometric Optics
What’s In
What I Need to Know
What’s New:
What Is It
What’s More:
What I Have Learned:
What I Can Do:
1
1
1
2
4
5
8
Lesson 2: Mirrors
What I Need to Know
What’s In
What’s New:
What Is It
What’s More:
What I Have Learned:
What I Can Do:
16
16
17
17
23
26
29
Assessment: (Post-Test)
38
Key to Answers
42
Module 3
Geometric Optics
What This Module is About
This module provides you with scientific knowledge and skills about
Geometric Optics. Geometric Optics is a model of optics that describes light
propagation in terms of rays. The ray in geometric optics is an abstraction useful for
approximating the paths along which light propagates under certain circumstances.
The lessons in this module are necessary in studying other concepts of motion in the
next modules.
The following are the lessons contained in this module:


Lesson 1- Geometric Optics
Lesson 2- Mirrors
What I Need to Know
After going through this module, you are expected to:
1. Explain image formation as an application of reflection,
refraction, and paraxial approximation
2. Relate properties of mirrors and lenses (radii of curvature,
focal length, index of refraction [for lenses]) to image and
object distance and sizes
3. Determine graphically and mathematically the type
(virtual/real), magnification, location, and orientation of
image of a point and extended object produced by a plane
or spherical mirror
4. Determine graphically and mathematically the type
(virtual/real), magnification, location/ apparent depth,
and orientation of image of a point and extended object
produced by a lens or series of lenses
5. Apply the principles of geometric optics to discuss image
formation by the eye, and correction of common vision
defects
STEM_GP12OPTIVd-22
STEM_GP12OPTIVd-23
STEM_GP12OPTIVd-24
STEM_GP12OPTIVd-27
STEM_GP12OPTIVd-28
How to Learn from this Module
To achieve the learning competencies cited above, you are to do the following:
•
Take your time reading the lessons carefully.
•
Follow the directions and/or instructions in the activities and exercises diligently.
•
Answer all the given tests and exercises.
5
Icons of this Module
6
What I Know
I.
DIRECTION: Choose the best answer from the choices given. Write the letter
only of your answer.
1. Find the focal length of the mirror that forms an image 6.2cm behind the mirror of an
object placed at 26 cm in front of the mirror.
a. -6.58
b.-8.41 cm
c.-7.8 cm
d.-9.6cm
2. The lens with a long focal length is
a. Not strongly curved only
c. Thick
b. Thin only
d. Not strongly curved and thin
3. The variation of focal length to form a sharp image on retina is called
a. Accommodation
b. Aperture
c. Retina Control
d. Sutter
4. In concave mirror, the size of the image depends upon
a. Size of the object
c. Area covered by an object
b. b. Position of object
d. Shape of object
5. When light travelling in a certain medium fall on the surface of another medium, a
part of it turns back in the same medium. This phenomenon is called
a. Reflection
b. Refraction
c. Diffraction
d. Acoustics
6. If the object towards the right side of the lens, the object distance will be
a. Zero
b. Constant
c. Positive
d. Negative
7. The power of magnifying glass given by
a. f + p
b. 1 + d/f
c. d + f
d. 1 + fd
8. In astronomical compare to eye piece, objective lens has
a. Negative focal length
c. Small focal length
b. Zero focal length
d. Large local length
9. The inability of eye to see the objects clearly is called
a. Clarity of image b. Defect of vision
c. Blur image
d. Small image
10. The size of the image is always smaller than the object in
a. Convex Mirror
b. Concave Mirror
c. Silver Mirror
d. Plane Mirror
11. To describe the change in speed of light in a medium, the term used is called index
of
a. Reflection
b. Refraction
c. Diffraction
d. Acoustics
12. The power of lens is
a. 1/p
b. 1/q
c. 1/f
d. 1/l
13. When a ray of light enters from a denser medium to rare medium it bends
a. Towards normal
c. Perpendicular to normal
b. b. Away from normal
d. Parallel to normal
14. The critical angle of water when refracted angle is 900 and refractive index of water
and air is 1.33 and is
a. 48.80
b. 49.10
c. 500
d. 510
15. A convex mirror is used to reflect light from an object placed 30com in front of the
mirror. If the focal length of the mirror is 20 cm then the location of the image should
be.
a. -27 cm
b. -37cm
c.-29cm
d.-47cm
7
Lesson
1
Geometric Optics
What’s In
The study of light is called optics. This area of physics dates to at least third
century BC. Eyeglasses were invented around AD 1300, and microscopes and
telescopes were first developed around 1600. These applications are based on the
ability of lenses and mirrors to focus light. Light is an electromagnetic wave, and our
discussion of optics in this lesson will need to account for the wave nature of light
concepts such as wave fronts and rays.
What I Need to Know
At the end of the module you will be able to:
1. Determine graphically and mathematically the type (virtual/real ) ,
magnification, location and orientation of image of a point and extended object
produced by a plane or spherical mirror
What’s New
In this lesson, we consider geometric optics, the regime in which light
travels in straight line paths and effects involving wave interference are
not important. In general, geometric optics describe cases in which the
wavelength of light is much smaller than the size of objects in the light’s path.
The wavelength of visible is less than 1µm (10-6 m) and is about one hundred
times smaller than the diameter of a human hair. Most object we work in
everyday life are much larger than that, so geometrical optics describes many
everyday applications of optics including the behavior of lenses and mirror.
RAY (GEOMETRICAL) OPTICS
8
Figure 1 rays of light passing through an opening
and onto a screen. A. A large opening casts an ordinary
shadow on the screen. This is the regime geometrical is
also called ray optics. B. If the size of an opening is
comparable to or smaller than the wavelength ‫ת‬, the light
spread out as it passes through the opening.
Figure 1 shows a light wave as it passes through
an opening that is large compared to the wavelength of
light. This figure also shows rays, indicating the path and
direction of propagation. Rays indicate the path followed
by a wave. For convenience, we will often say that light
ray “travels” or “propagate” from place to another. In
figure 1.1 A a rays pass through a wide opening and
make an ordinary shadow on the blocked regions of the
screen to a very good approximation, these rays follow
straight line that pass through the opening. If, however,
the width of the opening is made very small figure 1.1 B,
one would now observe that light “spreads out” after
passing through the opening. This spreading is due to the
nature of light.
CHECK YOUR UNDERSTANDING # 1
Instruction: Organize a collection of photographs from ocean around the world
during various season. Observe the color of the ocean water. Briefly answer
the following questions.
1.
Why does the ocean appear different colors?
2.
How does color affect water?
What Is It
WHAT IS WAVE FRONT?
Wave front surfaces are determined by the crest and trough of the waves and are
always perpendicular to the associated rays. The shape of a wave front thus depend on how
the wave is generated and the distance from the source. Wave front can be simple plane or
spherical. When a point source is very far away, the wave front from very large spheres,
which appear nearly flat to an observer and can be used to form approximate plane waves.
Although the behavior of light rays and wave fronts can be derived mathematically
from MAXWELL EQUATIONS.
Properties of light according to MAXWELL
1. The motion of light along a light ray is reversible
2. The perpendicular distance between two wave fronts is proportional to the speed of light
because of the way wave fronts are related to the crests and trough of a wave.
9
CHECK YOUR UNDERSTANDING #2
Instruction:
Describe the illustration based on the properties of light of MAXWELL
10
What’s More
FORMING AN IMAGE: RAY TRACING
Light that emanates from an object is used by
your eye to form an image of the object. Figure 3 shows
how rays originating from two different points on an
object reach the eye. When the eye combines these rays
to form an image, your brain extrapolates the rays back
to their point of origin. The method of following individual
rays as they travel from an object to your eye to some
other point is called ray tracing. Ray tracing involves the
extensive use of geometry; hence ray optics is also
called “geometrical optics”.
To understand how images are formed, we must
consider (1) what happens to light rays when they reflect
from a surface such as a mirror or a piece of glass and
(2) what happens to light rays when they pass across a surface separating one material from
another such as when they pass from air into a piece of glass.
In each case, we must distinguish between a flat surface and a curved surface.
CHECK YOUR UNDERSTANDING #3
Instruction: Choose the letter only and explain your answer.
11
What I Have Learned
RAY TRACING AND PROBLEM SOLVING:
We learned about mathematical equation relating the two angles (angles of incidence
and refraction) and the indices of refraction of the two material on each side of the boundary.
The equation is known as the Snell's Law equation and is expressed as follows.
ni • sine (Θi) = nr * sine (Θr)
where Θi ("theta i") = angle of incidence
Θr ("theta r") = angle of refraction
ni = index of refraction of the incident medium
nr = index of refraction of the refractive medium
As with any equation in physics, the Snell's Law equation is valued for its predictive ability. If
any three of the four variables in the equation are known, the fourth variable can be
predicted if appropriate problem-solving skills are employed. In this part of Lesson 2, we will
investigate several of the types of problems that you will have to solve, and learn the task of
tracing the refracted ray if given the incident ray and the indices of refraction.
Example Problem A
A ray of light in air is approaching the boundary with water at an angle of 52 degrees.
Determine the angle of refraction of the light ray. Refer to the table of indices of refraction if
necessary.
Solution to Problem A
The solution to this problem begins like any problem: a diagram is constructed to assist in
the visualization of the physical situation, the known values are listed, and the unknown
value (desired quantity) is identified. This is shown below:
Diagram:
Given:
Find:
ni = 1.00 (from table)
nr = 1.333 (from table)
Θr =??
Θi = 52 degrees
SOLUTION:
ni • sine (Θi) = nr * sine (Θr)
1.00 * sine (52 degrees) = 1.333 * sine (Θr)
0.7880 = 1.333 * sine (Θr)
0.591 = sine (Θr)
sine-1 (0.591) = sine-1 ( sine (Θr))
36.2 degrees = Θr
Proper algebra yields the answer of 36.2 degrees for the angle of refraction. In this problem,
the light ray is traveling from a less optically dense or fast medium (air) into a more optically
12
dense or slow medium (water), and so the light ray should refract towards the normal - FST.
Thus, the angle of refraction should be smaller than the angle of refraction. And indeed it is 36.2 degrees (theta r) is smaller than 52.0 degrees (theta i).
Example Problem B
A ray of light in air is approaching the boundary with a layer of crown glass at an angle of
42.0 degrees. Determine the angle of refraction of the light ray upon entering the crown
glass and upon leaving the crown glass. Refer to the table of indices of refraction if
necessary.
Solution to Problem B
This problem is slightly more complicated than Problem A since refraction is taking place at
two boundaries. This is an example of a layer problem where the light refracts upon
entering the layer (boundary #1: air to crown glass) and again upon leaving the layer
(boundary #2: crown glass to air). Despite this complication, the solution begins like the
above problem: a diagram is constructed to assist in the visualization of the physical
situation, the known values are listed, and the unknown value (desired quantity) is identified.
This is shown below:
Given:
Boundary #1
ni = 1.00 (from table)
nr = 1.52 (from table)
Θi = 42.0 degrees
Note that the angle of
refraction at boundary #1
is the same as the angle of
incidence at boundary #2.
Boundary #2
ni = 1.52 (from table)
nr = 1.00 (from table)
Find:
Θr at
boundary #1 and
#2
SOLUTION:
Boundary #1:
ni • sine (Θi) = nr * sine (Θr)
1.00 * sine (42.0 degrees) = 1.52 * sine (Θr)
0.669 = 1.52 * sine (Θr)
0.4402 = sine (Θr)
sine-1 (0.4402) = sine-1 ( sine (Θr))
26.1 degrees = Θr
13
Θr at
boundary
The value of 26.1 degrees corresponds to the angle of refraction at boundary #1. Since
boundary #1 is parallel to boundary #2, the angle of refraction at boundary #1 will be the
same as the angle of incidence at boundary #2 (see diagram above). So now repeat the
process to solve for the angle of refraction at boundary 2.
Boundary #2:
ni • sine (Θi) = nr * sine (Θr)
1.52 * sine (26.1 degrees) = 1.00 * sine (Θr)
1.52 * (0.4402) = 1.00 * sine (Θr)
0.6691 = sine (Θr)
sine-1 (0.6691) = sine-1 ( sine (Θr)
42.0 degrees = Θr
The answers to this problem are 26.1 degrees and 42.0 degrees.
There is an important conceptual idea that is found from an inspection of the above answer.
The ray of light approached the top surface of the layer at 42 degrees and exited through the
bottom surface of the layer with the same angle of 42 degrees. The light ray refracted one
direction upon entering and the other direction upon exiting; the two individual effects have
balanced each other, and the ray is moving in the same direction. The important concept is
this:
When light approaches a layer that has the shape of a parallelogram that is bounded on
both sides by the same material, then the angle at which the light enters the material is equal
to the angle at which light exits the layer.
If the layer is not a parallelogram or is not bound on both sides by the same material, then
this will not be the case. Knowing this concept will allow you to conduct a quick check of an
answer in a situation like this.
CHECK YOUR UNDERSTANDING #4
1.Perform the necessary calculations at each boundary to trace the path of the light ray
through the following series of layers. Use a protractor and a ruler and show all your work.
14
What I Can Do
IMAGE FORMATION BY A PLANE MIRROR
When you view an object through its reflection in a mirror, you are viewing an image
of the object. The image formed by a plane mirror is always virtual (meaning that the light
rays do not actually come from the image), upright, and of the same shape and size as the
object
it
is
reflecting.
A
virtual image is
a
copy
of
an
object formed at
the
location from which the light rays appear to come.
Observer’s eye
Extrapolate
rays back to
apparent
point of orgin
object
image
Figure 4
In figure 4 an infinite number of light rays emanate from each point on the object,
although we only show two rays emanating from the top and bottom. Some of the emanated
rays strike the mirror and are reflected to reach to the eye of the observer. To the observer’s
eye the location of the image- that is the location of the arrow- is the point from which these
rays appear to emanate. This point can be found by extrapolating the rays back in the
15
straight line to their apparent point of origin as shown in the dash line. This is an example of
ray tracing. Since each ray obeys the law of reflection.
In figure 4 also shows that the size of an image formed by a plane mirror is equal to
the size of an image formed by a plane mirror is equal to the size of the object; that is, the
height of the image hi is equal to the height of the object ho. Also, the image point is located
behind the mirror, so light does not actually pass through the image. For this reason, the
image is called a virtual image.
IMAGE CHARACTERISTICS:
APPARENT LEFT-RIGHT IMAGE REVERSAL
Besides the fact that plane mirror images are virtual, there are several other
characteristics that are worth noting. The second characteristic has to do with the orientation
of the image. If you view an image of yourself in a plane mirror (perhaps a bathroom mirror),
you will quickly notice that there is an apparent left-right reversal of the image. That is, if you
raise your left hand, you will notice that the image raises what would seem to be its right
hand. If you raise your right hand, the image raises what would seem to be its left hand. This
is often termed left-right reversal. This characteristic becomes even more obvious if you
wear a shirt with lettering. For example, a shirt displaying the word "NIKE" will read "EKIN"
when viewed in the mirror; a shirt displaying the word "ILLINOIS" will read "SIONILLI;" and a
shirt displaying the word "BOB" will read "BOB." (NOTE: Not only will the order of letters
appear reversed, the actual orientation of the letters themselves will appear reversed as
well. Of course, this is a little difficult to do when typing from a keyboard.) While there is an
apparent left-right reversal of the orientation of the image, there is no top-bottom vertical
reversal. If you stand on your feet in front of a plane mirror, the image does not stand on its
head. Similarly, the ceiling does not become the floor. The image is said to be upright, as
opposed to inverted.
To further explore the reason for this appearance of left-right reversal, let's suppose
we write the word PHYSICS on a transparency and hold it in front of us in front of a plane
mirror. If we look at the image of the transparency in the mirror, we will observe the expected
- SCISYHP. The letters are written reversed when viewed in the mirror. But what if we look
at the letters on the transparency? How are those letters oriented? When we face the mirror
and look at the letters on the transparency, we observe the unexpected - SCISYHP. When
viewed from the perspective of the person holding the transparency (and facing the mirror,
the letters exhibit the same left-right reversal as the mirror image. The letters appear
reversed on the image because they are actually reversed on the shirt. At least they are
reversed when viewed from the perspective of a person who is facing the mirror. Imagine
that! All this time you thought the mirror was reversing the letters on your shirt. But the fact is
that the letters were already reversed on your shirt; at least they were reversed from the
person who stands behind the T-shirt. The people who view your shirt from the front have a
different reference frame and thus do not see the letters as being reversed. The apparent
left-right reversal of an image is simply a frame of reference phenomenon. When viewing the
image of your shirt in a plane mirror (or any part of the world), you are viewing your shirt
from the front. This is a switch of reference frames.
16
OBJECT DISTANCE AND
IMAGE DISTANCE
Another characteristic of plane
mirror image pertains to the
relationship between the object's
distance to the mirror and the
image's distance to the mirror.
For plane mirrors, the object
distance (often represented by the symbol do) is equal to the image distance (often
represented by the symbol di). That is the image is the same distance behind the mirror as
the object is in front of the mirror. If you stand a distance of 2 meters from a plane mirror,
you must focus on a location 2 meters behind the mirror in order to view your image.
RELATIVE SIZE OF IMAGE AND OBJECT
Plane mirror images is that the dimensions of the image are the same as the dimensions of
the object. If a 1.6-meter tall person stands in front of a mirror, he/she will see an image that
is 1.6-meters tall. If a penny with a diameter of 18-mm is placed in front of a plane mirror, the
image of the penny has a diameter of 18 mm. The ratio of the image dimensions to the
object dimensions is termed the magnification. Plane mirrors produce images that have a
magnification of 1.
In conclusion, plane mirrors produce images with a number of distinguishable
characteristics. Images formed by plane mirrors are virtual, upright, left-right reversed,
the same distance from the mirror as the object's distance, and the same size as the
object.
CHECK YOUR UNDERSTANDING #5
1. Arrange the jumble word “ECNALUBMA”
2. If Suzie stands 3 feet in front of a plane mirror, how far from the person will her image
be located?
3. If a toddler crawls towards a mirror at a rate of 0.25 m/s, then at what speed will the
toddler and the toddler's image approach each other?
What I Still Want To Know
RAY DIAGRAMS
The line of sight principle suggests that in order to view an image of an object in a
mirror, a person must sight along a line at the image of the object. When sighting along such
a line, light from the object reflects off the mirror according to the law of reflection and travels
to the person's eye. This process was discussed and explained earlier in this lesson. One
useful tool that is frequently used to depict this idea is known as a ray diagram. A ray
diagram is a diagram that traces the path that light takes in order for a person to view a
point on the image of an object. On the diagram, rays (lines with arrows) are drawn for the
incident ray and the reflected ray. Complex objects such as people are often represented by
17
stick figures or arrows. In such cases it is customary to draw rays for the extreme positions
of such objects.
DRAWING RAY DIAGRAMS-A STEP-BY-STEP APPROACH
This section illustrates the procedure for drawing ray diagrams. Let's begin with the task of
drawing a ray diagram to show how Sam will be able to see the image of the green object
arrow in the diagram below. For simplicity sake, we will suppose that Sam is viewing the
image with her left eye closed. Thus, we will focus on how light travels from the two
extremities of the object arrow (the left and right side) to the mirror and finally to Sam right
eye as she sights at the image. The four steps of the process for drawing a ray diagram are
listed, described and illustrated below.
1. Draw the image of the object.
Use the principle that the object distance is equal to the image
distance to determine the exact location of the object. Pick
one extreme on the object and carefully measure the distance
from this extreme point to the mirror. Mark off the same distance
on the opposite side of the mirror and mark the image of
this extreme point. Repeat this process for all extremes on the
object until you have determined the complete location and shape
of the image. Note that all distance measurements should be
made by measuring along a segment that is perpendicular to the mirror.
2. Pick one extreme on the image of the object and draw the reflected ray that will travel to
the eye as it sights at this point.
Use the line of sight principle: the eye must sight along a line at
the image of the object in order to see the image of the object. It is
customary to draw a bold line for the reflected ray (from the mirror
to the eye) and a dashed line as an extension of this reflected ray;
the dashed line extends behind the mirror to the location of the
image point. The reflected ray should have an arrowhead upon it
to indicate the direction that the light is traveling. The arrowhead
should be pointing towards the eye since the light is traveling from
the mirror to the eye, thus enabling the eye to see the image.
4. Draw the incident ray for light traveling from the corresponding extreme on the object
to the mirror.
The incident ray reflects at the mirror's surface according to
the law of reflection. But rather than measuring angles, you can
merely draw the incident ray from the extreme of the object to the
point of incidence on the mirror's surface. Since you drew the
reflected ray in step 2, the point of incidence has already been
determined; the point of incidence is merely the point where the
line of sight intersects the mirror's surface. Thus draw the incident
ray from the extreme point to the point of incidence. Once more,
be sure to draw an arrowhead upon the ray to indicate its direction of travel. The arrowhead
should be pointing towards the mirror since light travels from the object to the mirror.
4. Repeat steps 2 and 3 for all other extremities on the object.
18
After completing steps 2 and 3, you have only shown how light travels from a
single extreme on the object to the mirror and finally to the eye. You will also have to show
how light travels from the other extremes on the object to the eye. This is merely a matter of
repeating steps 2 and 3 for each individual extreme. Once repeated for each extreme, your
ray diagram is complete.
EXAMPLE
Suppose that six students - Al, Bo, Cy, Di, Ed, and Fred sit in front of a plane mirror
and attempt to see each other in the mirror. And suppose the exercise involves answering
the following questions: Whom can Al see? Whom can Bo see? Whom can Cy see? Whom
can Di see? Whom can Ed see? And whom can Fred see?
The task begins by locating the images of the given students. Then, Al is isolated from the
rest of the students and lines of sight are drawn to see who Al can see. The leftward-most
student whom Al can see is the student whose image is to the right of the line of sight that
intersects the left edge of the mirror. This would be Ed. The rightward-most student who Al
can see is the student whose image is to the left of the line of sight that intersects the right
edge of the mirror. This would be Fred. Al could see any student positioned between Ed and
Fred by looking at any other positions along the mirror. However in this case, there are no
other students between Ed and Fred; thus, Ed and Fred are the only students whom Al can
see? The diagram below illustrates this using lines of sight for Al.
SOLUTION:
Bo can see Di, Ed, and Fred
Cy can see Cy, Di, Ed, and Fred
Di can see Cy, Di, Ed, and Fred
Ed can see Bo, Cy, Di, and Ed
Fred can see Al, Bo, Cy, and Di
CHECK YOUR UNDERSTANDING # 6
1.Six students are arranged in front of a mirror. Their positions are shown below.
Whom can Al see? Whom can Bo see? Whom can Cy see? Whom can Di see? Whom can
Ed see? And whom can Fred see?
19
What is Required
The use of ray diagrams were introduced and illustrated. Ray diagrams can be used
to determine where a person must sight along a mirror in
order to see an image of him/herself. As such, ray diagrams
can be used to determine what portion of a plane mirror must
be used in order to view an image. The diagram below depicts
a 6-foot tall man standing in front of a plane mirror. To see the
image of his feet, he must sight along a line towards his feet;
and to see the image of the top of his head, he must sight
along a line towards the top of his head. The ray diagram
depicts these lines of sight and the complete path of light from
his extremities to the mirror and to the eye. In order to view his image, the man must look as
low as point Y (to see his feet) and as high as point X (to see the tip of his head). The man
only needs the portion of mirror extending between points X and Y in order to view his entire
image. All other portions of the mirror are useless to the task of this man viewing his own
image.
The diagram depicts some important information about plane mirrors. Using a cmruler, measure the height of the man (the vertical arrow) on the computer screen and
measure the distance between points X and Y. What do you notice? The man is twice as tall
as the distance between points X and Y. In other words, to view an image of yourself in a
plane mirror, you will need an amount of mirror equal to one-half of your height. A 6-foot tall
man needs 3-feet of mirror (positioned properly) in order to view his entire image.
WHAT IS THE EFFECT OF VARYING OBJECT DISTANCE?
But what if the man stood a different distance from the mirror? Wouldn't that cause
the man to need a different amount of mirror to view his image? Maybe less mirror would be
required in such an instance? These questions can be explored with the help of another ray
diagram. The diagram below depicts a man standing different distances from a plane mirror.
Ray diagrams for each situation (standing close and standing far away) are drawn. To assist
in distinguishing between the two ray diagrams, they have been color-coded. Red and blue
light rays have been used for the situation in which the man is standing far away. Green and
purple light rays have been used for the situation in which the man is standing close to the
mirror.
20
The two ray diagrams above demonstrate that the distance that a person stands from the
mirror will not affect the amount of mirror that the person needs to see their image. Indeed in
the diagram, the man's line of sight crosses the mirror at the same locations. A 6-foot tall
man needs 3-feet of mirror to view his whole image regardless of where he is standing. In
fact, the man needs the exact same 3-feet of mirror.
A common Physics lab involves using a tall plane mirror to explore the relationship between
object height and the portion of mirror needed to view an image. A student stands a few
meters from a plane mirror and views her image. With the student standing upright and still
and staring at her feet, the lab partner moves a marker up and down the mirror until the sight
location on the mirror is identified. The partner then marks this location on the mirror with an
erasable marker. The process is repeated for the student staring at the tip of her head. Of
course, being a lab, the procedure is subject to a variety of procedural and measurement
error that may yield less than ideal results. The mirrors are occasionally mounted on a wall
that is not perfectly vertical. Or a student will lean forward a slight amount, thus reducing
his/her effective height. Or the mirror warps over the years leading to one that is concave or
convex rather than planar. Despite these potential complications, the 1:2 ratio between the
amount of mirror required to view the image and the height of the object is often observed.
CHECK YOUR UNDERSTANDING #7
Sam Azucena is 6-feet tall. He is the tallest person in his family. It just so happens
that Sam learned the important principle of the 2:1 relationship just prior to his family's
decision to purchase a mirror that was to be used by the entire family. Sam decided to put it
to good use. Sam convinced his parents that it would be a waste of money to buy a mirror
longer than 3 feet. "After all," Sam argued, "I'm the tallest person in the family and only three
feet of mirror would be required to view my image." Sam's parents conceded and they
purchased a 3-foot tall mirror and mounted it on the bathroom wall.
Comment on the wisdom behind the Azucena family decision.
How much did I learn?
CHECK YOUR UNDERSTANDING #8
Reflection of Light With Two Plane Mirrors—Double Mirrors Placed at a 90-Degree
Angle
Objectives:
The student will experiment with reflections of two plane mirrors placed at a 90degree angle to see what will be reflected.
21
Theory:
When you place two plane mirrors at a 90-degree angle, the image of the first mirror
is reflected in the second mirror so that the reversed mirror image is reversed again, and you
see a true image. The placement of images in the mirror will vary with the distance of the
person or object in front of the mirror
Materials:




1 protractor
2 plane mirror tiles 12 inches square (These mirrors should be backed with heavy
cardboard and sealed around the edges with thick tape. The mirrors should then be
taped together to form two to four hinges. You now have framed mirrors that can
stand alone.)
Cardboard
Tape
Procedures:
1. Place the mirrors at a 90-degree angle.
2. Place yourself in front of the mirrors.
3. Look into the mirror and follow the instructions.
All instructions should be followed while looking into the mirror, not at your body.
A. Raise the right hand that you see in the mirror.
B. Turn your head to the left.
C. Touch your right ear with your left hand.
D. Look into the mirror and wink your left eye.
E. Raise both hands with your palms facing the mirror.
F. Touch one little finger to the thumb on the other hand.
G. Bring both hands together until your fingers touch.
H. Raise the left hand with the palm facing the mirror and the right hand with the
palm turned away from the mirror.
I. Touch your right shoulder with your left hand.
J. Choose a partner and give five instructions of your own.
Observation, Data and Conclusion:
1. What did you observe during this activity?
2. What information did your eyes give you?
3. Why was this activity difficult?
4. What characteristic of light did this activity use or demonstrate?
22
Lesson
2
Mirrors
What I Need to Know
At the end of the module you will be able to:
a. Determine graphically and mathematically the type (virtual/real ) ,
magnification, location and orientation of image of a point and extended object
produced by a plane or spherical mirror
What I Know
DIRECTION: Choose the best answer from the choices given. Write the letter
only of your answer.
1. In concave mirror, the size of image depends upon
a. size of object
b. position of object c. area covered by object d. shape of object
2. The focus lies behind the mirror in
a. convex mirror b. concave mirror
c. silver mirror
d. plane mirror
3. The size of the image is always smaller than the object in
a. convex mirror b. concave mirror
c. silver mirror
d. plane mirror
4. [Radius of curvature (R)⁄2] is equal to
a. focal length(f) b. center of curvature c. vertex
d. pole
5. After reflection from a concave mirror, rays of light parallel to the principal axis converge
to a point which is called
a. Pole
b. center of curvature
c. focal length
d. principal focus
6. Mirrors having a curved reflecting surface are called as
a. plane mirror
b. spherical mirrors c. simple mirror
d. none of the above
7. How many types of spherical mirrors?
a. 2
b. 4
c. 5
d.3
8. Spherical mirror with reflecting surface curved inwards is called ……………
a. convex mirror b. concave mirror
c. curved mirror
d. none of the above
9. Centre of curvature is not a part of spherical mirror rather it lies _______the mirror
a. Boundary
b. inside
c. outside
d. none of the above
10. Pole lies on the surface of
a. spherical mirrors
b. simple mirror c. plane mirror
d. none of the above
11. Spherical mirror with reflecting surface curved outwards is called ……
a. spherical mirror
b. curved mirror c. convex mirror d. none of the above
12. The center of a sphere of which the reflecting surface of a spherical mirror is a part is
called ……
a. Pole
b. centre of curvature c. Radius of Curvature
d. Aperture
13. In the case of concave mirror centre of curvature lies in ………… of the reflecting surface
a. Boundary
b. inside
c. outside
d. front
14. The radius of a sphere; of which the reflecting surface of a spherical mirror is a part; is
called the………….
a. centre of curvature
b. The radius of Curvature
c. Poled
d. Aperture
15. The diameter of the reflecting surface of a spherical mirror is called ……
a. centre of curvature
b. The radius of Curvature
c. Poled
d. Aperture
What’s In
23
The image in a plane mirror has the same size as the object, is upright, and is the
same distance behind the mirror as the object is in front of the mirror. A curved mirror, on the
other hand, can form images that may be larger or smaller than the object and may form
either in front of the mirror or behind it. In general, any curved surface will form an image,
although some images make be so distorted as to be unrecognizable cause curved mirrors
can create such a rich variety of images, they are used in many optical devices that find
many uses. We will concentrate on spherical mirrors for the most part because they are
easier to manufacture than mirrors such as parabolic mirrors and so are more common.
What’s New
Curved Mirrors
We can define two general types of spherical mirrors. If the reflecting surface is the outer
side of the sphere, the mirror is called a convex mirror. If the inside surface is the reflecting
surface, it is called a concave mirror.
Symmetry is one of the major hallmarks of many optical devices, including mirrors and
lenses. The symmetry axis of such optical elements is often called the principal axis or
optical axis. For a spherical mirror, the optical axis passes through the mirror’s center of
curvature and the mirror’s vertex, as shown in Figure 1.
Figure 1
A spherical mirror is formed by cutting out a piece of a sphere and silvering either the
inside or outside surface. A concave mirror has silvering on the interior surface (think
“cave”), and a convex mirror has silvering on the exterior surface.
Consider rays that are parallel to the optical axis of a parabolic mirror, as shown in Figure
2a. Following the law of reflection, these rays are reflected so that they converge at a point,
called the focal point. Figure 2b shows a spherical mirror that is large compared with its
radius of curvature. For this mirror, the reflected rays do not cross at the same point, so the
mirror does not have a well-defined focal point. This is called spherical aberration and
results in a blurred image of an extended object. Figure 2c shows a spherical mirror that is
24
small compared to its radius of curvature. This mirror is a good approximation of a parabolic
mirror, so rays that arrive parallel to the optical axis are reflected to a well-defined focal
point. The distance along the optical axis from the mirror to the focal point is called the focal
length of the mirror.
Figure 2
CHECK YOUR UNDERSTANDING # 1
1. What happens when a ray parallel to the principal axis passes through a convex lens
Explain the illustration below in brief and concise.
A convex spherical mirror also has a focal point, as shown in Figure 3. Incident rays
parallel to the optical axis are reflected from the mirror and seem to originate from point F at
focal length f behind the mirror. Thus, the focal point is virtual because no real rays actually
pass through it; they only appear to originate from it.
25
Figure 3
Figure 3: (a) Rays reflected by a convex spherical mirror: Incident rays of light
parallel to the optical axis are reflected from a convex spherical mirror and seem to originate
from a well-defined focal point at focal distance f on the opposite side of the mirror. The focal
point is virtual because no real rays pass through it. (b) Photograph of a virtual image formed
by a convex mirror. (credit by: modification of work by Jenny Downing)
CHECK YOUR UNDERSTANDING # 2
1. How does the focal length of a mirror below
curvature?
Using Ray Tracing to
relate to the mirror’s radius of
Locate Images
To
find
the
location of an
image formed by a
spherical mirror,
we first use ray tracing,
which is the
technique of drawing
rays and using
the law of reflection to
determine
the
reflected rays (later, for
lenses, we use
the law of refraction to
determine
refracted rays). Combined with some basic geometry, we can use ray tracing to find the focal
point, the image location, and other information about how a mirror manipulates light. In fact,
we already used ray tracing above to locate the focal point of spherical mirrors, or the image
distance of flat mirrors. To locate the image of an object, you must locate at least two points
of the image. Locating each point requires drawing at least two rays from a point on the
object and constructing their reflected rays. The point at which the reflected rays intersect,
either in real space or in virtual space, is where the corresponding point of the image is
located. To make ray tracing easier, we concentrate on four “principal” rays whose
reflections are easy to construct.
Figure 4 shows a concave mirror and a convex mirror, each with an arrow-shaped object in
front of it. These are the objects whose images we want to locate by ray tracing. To do so,
we draw rays from point Q that is on the object but not on the optical axis. We choose to
draw our ray from the tip of the object. Principal ray 1 goes from point Q and travels parallel
26
to the optical axis. The reflection of this ray must pass through the focal point. Thus, for the
concave mirror, the reflection of principal ray 1 goes through focal point F, as shown in
Figure 4b. For the convex mirror, the backward extension of the reflection of principal ray 1
goes through the focal point (i.e., a virtual focus). Principal ray 2 travels first on the line going
through the focal point and then is reflected back along a line parallel to the optical axis.
Principal ray 3 travels toward the center of curvature of the mirror, so it strikes the mirror at
normal incidence and is reflected back along the line from which it came. Finally, principal
ray 4 strikes the vertex of the mirror and is reflected symmetrically about the optical axis.
Figure 4
Figure 4.
The four principal rays shown for both (a) a concave mirror and (b) a convex mirror.
The image forms where the rays intersect (for real images) or where their backward
extensions intersect (for virtual images).
Ray-Tracing Rules
Ray tracing is very useful for mirrors. The rules for ray tracing are summarized here for
reference:

A ray traveling parallel to the optical axis of a spherical mirror is reflected along a line
that goes through the focal point of the mirror.
27



A ray traveling along a line that goes through the focal point of a spherical mirror is
reflected along a line parallel to the optical axis of the mirror.
A ray traveling along a line that goes through the center of curvature of a spherical
mirror is reflected back along the same line.
A ray that strikes the vertex of a spherical mirror is reflected symmetrically about the
optical axis of the mirror.
What is Mirror Equation?
It is an equation relating object distance and
image distance with focal length is known as a
mirror equation. It is also known as a mirror
formula.
In a spherical mirror:
The distance between the object and the
pole of the mirror is called the object
distance(u).
 The distance between the image and the
pole of the mirror is called Image
distance(v).
 The distance between the Principal focus
and pole of the mirror is called Focal
Length(f).
In ray optics, The object distance, image
distance, and Focal length are related as,

1/ v + 1/ u = 1/f
Where,




u is the Object distance
v is the Image distance
f is the Focal Length given by f= R/2
R is the radius of curvature of the spherical mirror
Note: we have been very careful with the signs in deriving the mirror equation. For a
plane mirror, the image distance has the opposite sign of the object distance. Also, the real
image formed by the concave mirror is on the opposite side of the optical axis with respect to
the object. In this case, the image height should have the opposite sign of the object height.
To keep track of the signs of the various quantities in the mirror equation, we now introduce
a sign convention.
Sign convention for spherical mirrors
28
Using a consistent sign convention is very important
in geometric optics. It assigns positive or negative
values for the quantities that characterize an optical
system. Understanding the sign convention allows
you to describe an image without constructing a ray
diagram. This text uses the following sign convention:
1. The focal length f is positive for concave
mirrors and negative for convex mirrors.
2. The image distance d_i is positive for real
images and negative for virtual images.
New Cartesian Sign Convention is used to avoid
confusion in understanding the ray directions. Refer to the diagram for clear visualization.





For the measurement of all the distances, the optical center of the lens is considered.
When the distances are measured opposite to the direction of the incident light, they
are considered to be negative.
When the distances are measured in the same direction of the incident light, they are
considered to be positive.
When the heights are measured upwards and perpendicular to the principal axis,
they are considered to be positive.
When the heights are measured downwards and perpendicular to the principal axis,
they are considered to be negative.
CHECK YOUR UNDERSTANDING # 3
1. The radius of curvature of a spherical mirror can be positive or negative. What does it
mean to have a negative radius of curvature of a convex mirror?
Mirror Equation for concave mirror and Mirror Equation for a convex mirror
The mirror equation 1/v + 1/u = 1/f holds good for concave mirrors as well as convex
mirrors.
Example of Mirror Equation
29
1. The radius of curvature of a convex mirror used for rearview on a car is
4.00 m. If the location of the bus is 6 meters from this mirror, find the position
of the image formed.
Solution:
Given:
The radius of curvature (R)= +4.00 m
Object distance(u) = -6.00 m
Find:
Image distance(v) =?
Formula used:
f = R/2
1/ v + 1/u = 1/f
Calculation:
To calculate the Focal length of the given mirror, substitute the value of
Radius of Curvature (R) in
the f = R/2.
f= +4 m/2 = +2m
Since 1/v + 1/u = 1/f we can rearrange 1/v = 1/f – 1/ u
1/v = 1 / 2 m - 1/-6 m
1/v = 1/ 2 m + 1/6 m
1/v = 6 + 2 m/ 12
1/ v = 8/ 12 m
V = 1.5 m
The image is 1.5 meters behind the mirror.
CHECK YOUR UNDERSTANDING # 4
1. As you are analyzing a spherical mirror situation, you write an equation that states:
1/ f = 1 / +12 cm + 1 / + 24 cm. What is the value of 1/ f? What is f?
What’s More
THE MIRROR EQUATION-CONCAVE MIRRORS
Ray diagrams can be used to determine the image location, size, orientation and
type of image formed of objects when placed at a given location in front of a concave mirror.
Ray diagrams provide useful information about object-image relationship and it may help
one determine the approximate location and size of the image, thus it will not provide
numerical information about image distance and object size. To obtain this type of numerical
information, it is necessary to use the Mirror Equation and the Magnification Equation.
30
The mirror equation expresses the quantitative relationship between the object distance (do),
the image distance (di), and the focal length (f). The equation is stated as follows:
1/f = 1/ do + 1/ di
The magnification equation relates the ratio of the image distance and object
distance to the ratio of the image height (hi) and object height (ho). The magnification
equation is stated as follow:
M = hi / ho = - di / do
Example:
1. A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having
a focal length of 15.2 cm. Determine the image distance and the image size.
Solution:
Given : ho = 4.00 cm
do = 45.7 cm
f = 15.2 cm
Find : hi and di
Formula used:
f = 1/ do + 1/ di
hi / ho = -di / do
Calculation:
a. 1/f = 1/ do + 1/ di
1/15.2 cm = 1/45.7 cm + 1/ di
0.0658 cm-1 = 0. 0219 cm-1 + 1 / di
0.0658 cm-1 – 0. 0219 cm-1 = 1 / di
0. 0439cm-1 = 1/ di
di = 22.78 cm
b. hi / h0 = -di / do
hi do = -ho di
hi ( 45.70 cm ) = - (4. 00 cm ) (22.78 cm)
hI = - 91.12 cm2 / 45. 70 cm
hi = - 1.99 cm
31
The negative values for image height indicate that the image is an inverted image. As
is often the case in physics, a negative or positive sign in front of the numerical value for a
physical quantity represents information about direction. In the case of the image height, a
negative value always indicates an inverted image.
2.
A 4.0-cm tall light bulb is placed a distance of 8.3 cm from a concave mirror having a
focal length of 15.2 cm. (NOTE: this is the same object and the same mirror, only this
time the object is placed closer to the mirror.) Determine the image distance and the
image size.
Solution:
Given :
ho = 4.00 cm
do = 8.3 cm
Find : hi and di
f = 15.2 cm
Formula used:
f = 1/ do + 1/ di
hi / ho = -di / do
Calculation:
a. 1/f = 1/ do + 1/ di
1/15.2 cm = 1/8.3 cm + 1/ di
0.0658cm-1 = 0. 1205 cm-1 + 1 / di
0.0658 cm-1 – 0. 1205 cm-1 = 1 / di
-0.0547 cm-1 = 1/ di
di = -18.28 cm
b. hi / h0 = -di / do
hi do = -ho di
hi ( 8.3 cm ) = - (4. 00 cm ) (-18.28 cm)
hI = 73.12 cm2 / 8.3 cm
hi
= 8.81 cm
The negative value for image distance indicates that the image is a virtual image
located behind the mirror. Again, a negative or positive sign in front of the numerical value
for a physical quantity represents information about direction. In the case of the image
distance, a negative value always means behind the mirror. Note also that the image height
is a positive value, meaning an upright image. Any image that is upright and located behind
the mirror is considered to be a virtual image.
The +/- Sign Conventions
The sign conventions for the given quantities in the mirror equation and magnification
equations are as follows:






f is + if the mirror is a concave mirror
f is - if the mirror is a convex mirror
di is + if the image is a real image and located on the object's side of the mirror.
di is - if the image is a virtual image and located behind the mirror.
hi is + if the image is an upright image (and therefore, also virtual)
hi is - if the image an inverted image (and therefore, also real)
32
CHECK YOUR UNDERSTANDING #5
Direction: Solve the following problem and show your complete solution.
1. Determine the image distance and image height for a 5.00-cm tall object placed
45.0 cm from a concave mirror having a focal length of 15.0 cm.
2. A magnified, inverted image is located a distance of 32.0 cm from a concave
mirror with a focal length of 12.0 cm. Determine the object distance and tell
whether the image is real or virtual.
What I Have Learned
Light always reflects according
to the law of reflection, regardless of
whether the reflection occurs off a flat
surface or a curved surface. Using reflection laws allows one to determine the image
location for an object. The image location is the location where all reflected light appears to
diverge from. Thus to determine this location demands that one merely needs to know how
light reflects off a mirror. The image of an object for a concave mirror was determined by
tracing the path of light as it emanated from an object and reflected off a concave mirror. The
image was merely that location where all reflected rays intersected. The use of the law of
reflection to determine a reflected ray is not an easy task. For each incident ray, a normal
line at the point of incidence on a curved surface must be drawn and then the law of
reflection must be applied. A simpler method of determining a reflected ray is needed.
The simpler method relies on two rules of reflection for concave mirrors. They are:
Any incident ray traveling
parallel to the principal axis on
the way to the mirror will pass
through the focal point upon
reflection.
incident ray passing
 Any
through the focal point on the
way to the mirror will travel
parallel to the principal axis upon reflection.
These two rules of reflection are illustrated in the diagram .

RAY DIAGRAM – CONCAVE MIRRORS
The theme of this unit has been that we see an object because light from the object
travels to our eyes as we sight along a line at the object. Similarly, we see an image of an
object because light from the object reflects off a mirror and travel to our eyes as we sight at
the image location of the object. From these two basic premises, we have defined the image
location as the location in space where light appears to diverge from. Ray diagrams have
been a valuable tool for determining the path taken by light from the object to the mirror to
our eyes. In this section, we will investigate the method for drawing ray diagrams for objects
placed at various locations in front of a concave mirror.
33
To draw these diagrams, we will have to recall the two rules of reflection for concave
mirrors:


Any incident ray traveling parallel to the principal axis on the way to the mirror will
pass through the focal point upon reflection.
Any incident ray passing through the focal point on the way to the mirror will
travel parallel to the principal axis upon reflection.
In this diagram five incident rays are drawn along with their corresponding reflected
rays. Each ray intersects at the image location and then diverges to the eye of an observer.
Every observer would observe the same image location and every light ray would follow the
law of reflection. Yet only two of these rays would be needed to determine the image
location since it only requires two rays to find the intersection point. Of the five incident rays
drawn, two of them correspond to the incident rays described by our two rules of
reflection for concave mirrors.
Step-by-Step Method for Drawing Ray Diagrams
The method for drawing ray diagrams for concave mirror is described below. The method is
applied to the task of drawing a ray diagram for an object located beyond the center of
curvature (C) of a concave mirror. Yet the same method works for drawing a ray diagram for
any object location.
1. Pick a point on the top of the object and draw two incident rays traveling towards the
mirror.
Using a straight edge, accurately draw one ray so that it passes exactly through the focal
point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the
principal axis. Place arrowheads upon the rays to indicate their direction of travel.
2. Once these incident rays strike
according to the two rules of reflection for concave mirrors.
the
mirror,
reflect
them
The ray that passes through the focal point on the way to the mirror will reflect and travel
parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that
traveled parallel to the principal axis on the way to the mirror will reflect and travel through
34
the focal point. Place arrowheads upon the rays to indicate their direction of travel. Extend
the rays past their point of intersection.
3. Mark the image of the top of the object.
The image point of the top of the object is the point where the two reflected rays intersect. If
your were to draw a third pair of incident and reflected rays, then the third reflected ray
would also pass through this point. This is merely the point where all light from the top of the
object would intersect upon reflecting off the mirror. Of course, the rest of the object has an
image as well and it can be found by
applying
the
same three steps to another chosen point.
4. Repeat the process for the bottom of the object.
The goal of a ray diagram is to determine the location, size, orientation, and type of image
that is formed by the concave mirror. Typically, this requires determining where the image of
the upper and lower extreme of the object is located and then tracing the entire image. After
completing the first three steps, only the image location of the top extreme of the object has
been found. Thus, the process must be repeated for the point on the bottom of the object. If
the bottom of the object lies upon the principal axis (as it does in this example), then the
image of this point will also lie upon the principal axis and be the same distance from the
mirror as the image of the top of the object. At this point the entire image can be filled in.
35
CHECK YOUR UNDERSTANDING #6
The diagram below shows two light rays emanating from the top of the object and incident
towards the mirror. Describe how the reflected rays for these light rays can be drawn without
actually using a protractor and the law of reflection.
What I Can Do
IMAGE CHARACTERISTICS FOR CONCAVE MIRROR
Ray diagrams were constructed in order to determine the general location, size,
orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is
a definite relationship between the image characteristics and the location where an object is
placed in front of a concave mirror. The purpose of this portion of the lesson is to summarize
these object-image relationships - to practice the L•O•S•T art of image description. We wish
to describe the characteristics of the image for any given object location.
The L of L•O•S•T represents the relative location. The O of L•O•S•T represents the
orientation (either upright or inverted). The S of L•O•S•T represents the relative size (either
magnified, reduced or the same size as the object). And the T of L•O•S•T represents the
type of image (either real or virtual). The best means of summarizing this relationship
between object location and image characteristics is to divide the
possible object locations into five general areas or points:





Case 1: the object is located beyond the center of
curvature (C)
Case 2: the object is located at the center of curvature
(C)
Case 3: the object is located between the center of curvature (C) and the focal
point (F)
Case 4: the object is located at the focal point (F)
Case 5: the object is located in front of the focal point (F)
Case 1: The object is located beyond C
When the object is located at a location beyond the center of
curvature, the image will always be located somewhere in between the
center of curvature and the focal point. Regardless of exactly where
the object is located, the image will be located in the specified region.
In this case, the image will be an inverted image. That is to say, if the
object is right side up, then the image is upside down. In this case, the
image is reduced in size; in other words, the image dimensions are smaller than the object
dimensions. If the object is a six-foot tall person, then the image is less than six feet tall.
Earlier in Lesson 2, the term magnification was introduced; the magnification is the ratio of
the height of the image to the height of the object. In this case, the absolute value of the
magnification is less than 1. Finally, the image is a real image. Light rays actually converge
36
at the image location. If a sheet of paper were placed at the image location, the actual
replica of the object would appear projected upon the sheet of paper.
Case 2: The object is located at C
When the object is located at the center of curvature, the image will
also be located at the center of curvature. In this case, the image will
be inverted (i.e., a right side up object results in an upside-down
image). The image dimensions are equal to the object dimensions. A
six-foot tall person would have an image that is six feet tall; the
absolute value of the magnification is equal to 1. Finally, the image is a
real image. Light rays actually converge at the image location. As such, the image of the
object could be projected upon a sheet of paper.
Case 3: The object is located between C and F
When the object is located in front of the center of curvature, the
image will be located beyond the center of curvature. Regardless of
exactly where the object is located between C and F, the image will be
located somewhere beyond the center of curvature. In this case, the
image will be inverted (i.e., a right side up object results in an upsidedown image). The image dimensions are larger than the object
dimensions. A six-foot tall person would have an image that is larger than six feet tall; the
absolute value of the magnification is greater than 1. Finally, the image is a real image. Light
rays actually converge at the image location. As such, the image of the object could be
projected upon a sheet of paper.
Case 4: The object is located at F
When the object is located at the focal point, no image is formed. As
discussed earlier in Lesson 3, light rays from the same point on the
object will reflect off the mirror and neither converge nor diverge. After
reflecting, the light rays are traveling parallel to each other and do not
result in the formation of an image.
Case 5: The object is located in front of F
When the object is located at a location beyond the focal
point, the image will always be located somewhere on the opposite
side of the mirror. Regardless of exactly where in front of F the
object is located, the image will always be located behind the
mirror. In this case, the image will be an upright image. That is to
say, if the object is right side up, then the image will also be right
side up. In this case, the image is magnified; in other words, the
image dimensions are greater than the object dimensions. A six-foot tall person would have
an image that is larger than six feet tall; the magnification is greater than 1. Finally, the
image is a virtual image. Light rays from the same point on the object reflect off the mirror
and diverge upon reflection. For this reason, the image location can only be found by
extending the reflected rays backwards beyond the mirror. The point of their intersection is
the virtual image location. It would appear to any observer as though light from the object
were diverging from this location. Any attempt to project such an image upon a sheet of
paper would fail since light does not actually pass through the image location.
37
It might be noted from the above descriptions that there is a relationship between the
object distance and object size and the image distance and image size. Starting from a large
value, as the object distance decreases (i.e., the object is moved closer to the mirror), the
image distance increases; meanwhile, the image height increases. At the center of
curvature, the object distance equals the image distance and the object height equals the
image height. As the object distance approaches one focal length, the image distance and
image height approaches infinity. Finally, when the object distance is equal to exactly one
focal length, there is no image. Then altering the object distance to values less than one
focal length produces images that are upright, virtual and located on the opposite side of the
mirror. Finally, if the object distance approaches 0, the image distance approaches 0 and the
image height ultimately becomes equal to the object height. These patterns are depicted in
the diagram below. Nine different object locations are drawn and labeled with a number; the
corresponding image locations are drawn in blue and labeled with the identical number.
CHECK YOUR UNDERSTANDING #7
1. Identify the means by which you can use a concave and/or a plane mirror to form a
real image
2. Identify the means by which you can use a concave and/or a plane mirror to form a
virtual image.
3.
Are all real images larger than the object?
What I Still Want To Know
Convex Mirrors
The same set of rules in concave mirror, it’s just that things will look a bit different since
convex mirrors have their focal point and centre behind the mirror.

Convex mirrors also typically make a small, virtual image of the original object.
Let’s look at an example using the three rules covered in concave mirrors.
38

The only difference is that I will extend the rays behind the mirror as dotted lines to
be able to show how they “pass through the focal point” or “through the centre”.
Use Rule #2 (because it works well for this mirror) to draw the ray coming off of the tip of the
object parallel to the principle axis. When it hits the mirror, it bounces off so that a dotted line
drawn behind the mirror will pass through the focal point .
Rule #3 This ray comes off of the tip of the object aiming straight for the centre. Where it
hits the mirror it will bounce back, but I draw a dotted line behind the mirror to show where
the ray would have gone.
Notice that there is now a place where the two dotted lines hit. This is where the image will
appear.
39
The image is virtual (it’s behind the mirror), it is erect (right side up), and shrunken down a
bit.
The Formulas
There are a couple of formulas you can use to figure out where an image will appear, if it will
be inverted/erect, and magnified or shrunk.
The first is called the "Mirror Formula"...
Ultra-Special Notes for Using the Mirror Formula:
1. Concave Mirrors have positive focal length, and convex mirrors have a negative focal
length.
2. Distances are positive if they are in front of the mirror, and negative if they are
behind.
3. All measurements are made along the principle axis from the surface of the mirror.
The "Magnification Formula" is...
Ultra-Special Notes for Using the Magnification Formula:
As above plus...
40
1. Positive heights are above the principle axis, and negative heights are below the
principle axis.
2. Positive magnifications are above the principle axis, and negative magnifications are
below the principle axis.
3. Magnifications greater than 1 are bigger than the original object, magnifications less
than 1 are smaller than the original object.
Example 1: A convex mirror has a radius of 20 cm. An object is placed 30 cm in front of the
mirror. Determine where the image will appear.
Since the radius is 20 cm (which is the distance from the mirror to the centre),
and since the focal point is half ways in between and negative for a convex
mirror, f = -10 cm.
Since di is negative, it appears behind the mirror as a virtual image.
Example 2: For the same situation from Example 1, determine how tall the image is if the
object is 5.0cm tall. Also determine the magnification.
To calculate the magnification either distances or heights could be used. Since the distances
have been through less calculations, I trust them more.
41
The magnification is positive, since the image is erect. But it less than one, so the image is
smaller than the object... one quarter the size!
CHECK YOUR UNDERSTANDING # 8
DIRECTION: Draw ray diagram to locate the images in each of the following and
describe the images using LOST.
1.
2.
3.
4.
5.
Object is beyond C
Object is at C
Object is between C and F
Object is at F
Object is in front of F
CHECK YOUR UNDERSTANDING # 9
DIRECTION: Draw ray diagram to locate the images in each of the following and
describe the images using LOST.
1.
F
C
F
C
2.
42
How Much Did Learn
CHECK YOUR UNDERSTANDING #10
LABORATORY 1
TITLE: IMAGES IN CONCAVE MIRROR
OBJECTIVE: To explore how light rays reflect off of concave mirror.
MATERIAL:
1 pc tablespoon
PROCEDURES:
1. Have a partner to hold the front of the tablespoon close to your eyes.
Describe the image you see:
2. Observe the image carefully as your partner slowly moves the mirror farther from
your eyes.
Describe any changes you observe in the image.
ANALYSIS:
Analyze the result by answering the questions:
1. Use the characteristics of images to describe the image in each of the following:
POSITION OF
OBJECT
Close to the
mirror
Far from the
mirror
LOCATION
ORIENTATION
SIZE
TYPE OF
IMAGES
2. If the focal length of a mirror is 4 cm, how far is the center of curvature from the
vertex?
3. If the center of curvature is 25cm from the mirror, what is the focal length?
CHECK YOUR UNDERSTANDING # 11
LABORATORY 1
TITLE: IMAGES IN CONCAVE MIRROR
OBJECTIVE: To explore how light rays reflect off of convex mirror.
MATERIAL:
1 pc tablespoon
PROCEDURES:
1. Have a partner to hold the back of the tablespoon close to your eyes.
Describe the image you see:
43
2. Observe the image carefully as your partner slowly moves the mirror farther from
your eyes.
Describe any changes you observe in the image.
ANALYSIS:
Analyze the result by answering the questions:
1. Use the characteristics of images to describe the image in each of the following:
POSITION OF
OBJECT
Close to the
mirror
Far from the
mirror
LOCATION
ORIENTATION
SIZE
TYPE OF
IMAGES
3. e.
4.
5.
6.
2.
An What
image
measured
c.
d.
Find
Where
object
isthe
is1.50
is
to
the
radius
the
bemagnification?
cm
0.167
image?
of
high
curvature
cm
is held
high.3.00
of thecm
convex
from amirror
person’s
formed
cornea,
by the
and
cornea.
its reflected
image is measured to be 0.167 cm high. a. What is the magnification? b. Where is
the image? c. Find the radius of curvature of the convex mirror formed by the
cornea.
44
Assessment (Post-test 1)
I.
DIRECTION: Choose the best answer from the choices given. Write the letter
only of your answer.
1. Using the image about plane mirrors. Which one is a true fact?
a. Angle of incidence + angle of reflection
b. Angle of incidence = angle of reflection
c. Angle of incidence - angle of reflection
2. __________________ is an imaginary line perpendicular to the surface.
a. Normal line
b. Reflected line
c. Incident line
3. A virtual image is a copy of an object formed at the location from which the light rays
appear to come
a. True
b. False
4. A flat sheet of glass that has a smooth, silver colored coating on one side
a. Plane mirror
b. Convex mirror
c. Concave Mirror
5. According to the law of reflection, a light ray striking a mirror
a. Continues moving through the mirror in the same direction
b. Bounces off the mirror toward the direction it came form
c. Moves into the mirror at a slightly different angle
d. Bounces off the mirror at the same angle it hits
6. The fact are the two angles are the same is an example of
a. Refraction
b. Law of reflection c. Angle of incidence
d. Magninfication
7.
The following is a picture of what type of mirror
a. Convex Mirror
b. Plane mirror c. Concave Mirror d. Flat Mirror
8. The following is a picture of what type of mirror
a. Convex Mirror
b. Plane mirror c. Concave Mirror d. Flat Mirror
9. Plane mirror create real images
a. True
b. False
10. A material that reflect and absorb all the light that strikes it is called which of the
following
a. Transparent
b. Diffuse
c. Translucent
d. Opaque
45
Assessment (Post-test 2)
I.DIRECTION: Choose the best answer from the choices given. Write the letter only
of your answer.
1. What happens to the image produced by a pinhole camera when you move the back wall
farther from the pinhole? It becomes
a. larger and fainter.
b. smaller and fainter.
c. larger and brighter.
d. smaller and brighter.
2. The shortest mirror in which a creature from outer space can see its entire body is .......
its height.
a. twice
b. equal to
c. one half
d. It depends on how far away it stands.
3. A ray reflected from a retroreflector
a. has an angle of reflection equal to the angle of incidence.
b. passes through the focal point.
c. forms a right angle with an incident ray.
d. travels in the direction opposite that of the incident ray.
4. A ray of light parallel to the optic axis of a concave mirror is reflected back
a. through the center of the sphere.
b. through the focal point.
c. parallel to the optic axis.
d. as if it came from the focal point.
5. The back surfaces of automobile headlights are curved
a. because inverted, real images of filaments shine brighter.
b. to concentrate light in one direction.
c. for structural reasons not related to optics.
d. to get multiple images of the filament.
6. A ray of light passing through the focal point at an angle to the optic axis of a concave
mirror is reflected back
a. through the center of the sphere.
b. through the focal point.
c. parallel to the optic axis.
d. in the horizontal direction.
7. What type of image is formed when an object is placed at a distance of 1.5 focal lengths
from a convex mirror?
a. erect and virtual
b. inverted and virtual
c. erect and real
d. inverted and real
46
8. Where is the image located when an object is placed 30 cm from a convex mirror with a
focal length of 10 cm? Note: you should have to figure this out exactly. Only one answer
is in the right ballpark.
a. 7.5 cm in back
b. 15 cm in back
c. 30 cm in back
d. 7.5 cm in front
9. If the sun is 150 million km away from the earth, how long does it take sunlight to reach
the earth?
a. 0.5 s
b. 15 s
c. 45 s
d. 500 s
10. Some yellow objects actually absorb yellow light but reflect red and green light. If we
shine yellow light on such a yellow object, it will appear ...... to our eyes.
a. yellow
b. green
c. red
d. black
11. The sky appears blue because
a. that is its natural color.
b. the earth's atmosphere emits blue light.
c. the air away from the sun cools down and turns blue.
d. the earth's atmosphere scatters more blue light than red.
12. If a ray of light strikes a pane of glass at 45 degrees to the normal, it
a. passes straight through as if the glass were not there.
b. leaves the glass at a smaller angle to the normal.
c. leaves the glass at a larger angle to the normal.
d. leaves with the same angle to the normal, but is deflected to the side.
13. Two coins are at equal distances from your eye. One is under 40 cm of water, the other
under 40 cm of glass. Which coin appears closer?
a. The one under the glass.
b. The one under the water.
c. Neither, they both appear at the same distance.
14. The critical angle for total internal reflection at an air-water interface is approximately 48
degrees. In which of the following situations will total internal reflection occur?
a. light incident in water at 40 degrees.
b. light incident in water at 55 degrees
c. light incident in air at 40 degrees
d. light incident in air at 55 degrees
15. The dispersion of light when it passes through a prism shows that
a. the prism contains many narrow, equally spaced slits.
b. all colors in the light are treated the same.
c. different colors have different indices of refraction.
d. the speed of light in a vacuum is a constant.
16. For a converging lens, a ray arriving parallel to the optic axis
a. appears to come from the principal focal point.
b. passes through the principal focal point.
c. passes through the "other" focal point.
d. appears to come from the "other" focal point.
47
17. A converging lens is used to form a sharp image of a candle. If the lower half of the lens
is covered by a piece of paper, the
a. lower half of the image will disappear.
b. upper half of the image will disappear.
c. image will become dimmer.
d. image will not change.
18. A diverging lens has a focal length of 10 cm. Where is the image located when an object
is placed 30 cm from the lens?
a. 7.5 cm on the near side
b. 15 cm on the near side
c. 30 cm on the near side
d. 7.5 cm on the far side
19. A camera employs a .... lens to form..... images.
a. converging .... real
b. converging .... virtual
c. diverging .... real
d. diverging .... virtual
20. In most cameras the location of the image is adjusted to appear on the film by changing
the
a. position of the lens.
b. diameter of the diaphragm.
c. shape of the lens.
d. focal length of the lens.
21. A human eye employs a ..... lens to form ..... images.
a. converging .... real
b. converging .... virtual
c. diverging .... real
d. diverging .... virtual
48
For inquiries and feedback, please write or call:
Department of Education – Bureau of Learning Resources (DepEd-BLR)
DepEd Division of Cagayan de Oro City
Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro
Telefax:
((08822)855-0048
E-mail Address:
cagayandeoro.city@deped.gov.ph
FAIR USE AND CONTENT DISCLAIMER: This SLM (Self Learning Module) is for
educational purposes only. Borrowed materials (i.e. songs, stories, poems, pictures, photos,
brand names, trademarks, etc.) included in these modules are owned by their respective
copyright holders. The publisher and authors do not represent nor claim ownership over
them.
49