REFLECTION LESSON 4
Convex Mirrors
A convex mirror is sometimes referred to as a diverging mirror due to
the fact that incident light originating from the same point and will
reflect off the mirror surface and diverge. Subsequently they will
never intersect on the object side of the mirror. For this reason, convex
mirrors produce virtual images which are located somewhere behind
the mirror.
The task of determining the image location of an object is to determine
the location where reflected light intersects. Light rays originating at
the object location are shown approaching and subsequently reflecting from the
mirror surface. Each observer must sight along the line of a reflected ray to view the
image of the object. Each ray is extended backwards to a point of intersection - this
point of intersection of all extended reflected rays is the image location of the object.
The image is a virtual image. Light does not actually pass through the image
location. It only appears to observers as though all the reflected light from each part of
the object is diverging from this virtual image location.
Just like for Convex mirrors, there were two rules of reflection for convex mirrors that
make drawing the virtual image much easier. They are:

Any incident ray traveling parallel to the principal axis on the way to a
convex mirror will reflect in such a manner that its extension will pass
through the focal point.

Any incident ray traveling towards a convex mirror such that its
extension passes through the focal point will reflect and travel parallel to the
principal axis.
Ray Diagrams - Convex Mirrors
These two rules will be used to construct ray diagrams. A ray diagram is a tool which
is used to determine the location, size, orientation, and type of image formed by a
mirror.
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 travels
towards the focal point on the opposite side of the mirror; this ray
will strike the mirror before reaching the focal point; stop the ray at
the point of incidence with 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 the mirror, reflect them according to the two rules of
reflection for convex mirrors.
The ray that travels towards the focal point will reflect and travel
parallel to the principal axis. The ray which traveled parallel to
the principal axis on the way to the mirror will reflect and travel
in a direction such that its extension passes through the focal
point. Align a straight edge with the point of incidence and the
focal point, and draw the second reflected ray.
3. Locate and 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. Since the two
reflected rays are diverging, they must be extended
behind the mirror in order to intersect.
The point of intersection is the image point of the top
of the object. Both reflected rays would appear to
diverge from this point.
4. Repeat the process for 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 complete image
can be filled in.
What do you notice about the image in the diagram in comparison to the object?
Its located at a position behind the convex mirror, its upright, reduced in size and
virtual.
For the diagrams draw the rays and identify the image.
Image Characteristics for Convex Mirrors
Virtual, upright, reduced in size and located behind the mirror, will these always be
the characteristics of an image produced by a convex mirror? Can convex mirrors
ever produce real images? Inverted images? Magnified Images?
To answer these questions, we will look at three different ray diagrams for objects
positioned at different locations along the principal axis. The diagrams are shown
below.
The diagrams above shows that in each case, the image is




located behind the convex mirror
a virtual image
an upright image
reduced in size (i.e., smaller than the object)
What do you notice between the diagrams?
As the object distance is decreased, the image distance is decreased and the image
size is increased.
The Mathematics of Refraction
The Angle of Refraction
We have learned in the past that if a light wave passes from a medium in which it
travels slow (relatively speaking) into a medium in which it travels fast, then the light
wave will refract away from the normal.
In such a case, the refracted ray will be farther from the normal line
than the incident ray.
On the other hand, if a light wave passes from a medium in which it
travels fast (relatively speaking) into a medium in which it travels
slow, then the light wave will refract towards the normal. In such a
case, the refracted ray will be closer to the normal line than the
incident ray is.
HOW DOES REFRACTION WORK?
There are two conditions which are
required in order to observe the change in
direction of the path of the light or
soldiers:

The soldiers or light must change speed when crossing the boundary.
(must move from one medium to another with a different medium
characteristic)

The ssoldiers or light must approach the boundary at an angle;
refraction will not occur when they approach the boundary head-on (i.e.,
heading perpendicular to it).
The question iwe want to answer today is: "By how much does light refract when it
crosses a boundary?"
Snell's Law
To begin, consider a hemi-cylindrical dish filled with water. Suppose that a laser
beam is directed towards the flat side of the dish at the exact center of the dish. The
angle of incidence can be measured at the point of incidence.
This ray will refract, bending towards the normal (since the
light is passing from a medium in which it travels fast into one
in which it travels slow).
Once the light ray enters the water, it travels in a straight line
until it reaches the second boundary.
At the second boundary, the light ray is approaching along the normal to the
curved surface (this stems from the geometry of circles). The ray does not
refract upon exiting since the angle of incidence is 0-degrees. The ray of laser
light therefore exits at the same angle as the refracted ray of light made at the
first boundary. These two angles can be measured and recorded.
The equation relating the angles of incidence ("theta i") and the angle of refraction
("theta r") for light passing from air into water is given as
Observe that the constant of proportionality in this equation is 1.33 - the index of
refraction value of water.
But if the semi-cylindrical dish full of water was replaced by a semi-cylindrical disk
of Plexiglas, the constant of proportionality would be 1.51 - the index of refraction
value of Plexiglas.
Experimentally, it is found that for a ray of light traveling from air into some material,
the following equation can be written.
where nmaterial = index of refraction of the material
This study of the refraction of light as it crosses from one material into a second
material yields a general relationship between the sines of the angle of incidence and
the angle of refraction. This general relationship is expressed by the following
equation:
where
("theta i") = angle of incidence
("theta r") = angle of refraction
ni = index of refraction of the incident medium
nr = index of refraction of the refractive medium
This relationship between the angles of incidence and refraction and the indices of
refraction of the two medium is known as Snell's Law.
Using Snell's Law to Predict An Angle Value
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. This is illustrated
in the two examples below.
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.
Material
Vacuum
Index of Refraction
1.0000
<--lowest optical density
Air
Ice
Water
Ethyl Alcohol
Plexiglas
Crown Glass
Light Flint Glass
Dense Flint Glass
Zircon
Diamond
Rutile
Gallium phosphide
1.0003
1.31
1.333
1.36
1.51
1.52
1.58
1.66
1.923
2.417
2.907
3.50
<--highest optical density
Diagram:
Solution to Problem A
Given:
ni = 1.00 (from table)
Find:
=??
nr = 1.333 (from table)
= 52 degrees
1.00 * sine (52 degrees) = 1.333 * sine (theta r)
0.7880 = 1.333 * sine (theta r)
0.591 = sine (theta r)
sine-1 (0.591) = sine-1 ( sine (theta r))
36.2 degrees = theta r
Check your answer, does it make sense that the refractive angle is less than the
indicence angle?
In this problem, the light ray is traveling from a less optically dense or fast medium
(air) into a more optically dense or slow medium (water), and so the light ray should
refract towards the normal. Thus, the angle of refraction should be smaller than the
angle of refraction. And indeed it is!
Example Problem B
A ray of light in air is approaches a triangular piece of crown glass at an
angle of 0.00 degrees (as shown in the diagram at the right). Perform the
necessary calculations in order to trace the path of the light ray as it enters
and exits the crown glass. Refer to the table of indices of refraction if
necessary.
Solution to Problem B
Despite the complication of there being nonparallel boundaries, 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:
Diagram:
Given:
Find:
Boundary #1
Trace path of light.
ni = 1.00 (from table)
nr = 1.52 (from table)
= 0.0 degrees
That is, find
boundary #1
and
and
Boundary #2
at
at
boundary #2
ni = 1.52 (from table)
nr = 1.00 (from table)
Now list the relevant equation (Snell's Law), substitute known values into the
equation, and perform the proper algebraic steps to solve for the unknown. Begin the
process at boundary #1 and then repeat for boundary #2 until the final answer is
found.
Boundary #1:
1.00 * sine (0.0 degrees) = 1.52 * sine (theta r)
0.000 = 1.52 * sine (theta r)
0.000 = sine (theta r)
sine-1 (0.000) = sine-1 ( sine (theta r))
0.00 degrees = theta r
This problem is made easier if you draw upon your conceptual knowledge of what
occurs when a light ray approaches at an angle of incidence of 0-degrees. When
approaching along the normal, the light ray passes across the boundary without
refracting.
The next step demands that the light ray be traced through the triangular block until it
reaches the second boundary. At the second boundary, the normal line
must be drawn (labeled N) and the angle of incidence (between the
incident ray and the normal) must be measured. The measured angle of
incidence at the second boundary is 30.0 degrees.
This angle measurement now provides knowledge of three of the four
variables in the Snell's Law equation and allows for the determination of the fourth
variable (the angle of refraction) at the second boundary.
Boundary #2:
1.52 * sine (30.0 degrees) = 1.00 * sine (theta r)
1.52 * (0.5000) = 1.00 * sine (theta r)
0.7600 = sine (theta r)
sine-1 (0.7600) = sine-1 ( sine (theta r))
49.5 degrees = theta r
The refracted ray at the second boundary will exit at an angle of
49.5 degrees from the normal.
REFLECTION LESSON 4 HOMEWORK
1. The diagram below shows seven different object locations (in front of the convex
mirror) and their corresponding image locations (behind the convex mirror). Label
the image locations to match the objects that made them.
The diagram below shows a spherical surface which is silvered on both sides. Thus,
the surface serves as double-sided mirror, with one of the sides being the concave and
one being the convex side. The principal axis, focal point, and center of curvature are
shown. The region on both sides of the mirror is divided into eight sections (labeled
M, N, P, Q, R, S, T, and W). Five objects (labeled 1, 2, 3, 4, and 5) are shown at
various locations about the double-sided mirror. Use the diagram to answer the
questions #2-7.
2. The image of object 1 would be located in section ______.
M
N
P
Q
R
S
T
W
T
W
3. The image of object 2 would be located in section ______.
M
N
P
Q
R
S
4. The image of object 3 would be located in section ______.
M
N
P
Q
R
S
T
W
T
W
T
W
5. The image of object 4 would be located in section ______.
M
N
P
Q
R
S
6. The image of object 5 would be located in section ______.
M
N
P
Q
R
S
7. The double-sided mirror would cause virtual image to be formed of objects
________.
a. 1, 2, and 4
b. 1, 2, and 3
c. 3 and 5
d. 4 and 5
e. 3 only
8. How can a plane mirror, concave mirror, and/or convex mirror be used to produce
an image which has the same size as the object?
9. How can a plane mirror, concave mirror, and/or convex mirror be used to produce
an upright image?
10. How can a plane mirror, concave mirror, and/or convex mirror be used to produce
a real image?
11. The image of an object is found to be upright and reduced in size. What type of
mirror is used to produce such an image?
12. Determine the angle of refraction for the following two refraction problems.
Answer: 53.9 and 28.4 degrees
13. 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. Answer: 26.1 and 42.0 degrees
14. Perform the necessary calculations at each boundary in order 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. Answer: 18, 22, 12, 13, 30 degrees
15. A ray of light in crown glass exits into air at an angle of 25.0 degrees. Determine
the angle at which the light approached the glass-air boundary. Refer to the table of
indices of refraction if necessary. Answer: 16.1 degrees
16. A ray of light is traveling through air (n = 1.00) towards a lucite block (n = 1.40)
in the shape of a 30-60-90 triangle. Trace the path of the light ray through the lucite
block shown in the diagram below. Answer: 44 degrees
REFLECTION LESSON 4 HOMEWORK KEY
1.
2. Answer: S
When the object is located beyond C, the image is located between C and F and is
inverted.
3. Answer: R
When the object is located between C and F, the image is located beyond C and is
inverted. So if the object is positioned above the principal axis, then the image is
positioned below the principal axis.
4. Answer: P
The object is on the convex side of the mirror so the image is located on the other
side of the mirror and is upright.
5. Answer: M
When the object is located betwee C and F, the image is located beyond C and is
inverted. So if the object is positioned below the principal axis, then the image is
positioned above the principal axis.
6. Answer: W
The object is located in front of F for a concave mirror, so the image is upright,
virtual and on the opposite side of the mirror. So if the object is positioned below the
principal axis, the image is positioned below the principal axis.
7. Answer: Objects 3 and 5
Virtual images are always formed by convex mirrors and are formed by concave
mirrors when the object is placed in front of F.
8. Plane mirrors will always do this.
Concave mirrors will do this when the object is at C or when the object is right on
the mirror surface.
Convex mirrors will only do this when the object is right on the mirror surface.
9. Plane mirrors and convex mirrors will always produce an upright image. A concave
mirror will only produce an upright image if the object is located in front of the focal
point.
10. Plane mirrors and convex mirrors only produce virtual images. Only a concave
mirror is capable of producing a real image and this only occurs if the object is located a
distance greater than a focal length from the mirror's surface.
11. Only a convex mirror could produce such an image. The upright images produced by
concave mirrors (when object is in front of F) are magnified images. And the upright
images produced by plane mirrors have the same size as the object.
12. A: Answer: 53.9 degrees
Measure the angle of incidence - the angle between the normal and incident ray. It is
approximately 45 degrees.
Given:
ni = 1.52
nr = 1.33
theta i = 45 degrees
Find: theta r
Substitute into Snell's law equation and perform the necessary algebraic operations
to solve:
1.52 • sine(45 degrees) = 1.33 • sine (theta r)
1.075 = 1.33* sine (theta r)
0.8081 = sine (theta r)
53.9 degrees = theta r
B: Answer: 28.4 degrees
Measure the angle of incidence - the angle between the normal and incident ray. It is
approximately 60 degrees.
Given:
ni =1.33
nr = 2.42
theta i = 60 degrees
Find: theta r
Substitute into Snell's law equation and perform the necessary algebraic operations
to solve:
1.33 • sine(60 degrees) = 2.42 • sine (theta r)
1.152 = 2.42 • sine (theta r)
0.4760 = sine (theta r)
28.4 degrees = theta r
13. Answer: 26.1 degrees and 42.0 degrees
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:
Diagram:
Given:
Find:
Boundary #1
at
ni = 1.00 (from table)
boundary #1
nr = 1.52 (from table)
and
= 42.0 degrees
at
boundary #2
Boundary #2
Note that the angle of
ni = 1.52 (from table)
refraction at boundary #1
nr = 1.00 (from table)
is the same as the angle of
incidence at boundary #2.
Now list the relevant equation (Snell's Law), substitute known values into the equation,
and perform the proper algebraic steps to solve for the unknown. Begin the process at
boundary #1 and then repeat for boundary #2 until the final answer is found.
Boundary #1:
1.00 * sine (42.0 degrees) = 1.52 * sine (theta r)
0.669 = 1.52 * sine (theta r)
0.4402 = sine (theta r)
-1
sine (0.4402) = sine-1 ( sine (theta r))
26.1 degrees = theta r
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. So now repeat the process in order
to solve for the angle of refraction at boundary #2.
Boundary #2:
1.52 * sine (26.1 degrees) = 1.00 * sine (theta r)
1.52 * (0.4402) = 1.00 * sine (theta r)
0.6691 = sine (theta r)
sine-1 (0.6691) = sine-1 ( sine (theta r))
42.0 degrees = theta r
The answers to this problem are 26.1 degrees and 42.0 degrees.
INTERESTING: There is an important conceptual idea which 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 which 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.
14. First, draw normal and measure the angle of incidence at first boundary; it is
approximately 30 degrees. Then, use the given n values and Snell's Law to calculate
the theta r values at each boundary. The angle of refraction at one boundary
becomes the angle of incidence at the next boundary; e.g., the theta r at the air-flint
glass boundary is the theta i at the flint glass-water boundary. Here are the
calculated theta r values:
air - flint glass: 18 degrees
flint glass - water: 22 degrees
water - diamond: 12 degrees
diamond - zirconium: 13 degrees
cubic zirconium - air: 30 degrees
15. Answer: 16.1 degrees
Given:
ni = 1.52
nr = 1.000
theta r = 25.0 degrees
Find: theta i = ???
Substitute into Snell's law equation and perform the necessary algebraic operations
to solve:
1.52 • sine(theta i) = 1.00 • sine(25.0 degrees)
1.52 • sine(theta i) = 0.4226
sine(theta i) = 0.2780
theta i = 16.1 degrees
16. Answer: 44 degrees
Upon entering the lucite, the ray of light passes across boundary without bending
since the theta i = 0 degrees.
At second boundary, the theta i can be found geometrically or by measurement as 30
degrees
Now substitute known values into Snell's Law and carry out the necessary algebraic
operations:
1.40 • sine(30 degrees) = 1.00 • sine(theta r)
0.700 = sine(theta r)
44 degrees = theta r
See below diagram for the ray tracing.