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Lethbridge College
Applied Physics (PHY143)
Unit 2 – Optics – Light and Mirrors
The Nature of Light
White light is made up of a range of electromagnetic waves with varying frequencies (or
wavelengths). The electromagnetic spectrum may be divided into bands, as follows:
Light has a constant speed (c) of 3.00 x 108 m/s. This speed is equal to the frequency of
the electromagnetic wave (f with Hz as a unit, or seconds-1) times the wavelength (λ with
meters as a unit). As one variable increases the other decreases, because c (the speed of
light) is constant.
Equation 2-1:
c=fxλ
You might recognize the labels that describe the electromagnetic bands, with radio and
television waves to the far left of the spectrum (low frequency and large wavelength) to
gamma rays (high frequency and small wavelength) to the far right of the spectrum. The
visible spectrum is represented by a small band in the middle with a frequency range
from 4.29 x 1014 Hz to 7.50 x 1014 Hz.
The Law of Reflection
When light hits a mirror or any reflective surface, it will be reflected. If the surface is
perfectly flat, the light hitting the surface (the incident wave) will be reflected at the same
angle relative to the normal (reflected wave). The reference plane that we call ‘normal’ is
always perpendicular to the surface.
Equation 2-2:
Angle of Incidence = Angle of Reflection
Θi = Θr
If your surface is not perfectly flat, the light will diffuse. The light is still obeying the
Law of Reflection, but the normal plane is always changing as the surface changes.
It is important for specular reflection that the surface is not rough – the cost to produce a
quality reflective surface is related in the costs of good optical equipment.
Spherical Mirrors
Take a hollow sphere and cut a circle from the surface. If this surface is reflective, you
now have a spherical mirror.
If the reflective surface is on the inside, the spherical mirror is concave; and if the
reflective surface is on the outside, the spherical mirror is convex.
When light reflects off of a concave spherical mirror, the rays will converge onto a single
point – this is called the Focal Point. The focal point for a spherical mirror is always one
half of the radius of the sphere.
Equation 2-3
F=½C
Images Using Spherical Mirrors – Graphical Method (Ray Tracing)
If you were to look into a flat mirror, the image you see would be the same size as you, it
would be upright, and it would be the same distance behind the surface of the mirror as
you are standing in front of it. This is not true for spherical mirrors.
To determine the location of the image, the size of the image, and the orientation of the
image - whether the image is upright or inverted (up-side-down) - one can use either a
graphical or analytical method. In other words, to determine the characteristics of the
image in a mirror, you can either draw it or calculate it. It is extremely important,
however, that you take the time to use the graphical method before using the analytical
method - mainly because it is practically fool-proof.
We call the graphical method to determine the characteristics of an image in a spherical
mirror Ray Tracing. We use three rays (from the infinite number actually coming from
the visible object) to help solve mirror design problems: these are the Parallel ray, the
Focal-point ray, and the Centre ray.
For convex or concave mirrors, we must know the focal point (or the radius of the
sphere). To scale, we must draw the object that is being reflected by the mirror, using the
correct height of the object at the correct distance from the surface of the mirror. We may
then draw the three important rays – where they intersect indicates the location, size, and
orientation of the image.
Convex Mirrors
For example, note the following convex mirror:
The object stands in front of a convex mirror. Consider first the P-ray: it leaves the top of
the object as a parallel ray, and when it hits the convex mirror it is reflected. The
reflected ray passes through the focal point of the spherical mirror. Even though the ray
would be reflected up and away to the left in reality, there is a ‘virtual’ direction for the
ray which passes through the focal point.
Now consider the C-ray: it leaves the top of the object and travels straight through the
mirror towards the center of the sphere. In reality, this ray would reflect directly back on
the path that it arrived, as it is hitting the surface along the normal plane. The ‘virtual’
path of the ray will pass directly through the center of curvature for the mirror.
And finally, consider the F-ray: it leaves the top of the object and travels towards the
focal point of the mirror. When it hits the surface of the mirror it is reflect parallel to the
datum. In reality the ray is reflected back towards the object, but a ‘virtual’ ray passes
horizontally through the mirror.
The point where these three rays intersect is the top of the image. In this case, the image
is closer to the surface of the mirror than the object, it is smaller than the object, and the
image is upright. Because the image appears to be inside or behind the surface of the
mirror, it is called a Virtual Image.
The image from a convex mirror will always be upright, reduced in size, and virtual.
Summary
Object
Size
Orientation
Type
Anywhere
Smaller
Upright
Virtual
Concave Mirrors
Now, let’s trace the rays using a concave mirror.
As
with the convex mirror, the parallel ray travels horizontally to the surface of the mirror,
and is then reflected through the focal point of the mirror. The F-ray travels through the
focal point of the mirror to the surface where it is reflected parallel to the datum. And the
C-ray passes directly through the center of curvature for the mirror (the radius of the
sphere).
The image will be located at the point where the three rays intersect. In this case the
image is smaller than the object, closer to the surface of the mirror than the object, and
inverted. The image is called a Real Image because it is located on the same side of the
mirror as the object.
What would happen to the image if we moved it closer to the mirror – for example,
between the center and the focal point?
When the object is between the center of radius and the focal point, the image will be
larger than the object, inverted, and will appear farther away from the surface of the
mirror than the object. The image in this example is ‘real’.
If we move the object even closer to the mirror …
Self Check
How would you describe this image?
For a Concave Mirror, you can expect the following image characteristics depending
where the object is located:
Summary
Object
Size
Orientation
Type
Beyond C
At C
Between C & F
Just beyond F
Just inside F
Between F & surface
Smaller
Same
Larger
Nearing infinite
Nearing infinite
Larger
Inverted
Inverted
Inverted
Inverted
Upright
Upright
Real
Real
Real
Real
Virtual
Virtual
Images Using Spherical Mirrors – Analytical Method
There are two formulae that are used to solve optical problems analytically. These are:
Equation 2-4: The Mirror Equation
1 + 1 = 1
di
do
F
where: di
do
F
= distance from mirror to the image
= distance from the mirror to the object
= focal point
Equation 2-5: Magnification Equation
m
where: hi
ho
m
=
hi
ho
=
-di
do
= height of the image
= height of the object
= magnification
The reason we encourage you to use the graphical method to estimate the size, location
and orientation of the image is that it is quite easy to make errors using the analytical
method. The most common source of error is using the correct signs for the variables.
The following table outlines the sign rules.
Focal length
F is positive for concave mirrors
F is negative for convex mirrors
Magnification
m is positive for upright images
m is negative for inverted images
Image Distance
di is positive for real images
di is negative for virtual images
(Remember, real images are images that are located on the same side of the mirror as the
object, and virtual images appear to be located behind the surface of the mirror)
Object Distance
do is positive for objects in front of a mirror (real objects)
do is negative for objects behind the surface of the mirror (virtual objects)
(Note: do is typically found in multiple mirror systems in which one mirror is reflecting a
virtual image to another mirror)
EXAMPLE – Concave Mirror
Using a concave mirror with a focal point of 25 cm, we would like to create a virtual
image that is twice the size of the original object. The original object is 12 cm high.
First Step
– think about the problem …
Looking at the summary chart for a concave mirror, to create an enlarged virtual image
the object will be located between the focal point and the mirror. We know we can use a
concave mirror for this application.
Second Step
– extract the variables from the problem (there are only six possible
variables).
F
di
do
hi
ho
m
= 0.25 m
=?
=?
=?
= 0.12 m
=2
Third Step
- check signs
F is positive for concave mirrors.
m is positive for upright images (we know it is upright from the summary table).
di is negative for virtual images (we know it is virtual from the summary table).
do is positive because it is a real object.
Fourth Step
- solve analytically
m = -di/do
+2 = - di /do
… di = - 2do
Substitute into the Mirror Equation
1/do + 1/di = 1/F
1/do + 1/(-2do) = 1/(0.25)
… 1/do – 0.5/do = 4
0.5/do = 4
do = 0.125 m (the object is located half way between the focal point and the surface of
the mirror. This seem reasonable if you consider our thoughts in Step 1)
Fifth Step
- use the graphical method to check your answer.
By locating the object at 0.125 m (indicated by the analytical solution), the image will be
twice the size, upright and virtual. We can be confident that our solution is correct.
Self Check – try the following problems on your own.
1. You notice a convex mirror mounted on the ceiling at your local organic food
store. The clerk tells you that she cut the mirror from a sphere with a diameter of
2 m. The store is roughly 10m wide and 25 m long, and the ceiling is
approximately 4 m high. Your image in the mirror is about 10cm high and you are
1.85 m tall. How far away are you from the mirror. (Caution: you don’t need all
of the information I have provided to solve the problem).
2. An box that is 35 cm high, 20 cm wide and 15 cm long sits 28 cm from a concave
mirror. A real image is formed with a height of 47 cm. What is the focal point of
the mirror?
3. You are driving and look in the side mirror which states: “Objects appear closer
than they are”. The virtual image is upright and is approximately 5% of the size of
the actual object. It appears to be 15 m behind the surface of the mirror. How far
behind you is the real object?
Self Check Solutions
1. In the convex mirror your image is smaller, upright and virtual. Referring to the
summary table for a convex mirror, this seems reasonable.
F
do
di
hi
ho
m
=½C
=?
=?
= 0.10 m
= 1.85 m
=?
= ½ (2 m / 2) = 0.5 m
Checking signs:
F is negative for convex mirrors.
m is positive for upright images
di is negative for virtual images
do is positive for real objects
Using the Magnification Equation:
hi/ho = -di/do
… do = -di * ho / hi
do = - di (1.85) / (0.1)
… do = -18.5 di
Using the Mirror Equation:
1/do + 1/di = 1/F
1/ (-18.5 di) + 1/ di = 1/(-0.5)
… 0.946 / di = -2
di = -0.47 m (a negative answer is what we would expect from the sign rules)
do - -18.5 * (-0.47) = 8.75 m
You are standing roughly 8.8 m from the convex mirror.
Check by drawing this scenario.
The graphical solution shows a virtual, upright image at approximately 10 cm
compared to the original object with a height of 1.85 m. The virtual image is located near
the focal point at about 0.5 m behind the surface of the mirror.
2. A concave mirror that produces a larger, real image will require that the object be
located between the C and F for the mirror. The image will be inverted according
to the summary table for a concave mirror.
F
do
di
ho
hi
m
=?
= 0.28 m
=?
= 0.35 m
= 0.47 m
=?
Checking signs:
F is positive for concave mirrors.
m is negative for inverted images (therefore hi will have a negative sign)
di is positive for real images
do is positive for real objects
Using the Magnification Equation:
hi/ho = -di/do
… di = -do * hi / ho
di = - (0.28) * (-0.47) / (0.35)
… di = +0.376 m
Using the Mirror Equation:
1/do + 1/di = 1/F
1/ (0.28) + 1/ (0.376) = 1/F
… 6.231 = 1/F
F = 0.16 m (a positive answer is what we would expect from the sign rules)
The mirror you require will have a focal point of 16 cm.
Check by drawing this scenario.
The image shown is real, inverted, and larger than the original object, as expected.
3. In the convex mirror your image is smaller, upright and virtual. Referring to the
summary table for a convex mirror, this seems reasonable.
F
do
di
hi
ho
m
=?
=?
= 15 m
=?
=?
= 5% = 0.05
Checking signs:
F is negative for convex mirrors.
m is positive for upright images
di is negative for virtual images
do is positive for real objects
Using the Magnification Equation:
m = -di/do
… do = -di / m
do = - (-15) / (0.05)
… do = 300
The real object is 300 m in front of the surface of the mirror.
Check by drawing this scenario.
The graphical solution shows a virtual, upright image at considerably smaller than the
original object. If you were to draw this to scale the results would closely match the
solution found using the analytical method.