Lethbridge College
Applied Physics (PHY143)
Unit 2 – Optics – Lenses
Lenses are typically glass or plastic that are transparent to allow the light to pass through
them. There are many types of spherical lenses that are used in designing optical
equipment, including meniscus lenses, plano- lenses that have one flat side, and doublelenses that are symmetrical. Like mirrors there are concave and convex lenses, as shown
below.
We will work with double-convex and double-concave in this lesson.
A convex lens will focus light, like a magnifying glass, onto a single point (which is the
focal point for the lens). A parallel ray, will always refract to the focal point.
Light will diverge when passing through a concave lens. If the ray is extended backward,
however, the rays will converge on the focal point of the lens, as shown.
Images Using Spherical Lenses –Graphical Method
Convex Lenses
Like a spherical mirror, lenses can produce real or virtual images. And like mirrors, the
image can be determined graphically or analytically. We will begin with the graphical
method of Ray Tracing.
In the graphical method, there are three rays that may be used to determine the location,
size and orientation of the image. These are the parallel ray (P-ray), the focal ray (F-ray),
and the midpoint ray (M-ray). Note that the P-ray and F-ray are the same as described for
mirrors – the M-ray is used for lenses.
When using the graphical method, the object and lens must be drawn to scale. Unlike
mirrors, however, the only the mid-point of the lens is important (so you don’t have to
draw the lens to scale).
As indicated on the ray diagram, the parallel ray leaves the object horizontally, and at the
midpoint of the lens, is refracted to the focal point on the other side of the lens. The focal
ray leaves the object and passes through the focal point on the same side of the lens, and
at the midpoint of the lens is refracted parallel to the datum. Finally, the midpoint ray,
passes straight through the lens midpoint and intersects the other two rays.
The image location, size and orientation may be measured from the point of intersection
of these three rays. If the image is located on the other side of the lens from the object, it
is called a REAL image. If it is located on the same side of the lens as the object it is
called VIRTUAL. (Note that this is opposite to mirrors).
Convex lenses produce REAL and INVERTED images (if the object is beyond the focal
point, which is typical). The size of the image will depend upon the location of the object
relative to the focal point of the convex lens.
Concave Lenses
To draw a ray tracing diagram for concave lenses, the same three rays are used, as shown
below.
The image for a concave lens is always VIRTUAL, UPRIGHT, SMALLER, and located
between the object and the lens.
Images Using Spherical Lenses – Analytical Method
There are two formulae that are used to solve optical problems analytically. These are:
Equation 2-9: The Thin-Lens 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 2-10: Magnification Equation
m
where: hi
ho
m
=
hi
ho
=
-di
do
= height of the image
= height of the object
= magnification
You might have noticed that these are the same equation used for mirrors. The difference
is in the sign-rules.
As with mirrors, we encourage you to use the graphical method to estimate the size,
location and orientation of the image in 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 convex lenses
F is negative for concave lenses
Magnification
m is positive for upright images
m is negative for inverted images
Image Distance
di is positive for real images (convex lenses)
di is negative for virtual images (concave lenses)
(Remember, real images are images that are located on the opposite side of the lens as the
object, and virtual images appear to be located on the same side of the lens as the object)
Object Distance
do is positive for real objects
do is negative for virtual objects
EXAMPLE – Convex Lens
There is an object that is 3.00 m high, 2.50 m wide and 1.25 m long. It is located 12 m in
front of a convex lens with a focal length of 3.25 m. Describe the image produced.
Step 1: Thinking
Using a convex lens, we would expect a real, inverted image. The size and location will
have to be determined.
Step 2: List the variables – like mirrors, there are only six of them.
F
di
do
hi
ho
m
= 3.25 m
=?
= 12 m
=?
= 3.00 m
=?
Step 3: Check signs
F is positive for convex lenses.
m is negative for inverted images (we know it is inverted, because it is a convex lens).
di is positive for real images (we know the image is real for convex lenses).
do is positive because it is a real object.
Step 4: Analytical Method
Using the Thin-Lens Equation
1/do + 1/di = 1/F
1/12 + 1/di = 1/(3.25) … 1/di = 1/(3.25) – 1/12
1/di = 0.224
di = 4.457 m (the object is located beyond the focal point, it is a real image and should
be positive)
Using the Magnification Equation:
hi / ho = -di / do
hi = 3.00 (-4.457) / 120
hi = - 1.114 m (the negative sign indicated an inverted image, which was expected)
Step 5: Check results using graphical method.
The ray diagram shows a real, inverted image at a location of about 4.5 m from the
midpoint of the lens. The height of the image is about one-third of the original object, so
approximately 1 m. The analytical results are correct.
Optical Instruments:
Now that we have studied lenses, let’s look at a few practical applications.
In both these instruments, there is an objective lens, which creates a real, inverted image
inside the instrument. The focusing lens is used to adjust the location of the image so that
it is a clear image for the eyepiece. The transit includes an erector, which are convex
lenses that take the inverted image and makes it upright for the eyepiece. The eyepiece,
like a microscope, views the real image made by the objective lens and magnifies it as a
virtual image.
Note the convex and concave lenses that are placed next to each other. Why do the
optical designers do this?
Remember that when light refracts through a medium, the blue to red wavelengths tend to
spread out into a rainbow. The convex lens will cause this splitting of the visible
spectrum, while the concave lens bends them back together. This provides the instrument
with a sharp image. Without this correction, the image would be blurry like watching a 3D movie without the special glasses – we call this distortion a Chromatic Aberration.
Self-Check
1. In designing a slide projector, you want to take a slide with a height of 10mm and
a width of 15mm, and magnify by 100 times onto a screen. The object is located
0.33 m from the lens. What focal length of a lens is required, and where should
the screen be located?
2. An real image with a height of 2.50 m us produced 4.45 m from a lens with a
focal point of 0.55 m. Where is the object located, and what was the original size
of the object?
3. A virtual image is produced by a lens that is one half the size of the original
object. The focal point of the lens is 0.35 m. What is the location of the object and
the image relative to the lens?
4. An object is set 1.20 m from a lens, and an image is created 2.40 m on the other
side of the lens. What is the focal point of the lens, and the magnification?
Self-Check Solutions:
1. Since we are projecting a real image using a slide projector, we know it is a
convex lens. For convex lenses, the images are real and inverted.
F
di
do
hi
ho
m
=?
=?
= 0.33 m
=?
= 0.01 m
= - 100
F is positive for convex lenses.
m is negative for inverted images (we know it is inverted, because it is a convex
lens).
di is positive for real images (we know the image is real for convex lenses).
do is positive because it is a real object.
Using the Magnification Equation:
m = hi / ho = -di / do
-100 = - di / 0.33
di = 33 m (the image will be located 33 m from the lens)
Using the Thin-Lens Equation
1/do + 1/di = 1/F
1/0.33 + 1/33 = 1/F
1/F = 3.06
F = 0.327 m
(the focal length of the convex lens will be 0.327 m)
It is recommended that you draw a ray diagram to check these results.
2. Since we are projecting a real image, we know it is a convex lens. For convex
lenses, the images are real and inverted.
F
di
do
hi
ho
m
= 0.550 m
= 4.45 m
=?
= -2.50 m
=?
=?
F is positive for convex lenses.
m is negative for inverted images (we know it is inverted, because it is a convex
lens).
di is positive for real images (we know the image is real for convex lenses).
do is positive because it is a real object.
Using the Thin-Lens Equation
1/do + 1/di = 1/F
1/do + 1/4.45 = 1/0.55
… 1/do = 1.818 – 0.225
1/do = 1.593
do = 0.628 m (the object is located 0.628 m from the lens)
Using the Magnification Equation:
m = hi / ho = -di / do
-2.50 / ho = -4.45 / 0.628
ho = 0.353 m (the object is 0.353 m high)
3. Since we are projecting a virtual image, we know it is a concave lens. For
concave lenses, the images are virtual, upright, and reduced in size.
F
di
do
hi
ho
m
= - 0.35 m
=?
=?
=?
=?
= 0.5
F is negative for concave lenses.
m is positive for upright images (we know it is upright, because it is a concave
lens).
di is negative for virtual images (we know the image is virtual for concave
lenses).
do is positive because it is a real object.
Using the Magnification Equation:
m = hi / ho = -di / do
0.5 = - di / do
di = - 0.5 do
Using the Thin-Lens Equation
1/do + 1/di = 1/F
1/do + 1/di = 1/(-0.35 )
Substituting di - -0.5 do into the Thin-Lens Equation
1/do + 1/(-0.5 do) = 1/(-0.35)
… 1/do – 2/do = -2.857
1/do = 2.857
do = 0.35 m
(the object is located on the focal point)
di = - 0.5 (0.35) = -.18 m
(the image is located half way between the lens and
focal point)
The ray diagram supports the solution using the analytical method.
4. Since the image is on the other side of the lens, we know it is a convex lens. For
convex lenses, the images are real and inverted.
F
di
do
hi
ho
m
=?
= 1.20 m
= 2.40 m
=?
=?
=?
F is positive for convex lenses.
m is negative for inverted images (we know it is inverted, because it is a convex
lens).
di is positive for real images (we know the image is real for convex lenses).
do is positive because it is a real object.
Using the Magnification Equation:
m = hi / ho = -di / do
m = - 2.40 / 1.20
m = -2.0 (the image half the size as the object, the negative sign was predicted)
Using the Thin-Lens Equation
1/2.40 + 1/1.20 = 1/F
0.417 + 0.833 = 1/F
1/F = 1.25
F = 0.80 m
(the focal length of the convex lens will be 0.80 m)