SPH3UW
Notes
Unit 7.7 Concave Lens
Page 1 of 7
Physics Tool box



Thin Lens – is an optical system with two refracting surfaces. The most
simplest thin lens contain two spherical surfaces that are close enough
together that we can neglect the distance between them.
Converging Lens – a lens that has the rays converge to a single point, such
as a double convex lens.
Diverging Lens – a lens that has parallel rays diverge, thus giving the lens
a negative value, such as a double concave lens.
1 1 1
 
so si f
h
s
s
 Lateral Magnification - m  i   i  
ho
so
s

Object-Image Relation -
Properties of a Converging Lens
The properties of a double concave lens (called a diverging lens or negative lens) allow a beam of
parallel rays that pass through the lens to refract away so that the rays when projected backward
would come of a point called the focus, f which is one the same side as the original parallel rays.
The same equation that works with the converging lens is used with a diverging lens.
1 1 1
 
s s f
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Unit 7.7 Concave Lens
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Sign convention
The sign convention for lenses is similar to that for mirrors. Again, take the side of the lens
where the object is to be the positive side. Because a lens transmits light rather than
reflecting it like a mirror does, the other side of the lens is the positive side for images. In
other words, if the image is on the far side of the lens as the object, the image distance is
positive and the image is real. If the image and object are on the same side of the lens, the
image distance is negative and the image is virtual.
For converging mirrors, the focal length is positive. Similarly, a converging lens always has
a positive f, and a diverging lens has a negative f.
The signs associated with magnification also work the same way for lenses and mirrors. A
positive magnification corresponds to an upright image, while a negative magnification
corresponds to an inverted image. As usual, upright and inverted are taken relative to the
orientation of the object.
Note that in certain cases involving more than one lens the object distance can be
negative. This occurs when the image from the first lens lies on the far side of the second
lens; that image is the object for the second lens, and is called a virtual object.
Real object (in front of lens)
Virtual object (behind lens)
Real image (behind lens)
Virtual image (in front of lens)
Centre of curvature (back of lens)
Centre of curvature (in front of lens)
Diverging lens
Converging lens
s is positive
s is negative
s' is positive
s' is negative
r1, r2 is positive
r1, r2 is negative
f is negative
f is positive
SPH3UW
Unit 7.7 Concave Lens
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Step-by-Step Method for Drawing Ray Diagrams
The method of drawing ray diagrams for a double concave lens is described below.
1. Pick a point on the top of the object and draw three incident rays traveling
towards the lens.
Using a straight edge, accurately draw one ray so
that it travels towards the focal point on the
opposite side of the lens; this ray will strike the lens
before reaching the focal point; stop the ray at the
point of incidence with the lens. Draw the second
ray such that it travels exactly parallel to the
principal axis. Draw the third ray to the exact
centre
of the lens. Place arrowheads upon the rays to
indicate their direction of travel.
2. Once these incident rays strike the lens, refract
them according to the three rules of refraction for double concave lenses.
The ray that travels towards the focal point will
refract through the lens and travel parallel to the
principal axis. Use a straight edge to accurately
draw its path. The ray which traveled parallel to the
principal axis on the way to the lens will refract and
travel in a direction such that its extension passes
through the focal point on the object's side of the
lens. Align a straight edge with the point of
incidence and the focal point, and draw the second
refracted ray. The ray which traveled to the exact
center of the lens will continue to travel in the same
direction. Place arrowheads upon the rays to
indicate their direction of travel. The three rays
should be diverging upon refraction.
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 three refracted rays intersect. Since the
three refracted rays are diverging, they must be
extended behind the lens in order to intersect.
Using a straight edge, extend each of the rays
using dashed lines. Draw the extensions until they
intersect. All three extensions should intersect at
the same location. The point of intersection is the
image point of the top of the object. The three
refracted rays would appear to diverge from this
point. This is merely the point where all light from
the top of the object would appear to diverge from
after refracting through the double concave lens. 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. See note below.
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Unit 7.7 Concave Lens
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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 which
is formed by the double concave lens. 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 lens as the image of the top of the object.
At this point the complete image can be filled in.
Some students have difficulty understanding how
the entire image of an object can be deduced once
a single point on the image has been determined. If
the object is merely a vertical object (such as the
arrow object used in the example below), then the
process is easy. The image is merely a vertical line.
This is illustrated in the diagram below. In theory, it
would be necessary to pick each point on the object
and draw a separate ray diagram to determine the
location of the image of that point. That would
require a lot of ray diagrams as illustrated in the
diagram below.
A diverging lens will always produce
a virtual image (Remember an eye
cannot distinguish the difference
between a virtual image or a real
image).
As the object moves closer and closer to the diverging lens, the image (which starts at the
focus as a dot) also moves closer and closer to the lens (it also increases in size as it
approaches the lens).
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Unit 7.7 Concave Lens
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Example
A 4 cm object is placed 30.0 cm from a diverging lens of focal length 10.0 cm. Where is
the image of the object produced? What is the size and orientation is the image?
Solution:
1 1 1
 
f so si
1
1
1


10cm 30cm si
1
1
1


si
10cm 30cm
4
cm
30
si  7.5cm

Therefore the image is a virtual image located on the same side as the object. The image
is 7.5 cm from the lens.
Now for the magnification.
m
si
so
7.5cm
30cm
 0.25

The positive value in the magnification indicates that the image is in the same orientation
as the original, but is ¼ the size or 1 cm in height.
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Unit 7.7 Concave Lens
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Example
A 4 cm object is placed at the focal point a diverging lens of focal length 10.0 cm. Where
is the image of the object produced? What is the size and orientation is the image?
Solution:
1 1 1
 
f so si
1
1
1


10cm 10cm si
1
1
1


si
10cm 10cm
2
cm
10
si  5.0cm

Therefore the image is a virtual image located on the same side as the object. The image
is 5.0 cm from the lens.
Now for the magnification.
m
si
so
5.0cm
10cm
 0.5

Notice that if this was a converging
lens, we would have an image located
at infinity.
The positive value in the magnification indicates that the image is in the same orientation
as the original, but is 1/2 the size or 2 cm in height.
SPH3UW
Unit 7.7 Concave Lens
Extra Notes and Comments
Page 7 of 7