
In this section we deal with a very important aspect of concave lenses, namely how and where they form images. First, a concave lens is thinner in the center than it is near the edges as shown in this diagram:DEFINITION 1:
The focal point of a concave lens is the point where light rays parallel to the axis seem to diverge from after passing through the lens. The distance from the lens to this point is called the focal length of the lens. Because the light rays never really diverge from this point, this is often referred to as a virtual focal point.
DEFINITION 2:
Light rays originally headed towards the focal point on the other side of a concave lens are bent, emerging parallel to the axis. This is just the reverse of the definition above, but will be very important in determining how and where images are formed by a concave lens.
BIG NOTE:
A concave lens has two focal points  one on each side. They are equal distances from the lens. The lens does not need the same curvature on both sides for this to be true, and it doesn't depend on the direction the light takes entering the lens. The combined curvature determines the focal point.
ON TO IMAGES:
In our diagram, we see an object arrow placed some distance in front of a concave lens. The two focal points are indicated as f' since they are virtual focal points. We now consider two of the many light rays being emitted by the tip of the object arrow.
The light ray which was originally parallel to the axis is bent and appears to now be coming from the first virtual focal point. If we are on the right side of the lens, that is how it appears to us. The dotted line is the apparent path while the solid line is the real path.
In this diagram we see the light ray that was originally headed towards the far focal point. It is bent and emerges parallel to the axis. The next diagram puts those two rays together in the same diagram
From our diagram we see that the two light rays do not come together, they do not converge. Therefore there is no place where we could put a screen and see a real image. Let's imagine that we are on the righthand side of this lens. Looking through it towards the object, the light we see is represented by the solid lines. In our minds we trace them both back to where they seem to have originated, giving us this diagram:
The apparent source of the two rays we traced is this point behind the lens as shown by the hand. In fact, if we were to take all of the light from the tip of the arrow that went through the lens, all of the rays would appear to have originated in the same location. This is shown here, where colors help to sort out real rays (red) from apparent rays (green).
All the rays from the tip of the arrow appear to be coming from the original intersection point. Likewise, if we were to keep track of all the rays from the middle of the object they would appear to come from a point halfway between the axis and the point above. And all the rays from the tail of the object arrow would appear to come from the axis. Therefore we conclude that the image of our original object appears to be located as shown in this diagram:
So what are the characteristics of our perceived image?
 It is smaller than the original object, or we say it is reduced.
 It is closer to the lens than the original object, but it's on the same side.
 It is not a real image as it cannot be focused onto a screen.
 We call it a virtual image to represent that it is an apparent image.
 The image is rightside up. We call this upright as opposed to inverted.
GEOMETRY TIME:
It might surprise us to find that the same equation that governs the images in convex lenses works for concave lenses. But this is PHYSICS, so nothing should be too surprising. The diagrams which preceded are redrawn to show first the quantities we would typically measure. Then several triangles are drawn which turn out to be similar. When we are done, we will consider what to do with positive and negative quantities.
Quantity Meaning D_{o} Distance from lens to object
H_{o} Size of object (height)
D_{i} Distance from lens to image
H_{i} Size of image (height)
f Focal Length
In the two triangles formed above, the larger one being the lighter shading plus the darker, we see similar triangles if the object and image are perpendicular to the axis. From the geometry of similar triangles, we get the following relationship:
Hi / Ho = Di / Do We recognize this as the magnification equation stating the ratio of image size to object size. Because we get reduced images, the magnification will always be less than 1. There is a second problem, however. The image location is on the same side as the object, and is determined to be negative. The end result is a slight modification of our equation:
Hi / Ho = Di / Do
Now we see similar triangles if we extend the light ray heading toward the far focal point. The basic relationship will be Ho/Hi = (Do + f)/f. But the focal length of a concave lens is negative because it is a virtual focal point rather than a real one. Therefore our measurement for f must have f substituted, becoming:
Ho / Hi = (Do  f) / f In these two similar triangles, the larger being the lighter plus the darker shading, we get Ho/Hi = f/(fDi). The focal length of a concave lens is negative because it is a virtual focal point rather than a real one. Therefore our measurement for f must have f substituted. Additionally, the image is formed on the same side of the lens as the object, so Di is a negative number, making the measurement we want (f + Di) as:
Ho / Hi = f / (f + Di) We set the two righthand quantities equal to each other since they are equal to the same lefthand quantity:
(Do  f) / f = f / (f + Di) Cross multiply and expand the binomials:
f^{2} = (Do  f) (f + Di) = f Do  f Di + f^{2} + Do Di Subtract the f^{2} terms, move the DoDi term to the left while factoring out f
Do Di = f (Di + Do) We now do some mathematical magic. We take the negative of both sides, divide both sides by f and also by Do Di. Then we simplify the remaining terms:
1/f = (Di + Do) / DoDi = Di/DoDi + Do/DoDi = 1/Do + 1/Di 1/f = 1/Do + 1/Di
This equation is exactly the same as the one for convex lenses. However, when applying it to concave lenses, we note that the focal length must always be considered negative, and the arithmetic will generally result in Di being negative also. Note the next section which discusses signs for lenses forming images. It is very important.
SIGNS:
For lenses, we develop a sense of signs in our mathematics. It comes from the direction that light travels through the lens. In the diagram which follows, note that the positive sense of things occurs when light starts on one side and converges on the other.
Object distance is positive
Object distance is negative (very rare)
Image distance is negative
Image is upright
Image is reduced (concave)
or enlarged (convex)
Image is virtualImage distance is positive
Image is inverted
Image is real
Image can be enlarged or reducedFocal point is negative
Light rays are diverged by lensFocal point is positive
Light rays are converged by lensSAMPLE PROBLEMS:
Problem 1
Joe bought a concave lens having a focal length of 10 cm. He placed a lighted object 30 cm in front of the lens. Where will the image be formed? What type of image will it be?
Givens:f = 10 cm
Do = 30 cmUnknown:
Di = ?Equation:
1/f = 1/Do + 1/DiSolve:
1/Di = 1/f  1/Do = (1 / 10 cm)  1 / 30 cm = 4 / 30 cm = 0.1333 cm^{1}Di = 7.5 cm
The image is upright, reduced and virtual
Problem 2
The image formed by a concave lens for an object 40 cm away is reduced to one fourth its size. What is the focal length of the lens?
Givens:Do = 40 cm
Hi = 0.25 HoUnknown:
f = ?Equations:
1/f = 1/Do + 1/Di
Hi / Ho = Di / DoSolve:
Hi / Ho = 0.25 Ho / Ho = 0.25 = Di / Do ..... Di = 0.25 DoDi = 0.25 x 40 cm = 10 cm
1/f = 1/Do + 1/Di = 1/40 cm + 1/10 cm = 0.075 cm1
f = 13 cm
Problem 3
Jessica looks through a concave lens with a focal length of 20 cm at a tree that's a long distance away. We assume the distance is very large, or GBN. (Ask your math teacher.) Where is the image of Jessica's tree formed?
Givens:Do = GBN or °
f = 20 cmUnknown
Di = ?Equation:
1/f = 1/Do + 1/DiSolve:
1/Di = 1/f  1/Do = 1/(20cm)  1/° = 1/(20 cm)  0 = 1/(20 cm)Di = 20 cm, or the image is formed at the focal distance from the lens on the same side as the tree. Jessica will be looking through the lens and focusing her eye 20 cm on the other side.