Thin Lenses and Mirrors

One equation relates object distance d_o, image distance d_i, and focal length f for all types of lenses and mirrors, but you need to know the sign convention: virtual means negative. The equation is 1/d_i +1/d_o = 1/f. This is derived most easily using similar triangles. If you want to find d_i, this might be more convenient: d_i = d_of/(d_o-f).

First, what is f? Shine parallel rays at an optical element (lens or mirror) and see what happens. If the light converges to a point, the distance from the center of the lens or mirror to that point is f, the focal length. If the light diverges after reflection off a mirror or after passing through a lens, f is defined to be negative, and you need to trace the rays backwards (red dotted lines in the sketches below) to the point where they appear to be coming from.

Converging types, f positive, convex lens, then concave mirror:

Diverging types, f negative, concave lens, convex mirror (in the red means a negative balance in finance):

Neither converging nor diverging: f of a flat mirror is infinite.

What if the rays of light are not parallel? Then we do not call it f. Call it an image distance (not a very interesting image, but …):

Next, a little discussion about images. Unless you are blind, you have experienced more images (on the retinas in your eyes) than we can count. But let us consider manufactured optical devices. A real image of an object is formed when light from the object focuses on a surface (even if there is not a physical surface there- if putting a screen there would reveal an image, then we say the real image is there).

If light through a lens or reflection off a mirror enables you to see what appears to be an object, you are looking at the image of the object. It could be real or virtual, depending on which side of the lens or mirror it is on. If real focusing occurs, it is after the light hit the optical element, so the real image side of the lens is on the side away from the object, and the real image side of the mirror is the object side. Usually images that you see directly (without a screen) are virtual, but under the right conditions you can see a real image in space.

Other ways to tell if it is real or virtual:
1. If there is one lens or mirror, an upright image is virtual. Real images are always inverted compared to the object.
2. Again for the single lens or mirror:
a) If it is a diverging type (concave lens or convex mirror), the image is virtual.
b) If it is a converging type (convex lens or concave mirror), the image is real if d_o > f and virtual if d_o < f.

Locating images by diagram: Draw two rays from a point on the object-
1. A ray parallel to the optical axis to the lens or mirror. (The optical axis is the perpendicular bisector of the lens or mirror.) This ray will then go through the principal focus if it is a converging type (positive f). If it is a diverging type (negative f), draw a dotted line to the virtual principal focus; the ray goes in the opposite direction.

2. A ray to the center of the lens goes straight through (not quite, but close enough). If it is a mirror, a ray through the center of curvature (distance 2f) or toward the center of curvature if it is convex, reflects back on the same line.

Here are some examples (and note that if the object is too tall we still draw a ray parallel to the axis even though this ray misses the lens or mirror. It is still a good gimmick for locating the image):

You might say that you see yourself in a mirror. That is fine for everyday life, but to clarify the physics of it, say that you see the virtual image of yourself which is located an equal distance behind the mirror (d_i = -d_o if it is a plane mirror, negative because it is virtual).

How to locate a virtual image experimentally:
Method 1, the parallax method: Point to where the image seems to be with a vertical pencil or other thin object above or below the image. Move your head from side to side, and observe the image and the pencil (not the image of the pencil). If one has more apparent motion across your field of view, it is closer. Adjust the location of the pencil until they move together, then the pencil is pointing at the object.

Method 2.Put in a convex lens and a screen to get a real image of the virtual image, then remove the lens or mirror responsible for the virtual image and find the where to put the object such that the real image will be in the same place. Then the object is where the virtual image was.

Two Lenses:
If they are very close together, you can treat the pair as one lens with focal length equal to the product divided by the sum (and note that if your memory fails and you think it might be sum/product, the units will tell you which is correct).
For all other cases, determine the location of the first image as if the 2^nd lens did not exist. Then treat the first image as an object. Say the light is traveling left to right. If the 1^st image is left of the 2^nd lens, then d_o2 is positive, and if it is to the right of the 2^nd lens, define d_o2 to be negative. Then the lens equation will work. Here are a couple of examples. In this first one, the first image is that inverted black arrow just beyond the 2^nd lens. The d_o2 is that small distance from there to the 2^nd lens, but make it negative. The 2^nd lens is a diverging lens, hence f₂ is negative, (the distance to the tiny square), and the lens equation will tell you that d_i2 is positive and therefore real.

In the one below, the final image is virtual. In these cases it is hard to make sense out of the diagrams. And it doesn't help that I messed up the colors. Above, the first rays drawn are black, below they are red. After locating the first image, draw two rays, one through the center of the lens straight toward the first image tip. The final image is somewhere on that straight line. Next, draw a ray going toward the 2^nd lens parallel to the axis toward the first image tip. It bends down in a direction such that the extension of it to the upper left passes thru the principal focus (the little square) of the 2^nd lens. The image is on that slanted line.

The above picture shows that you can sometimes see two images. The light that goes through lens 1 but misses lens 2 will form the real inverted image on the right. The light that goes thru both will result in the virtual upright image.

You are with your lover and say, "Ooh the thrill of experiencing the images of you on my retinas…." Oops, former lover.

Let me outa here (back to waves, that is), this guy is loony.

Other main physics pages:

mechanics
fluids, heat, electricity and magnetism
quantum
index

Comments, questions: fredrick.gram at tri-c.edu (replace at with @ and no spaces)