# Geometric Optics

## Contents

#### Problems on Mirrors and Lenses

Problem : Where is the image of an object placed 7 centimeters away from a 5 centimeter focal length convex lens? Concave mirror? Are the images real or virtual in each case?

A convex lens has a positive focal length. We can apply the lens equation: 1/f = 1/s o +1/s i . This gives: 1/5 = 1/7 + 1/s i . Solving for s i we find s i = 17.5 centimeters. The image is real since s i is positive. A concave mirror also has a positive focal length so we get the same result (the image is real and 17.5 centimeters in front of the mirror).

Problem : Determine whether the images of the objects placed at the following positions will be upright or inverted. Concave mirror, s o = 2f ; convex mirror, f < s o < 2f ; concave lens, f > s o .

The orientation of the image is determined by the sign of the transverse magnification M T = . For a concave mirror the focal length is positive. Thus 1/f = 1/2f + 1/s i , and since 2f > f we have 1/f > 1/2f and s i must be positive, so the magnification must be negative (the object and image distances are both positive). Hence the first image is inverted. For a convex mirror the focal length is negative, thus -1/f = 1/s o +1/s iâá’1/s i = - (1/f + 1/s o) , where f and s o are now considered positive quantities. Thus s i must be negative, the magnification positive and the image upright. In the third case, concave lenses have a negative focal length, thus -1/f = 1/s o +1/s i , and we have the same situation as before, so the image is upright.

Problem : Determine whether the images of the objects placed at the following positions will be enlarged or diminished. Concave mirror, s o = f ; convex mirror, f < s o < 2f ; concave lens, s o > 2f . Also determine whether the images are real or virtual.

The magnification of the image is determined by the magnitude of the transverse magnification M T = . For a concave mirror the focal length positive, thus: 1/f = 1/f + 1/s iâá’1/s i = 0 . This implies that the image forms at an infinite distance--in other words there is effectively no image. A convex mirror has a negative focal length, so -1/f = 1/s o +1/s iâá’ - (1/f + 1/s o) = 1/s i . Since s o > f , 1/f > 1/s o , thus the left hand side is positive. But this clearly implies that | 1/s I| > | 1/s o| and that | s o| > | s i| and thus the image must be diminished. s i is negative so the image is virtual. In the last case, concave lenses have negative focal lengths so -1/f = 1/s o +1/s iâá’ - (1/f + 1/s o) = 1/s i . The term in parentheses is clearly greater than 1/s i , so | s i| < | s o| so the image is diminished. Once again s i is negative so the image is virtual.

Problem : Use what you know about geometric optics to decide what type of lens is used in a magnifying glass and describe how it works.

The type of lens used is a convex lens, usually with a fairly short focal length. A concave mirror would be no good since it always produces diminished images. From common experience with magnifying glasses you should know that the image appears on the same side of the glass as the object (the opposite side to your eye). Hence the image produced is virtual. The only regime in which a convex lens produces a virtual image is when s o < f . In this case 1/f = 1/s o +1/s iâá’1/f - 1/s o = 1/s i and since we know the right hand side to be negative (the image is virtual), | s i| > s o and the image is magnified, as required. A ray diagram should convince you of this result.

Problem : A diagram of a Keplerian telescope is shown in the figure below. Note that the objective lens and eyepiece lens share a focal plane. The image formed by the objective must be in the region beyond the focal plane of the eyepiece lens. Compute the transverse magnification of such a telescope in terms of the focal lengths of the two lenses.

Figure %: A ray diagram of a Keplerian telescope.

If we define x i for a convex lens as the distance between the image and the focus on the near side of the lens, and x o as the distance between the object and the nearest focus (there are foci on both sides of a convex lens), then | M T| = = = (the second equality holds because triangles F o OB and S 1 S 2 F o are similar.
Figure %: Geometry of a convex lens.
Also, however, we can write | M T| = = = since triangles AOF e and F e P 1 P 2 are similar. The magnification due to the objective lens is then M o = x i f o , and the magnification due to the eyepiece lens is M e = f e x o . But in this case it is clear from the diagram that x i = x o thus the total magnification is M = M e M o = .