Mirrors and lenses are the basis of many optical instruments. In this chapter we will look at how mirrors and lenses form images, and the properties of these images. Here we will establish the basics of optics, more complex optics are discussed in Chapter 25.
This is not intended to be a philosophical question, but a practical one. For the study of optics that we are doing now, what are the characteristics of an "image"?
We will use a vertical arrow to represent an arbitrary object. An image of the arrow is formed when several rays trace straight paths from the arrow to your eye. [Drawing] Obviously, these rays must be at rather small angles to each other in order to enter your pupil. However, in the interest of clarity, we can exaggerate the size of an eye, and the angles of the rays -- in fact, the presence of an eye is not really necessary for the definition, but is just a convenience.
We use an arrow because we want to know not only the position of the image, but its magnification (height of the image divided by height of the object), and its relative orientation (up or down). An arrow allows us to easily see these last two points.
Let's begin with the simplest optics arrangement possible. Consider what happens when the arrow is placed in front of a flat mirror. Light rays originating at the top of the arrow move in a straight line until striking the mirror, where they reflect following the law of reflection.
The rays coming from the mirror seem to originate from a point behind the mirror. That is, if we trace the rays backward in straight lines, they converge at a point behind the mirror. This is the image of the tip of the arrow, and leads to a slight modification of the definition of "seeing": an image of the arrow is formed when several rays seem to follow straight paths from the "image" to your eye. The "image" of an object is the apparent origin of the rays. If the rays actually pass through the image location, then we call it a real image. If the rays don't actually pass through the image location, as in this example with a flat mirror, then we call it a virtual image. Real images can be displayed on a screen, like the image of my transparency that is projected on the screen in lecture.
Carrying out the same procedure for the bottom of the arrow, we find that it too has an image located behind the mirror, directly below the image of the top of the arrow. We could do this for all the points in-between, and we would find that the image is a faithful reproduction of the actual arrow.
Where does the image appear to be located? From a construction of light rays as shown in Figure 23.2, we can show that the distance from the image to the mirror is the same as the distance from the object to the mirror. And the same construction leads to the conclusion that the image is the same size as the object. The relative sizes of the image and object is called the magnification, M. For a flat mirror, M = 1.
It does seem as if the image is identical to the object. But there is an important difference -- left and right are reversed. To see this, simply look at yourself in a mirror, and move your right hand -- your image moves his/her left hand! This fact doesn't have much direct use for the course, but it is made use of in quantum physics, and goes by the name "parity transformation".
A concave mirror is just a reflective surface on the inside of a curved shell. The curved shell is approximately spherical, with radius R. The most desirable shape is a "parabaloid of revolution", or parabolic for short. You may be familiar with several common examples of parabolic mirrors: parabolic satellite TV dishes, parabolic antennas for radar, or the parabolic reflectors used in expensive telescopes. Although only the telescope reflects visible light, all of these are examples of parabolic mirrors. For small angles, the parabolic shape is very, very close to the spherical shape, and it is much easier to cut or grind a spherical shape, resulting in the common use of spherical mirrors for focusing lamp light into beams, magnifying mirrors, etc.
Let's consider the geometry of a concave mirror. We'll align the mirror vertically. Concave mirrors are designed such that a beam of horizontal light rays, when reflected, pass through a common point called the focal point.
If we pick one of these "focused" rays and simply reverse its direction, it must reflect off the mirror and follow a horizontal path back out. So, a ray drawn horizontally from the object to the mirror reflects and passes through the focal point of the mirror, and a ray drawn from the object through the focal point will reflect from the mirror and travel horizontally. The image is located where the two reflected rays intersect or appear to intersect.
These two facts allow us to determine the location of an image in a concave mirror. But first, we'll add a third condition which allows us to check the result. The third condition uses the fact that a ray with a 0° angle of incidence will reflect at a 0° angle. Thus, the incident and reflected ray lie on the same line, and both are normal to the surface of the mirror. Using the spherical mirror approximation, we can use the fact that any line perpendicular to the surface of a sphere (circle) passes through the center of the sphere. So, our third condition is that a line drawn from the object through the center of curvature reflects off the mirror and travels back along the same path.
A line drawn through the center of curvature and the focal point is called the "principal axis". When the mirror is aligned vertically, the principal axis is horizontal. The focal point lies halfway between the radius of curvature and the mirror.
In geometric optics, distances are measured from the mirror or lens -- to be precise, from the intersection of the principal axis with the mirror or lens (center). The distance from the mirror or lens to the focal point is called the focal length, f. If the radius of curvature is R, then f = R/2 for a concave mirror. The distance to the object is called p, and the distance to the image is called q.
As derived in section 23.2, the magnification of the image is given by:
M = h'/h = -q/p
.
This means simply that the size of the image is proportional to its apparent distance.
The negative sign means that the image is inverted relative to the object if both p and q are positive.
The position of the image is given by:
1/p + 1/q = 1/f.
For a concave mirror, f > 0, and p > 0 for a real object.
Since 1/q = 1/f - 1/p, q > 0 if p > f, but q < 0 if p < f.
A convex mirror, sometimes called a diverging mirror, is a reflecting surface on the outside of a sphere. For convex mirrors, we can use the results for concave mirrors, with the following "rule": anything located behind the mirror is at a negative distance from the mirror. We already used this rule with the concave mirror in the case when the object is located between the focal point and the mirror.
Applying this rule results in the following:
Thin lens is the term used to describe common optical elements made by grinding glass or plastic to produce a shape with one or two spherical surfaces. Thin lenses come in a variety of shapes, as shown in Figure 23.22. We classify lenses as converging or diverging depending on their shape. Converging lenses are thicker in the center than at the edges, and diverging lenses are thinner in the center than at the edges.
We define the focal point for lenses in a manner similar to the definition for mirrors. If a lens is aligned vertically, then a horizontal beam of light rays will converge, or appear to converge, at the focal point of the lens. But, we make one change in the sign convention, due to the fact that light passes through a lens rather than reflecting from it: if the focal point is on the same side of the lens as the outgoing rays, then the focal length is positive, and if the focal point is on the opposite side as the outgoing rays, then the focal length is negative.
The same sign convention holds for image distance, q. The sign convention for object distance is the opposite from above, but generally speaking, we will be working with real objects for which p is always positive.
One further difference between mirrors and lenses is that lenses have two focal points, one on each side of the lens. You can think of one point as the focus of horizontal rays incident from the left, and the other as the focus of horizontal rays incident from the right.
Ray diagrams for lenses are drawn in a manner very similar to those for mirrors. Thin lenses really are thin, and when making ray diagrams, we will replace the vertically aligned lens by a vertical line. Rays will travel straight until they reach this line, at which point the refraction occurs and the rays bend.
Begin by aligning the lens vertically, and draw a vertical lens line through the center of the lens. Draw a horizontal line also through the center of the lens; this is the principal axis of the lens. There are three rays that are easily drawn to find the location of the image:
Applying this to thin lenses, we find that the magnification is:
M = h'/h = -q/p
and the position of the image is given by:
1/q + 1/p = 1/f
exactly as we had for mirrors.
A dentist uses a mirror to examine a tooth. The tooth is 1.00cm in front of the mirror, and the image is formed 10.0cm behind the mirror. Determine (a) the mirror's radius of curvature and (b) the magnification of the image.
(a) Since this is a spherical mirror, p = 1.00cm, q = -10.0cm. Therefore, 1/f = 1/q + 1/p = 1/1 - 1/10 = 9/10 such that f = 10/9 cm = 1.11 cm. R = 2f = 2.22cm.
(b) M = -q/p = 10/1 = 10.0. These results can be checked with the aid of a ray diagram.
A diverging lens has a focal length of 20.0cm. Locate the images for object distances of (a) 40.0cm, (b) 20.0cm, and (c) 10.0cm. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.
Since this is a diverging lens, f = -20.0cm [drawing].
(a) For p = 40.0cm, then 1/q = 1/f - 1/p = -1/20 - 1/40 = -3/40, or q = -40/3 cm = -13.3 cm. M = -q/p = 13.3/40.0 = 0.333. Since q is negative, the object is virtual, and since M is positive it is upright.
(b) For p = 20.0cm, 1/q = -1/20 -1/20 = -1/10, so q = -10.0cm. M = -q/p = 10.0/20.0 = 0.500. Since q is negative, the object is virtual, and since M is positive, it is upright.
(c) For p = 10.0cm, q = -6.67cm, M = 0.667, the object is virtual and upright.
Where must an object be placed to have no magnification, |M| = 1.0, (a) for a converging lens of focal length 12.0cm, and (b) for a diverging lens of focal length 12.0cm?
(a) For a converging lens, f = +12.0cm. |M| = 1.0 means that q = ± p. We can eliminate the minus sign since then 1/q + 1/p = 0, unless p = 0! Thus, q = p, and 2/p = 1/f, so p = 2f = 24.0cm.
(b) For a diverging lens, f = -12.0cm. For same reason as above, take q = p, such that p = 2f = -24.0cm. Note that this is a virtual object!
| type | focal length | equation | magnification | image type |
|---|---|---|---|---|
| flat mirror | n/a | q = -p | 1 | virtual |
| concave mirror | f = R/2, positive | 1/q + 1/p = 1/f | -q/p | real, inverted if p > f, virtual, upright if p < f |
| convex mirror | f = R/2, negative | 1/q + 1/p = 1/f | -q/p | virtual and upright |
| converging lens | f is positive | 1/q + 1/p = 1/f | -q/p | real, inverted if p > f, virtual, upright if p < f |
| diverging lens | f is negative | 1/q + 1/p = 1/f | -q/p | virtual and upright |