W9_parallel_resonance.eps

Week 12: Lenses and Mirrors 401

1

3

2

object

image

s

s’

f

Figure 160: Converging mirror withs= 25> f= 10.

α

y’ α

y

s’

s

Figure 161: Transverse magnification can be determined from the two right triangles formed with
the central ray as a hypoteneuse.

To compute the magnification of the image formed above, we note that:

tan(α) =−

y

s

=

y′

s

(980)

(where we rigorously follow the convention that counterclockwise rotation is positive to

assign the signs). We define the transverse magnificationmof a simple mirror (or lens)

is defined by the ratio of the image heighty′to the object heighty. If we rearrange the

terms in this expression, we obtain:

m=

y′

y

=−

s′

s

(981)

This expression is valid for all images obtained for any ideal lens or mirror.

Note that in this case, the image formed is real (because the light rays pass through the

actual object), inverted, and that the image formed is smaller than the original object.

Let’s look at two more possibilities for converging/concave mirrors.In figure (162), we

see an (upside down) object at a position betweenfand 2f. This range is the second

possibility for this kind of mirror, one that leads to amagnifiedreal image larger than

the object.

As before, 1/s′= 1/ 10 − 1 /15 = 1/30 sos′= 30 cm. The magnification ism=−′s/s=

− 30 /10 =−3. The image is again real and inverted (relative to the object), butin this

case the image is larger than the object.

Note that fors > f there is asymmetrybetween solutions with s > 2 f > s′ and

solutions withs′> 2 f > s, emphasized in the figure above by deliberately drawing the

object upside down so that it looks very much like figure (162). In factanyray diagram

involving real images can work both ways, withsands′(and the role of the object and

image) interchanged because 1/sand 1/s′appear symmetrically in the mirror/thin lens

equation.