An Introduction To The Geometrical Optics Of Spherical Mirrors

 As in the case of convex and concave lenses, light acted upon by any spherical mirror will conform to the laws of geometrical optics, thus we note the following in summary: i) any light ray which passes through the center of curvature (C) of a spherical mirror en route to the mirror surface, will be reflected back upon itself; ii) all light rays which approach the mirror in paths parallel to the optical axis are reflected through a common point on the optical axis known as the principal focus; iii) any light ray which passes through the focal point on its way to the mirror will be reflected parallel to the principal axis.

The basic law for image formation is:  1/s  + 1/s’ = 1/f.

Since the radius of curvature R = 2f, this can also be written as:

1/s  + 1/s’ = 2 /R

Which is called “the Mirror equation.”

The summary diagrams applicable to both concave and convex mirrors and their actions are given below:

                                            Image formation for a concave (converging) mirror.

                                                  Image formation for a convex (diverging) mirror.

Example Problem:

 

An illuminated object is placed 30 cm from a concave mirror. The image is found to be formed at a distance of 120 cm away from the mirror and to have a magnification of 4x. (E.g. h’ = 4h). Using this data, find: a) the focal length of the mirror and b) the radius of curvature. Also, (c), sketch a diagram showing the image formation based on the given data.

Solution:

We can use the mirror equation such that:

1/s  + 1/s’ = 2 /R

Where s = 30 cm and s’ = 120 cm, then:

1/30 + 1/ 120 = 2/R,  or:

R/2 =  (120 cm) (30cm)/ [120 cm  + 30 cm]

R/ 2 = 3600 cm²/ 150 cm = 24 cm

Since R = 2f then R/2 = f = 24 cm

So: R = 2 (24 cm) = 48 cm

A sketch of the arrangement in terms of a graphical solution (to scale) is shown below:

Note in particular that: 4h/h’ = 4 = s’/ s.

If an object is very far from the concave mirror, say effectively at infinity (s = ¥) then we have a situation peculiar to that for reflecting astronomical telescopes:

                                                          

For the case depicted here:

1/f = 1/s  + 1/s’ Þ 1/s =  1/ ¥  =  0  and s’ =  R/2

In the case of the convex or diverging mirror (see earlier ray diagram) light is reflected from an outer convex surface and the image produced is always erect, virtual and smaller than the real object. 

The sign conventions for all the possible object- image mirror problem combinations are summarized below:

                                                                          

Suggested Problem:

A small object lies 4 cm to the left of the vertex of a concave mirror who radius of curvature is 12 cm. Find the magnification of the image.  Include a basic sketch to show the relation of the image to the object position relative to the mirror.