CK-12-Physics - Intermediate

http://www.ck12.org Chapter 20. Geometric Optics

To the eye in the figure, the rays appear to converge to a point behind the mirror shown by the dashed lines.

Ray 1: A ray that is parallel to the principal axis and is reflected through the focal point.

Ray 2: A ray continued backward from the object, is in line with the focal point, and so reflects off the mirror parallel
to the principal axis.

Ray 3: A ray continued backward from the object, is in line with the radius of curvature, and so reflects back upon
itself.

Sign conventions for spherical mirrors

There are several sign conventions which must be followed when working with spherical mirrors.

1. All distances are measured with respect to the mirror.


2. The object heighthois always positive.


3. The image heighthiis positive if it is right-side-up (upright) and negative if it is inverted (relative to the
   object).


4. An object or image on the reflecting side of the mirror has positive distance and is real.


5. An object or image behind the mirror has negative distance and is virtual. The lateral magnificationmof an
   object is defined as the ratio of the image heighthiof the object to the actual heighthoof the object, and, it
   can be shown, equals the negative of the ratio of the image distancedito the object distancedo

m=hhoi=−ddoi

InFigure20.11, the image is on the reflecting side of the mirror, reduced, and inverted. As a consequence of these
results,hiis negative anddiis positive. Thus, the magnification equation yields consistent results, regardless of
whether the heights or the distances are used.

Concluding remarks for concave mirrors

An object placed closer to the mirror than the focal point always produces an upright, enlarged, and virtual image.

An object placed beyond the focal point always produces an inverted, real image.

http://demonstrations.wolfram.com/RayDiagramsForSphericalMirrors/

Illustrative Example 21.2.1

An object of height 10.0 cm is positioned 30.0 cm from a concave mirror with a focal length 8.0 cm.

a. Find the image position.

Solution:

We are givenho= 10. 0 cm,do= 30. 0 cm,and f= 8. 0 cm

Using the mirror equation:

1

do

+

1

di

=

1

f

→

1

30. 0 cm

+

1

di

=

1

8. 0 cm

→

1

di

=

1

8. 0 cm

−

1

30. 0 cm

=

30. 0 − 8. 0

240 cm

→

di= 10. 91 → 10. 9 cm;

di= 10. 9 cm

Since the image distance is positive, the image is real.