Principles of Technology
CH 13 LIGHT AND GEOMETRIC OPTICS
2
Name________
KEY OBJECTIVES
• Describe how images are produced by spherical mirrors, and use the mirror equations to solve problems
relating to these images.
• Draw ray diagrams that illustrate image formation by plane and curved mirrors.
The principal applications of reflected light involve the use of plane and curved mirrors.
CURVED MIRRORS
When parallel rays of light strike the surface of a curved mirror, the light may either converge or diverge, as
illustrated. Concave mirrors cause the light rays to converge to a point known as the principal focus. Convex
mirrors cause light rays to diverge. If the diverging rays were projected backward, they would meet at a point
called the virtual focus.
Spherical Concave Mirrors
If an object is placed at various distances from a curved mirror, the mirror will produce different types of
images.
The diagram above represents an object placed in front of a spherical con cave mirror. Four special rays
emerging from the top of the object may be used to locate the image produced by the mirror.
1. Which statement is false?
a. When parallel rays of light strike the surface of a curved mirror, the light may either converge or diverge,
as illustrated.
b. Concave mirrors cause the light rays to converge to a point known as the principal focus. Convex mirrors
cause light rays to diverge.
c. If the diverging rays were projected backward, they would meet at a point called the virtual focus.
d. If an object is placed at various distances from a curved mirror, the mirror will always produce the
same exact image.
If a ray parallel to the principal axis (the line passing perpendicularly through the center of the mirror) strikes
the mirror, as illustrated above, the reflected ray will pass through the focus of the mirror. The distance
between the surface of the mirror and the focus is known as the focal length of the mirror (f).
The center of curvature (C) is located at a point one radius from the surface of the mirror. This distance is
equal to two focal lengths.
C=2f
2. Which statement is false?
a. If a ray parallel to the principal axis (the line passing perpendicularly through the center of the
mirror) strikes the mirror, as illustrated above, the reflected ray will pass through the focus of the
mirror.
b. The distance between the surface of the mirror and the focus is known as the focal length of the
mirror (f).
c. The focal length is always greater than the center of curvature.
d. The center of curvature (C) is located at a point one radius from the surface of the mirror. This
distance of the center of curvature is equal to two focal lengths.
If a ray passes through the center of curvature of the mirror, it will strike the mirror perpendicularly to its
surface and will be reflected back on itself.
We note in the diagram above that, since the center of curvature lies on the principal axis, the principal axis is
used as a ray to locate the base of the image.
If a ray passes through the principal focus, as shown below, the reflected
ray will emerge parallel to the principal axis.
The following diagram shows that, if a ray from the object strikes the mirror at the principal axis, the principal
axis acts as the normal and the ray will emerge at the same angle.
3. Which statement is false?
a. Any two rays along with the principal axis can be used to locate the image formed by a concave
mirror.
b. All images that can be produced by a spherical concave mirror are identical in all traits.
c. Many tables and diagrams are used to summarize the various types of images that can be produced by
a spherical concave mirror.
d. If a ray passes through the principal focus, the reflected ray will emerge parallel to the principal axis.
If a ray from the object strikes the mirror at the principal axis, the principal axis acts as the normal and
the ray will emerge at the same angle.
Note that real, as well as virtual, images can be produced by this type of mirror.
Spherical Convex Mirrors
In every case, a spherical convex mirror produces an erect virtual, image that is located behind the
mirror. The image is always smaller than the object. The diagram below shows how such an image is
formed.
We note the following:
1. When a ray parallel to the principal axis strikes the surface of a convex mirror, it is reflected from the
surface of the mirror in such a way that, if it is projected back behind the mirror, it passes through a virtual
focus.
2. If a ray strikes the mirror at an angle such that, if it were projected as a straight line, it would pass through
the virtual focus, it is reflected parallel to the principal axis.
3. If a ray strikes the mirror in such a way that, if it were projected as a straight line, it would pass through the
center of curvature, the ray is reflected back on itself.
4. For a ray that strikes the surface of the mirror at the principal axis, the principal axis serves as a normal to
the surface, and the ray is, therefore, reflected at the same angle.
In all cases involving single curved mirrors, real images are inverted (upside down and left-right
reversed), and virtual images are erect.
4. Which statement is false?
a. Real, as well as virtual, images can be produced by concave mirrors.
b. Real, as well as virtual, images can be produced by convex mirrors.
c. In every case, a spherical convex mirror produces an erect virtual, image that is located behind the
mirror. The image produced by a spherical convex mirror is always smaller than the object.
d. In all cases involving single curved mirrors, real images are inverted (upside down and left-right
reversed), and virtual images are erect.
MIRROR EQUATIONS
We can calculate the sizes and distances of the images formed by curved mirrors by means of the following
equations,
where h and h represent the respective size (heights) of the image and object, and d and d represent the
respective distances of the image and object from the mirror.
The following sign conventions are used for these equations: When the object, image, or focal point is “real”
(i.e., is on the reflecting side of the mirror), the corresponding distance is considered positive; when the
object, image, or focal point is “virtual” (i.e., behind the mirror), the distance is considered negative. (Virtual
objects occur only in problems in which more than one mirror is used. Only then can an object be behind the
mirror, that is, if it is the image produced by another mirror.)
5.
Calculate the distance of the image from the mirror (di) if an object whose height is 0.15 meter (ho)
is placed 0.60 meter (do) from a concave mirror and the image has a height of 0.050 m (hi).
(hi / ho) = (di / do)
a. 0.1 m
b. 0.2 m
c. 0.5 m
d. 1.0 m
6.
Calculate the focal length (f) if the distance of the image from the mirror is 0.25 m (di) if an object is
placed 0.75 meter (do) from a concave mirror.
(1 / f) = (1 / do) + (1 / di)
a. 0.19 m
b. 0.55 m
c. 1.0 m
d. 5.3 m
We can use the following equation to calculate the magnification (m) of a mirror:
The magnification is the ratio of the height of the image to the height of the object. The negative sign is
inserted as a convention. Object and image heights are considered positive if they are above the axis and
negative if below the axis. Thus the magnification is positive for an erect image and negative for an inverted
image.
7. Which statement is false?
a. When the object, image, or focal point is “real” (i.e., is on the reflecting side of the mirror), the
corresponding distance is considered positive; When the object, image, or focal point is “virtual” (i.e.,
behind the mirror), the distance is considered negative.
b. Virtual objects occur only in problems in which more than one mirror is used. Only then can an object
be behind the mirror, that is, if it is the image produced by another mirror.
c. The magnification is the ratio of the height of the image to the height of the object. Object and image
heights are considered positive if they are above the axis and negative if below the axis. Thus the
magnification is positive for an erect image and negative for an inverted image.
d. Magnification is always positive for all images, virtual or real, above or below the axis.
PROBLEM
An object whose height is 0.15 meter (ho) is placed 0.60 meter (do) from a concave mirror whose focal
length is 0.20 meter (f).
(a) Where is the center of curvature of the mirror?
(b) Where is the image located?
(c) What is the height of the image?
The image is real, is located 0.30 meter from the surface of the mirror, and is 0.075 meter in height, below the
axis, inverted.
PROBLEM
Describe the image and its placement when a 0.10-meter object is placed
0.50 meter in front of a convex mirror whose focal length is 1.0 meter.
The image is located 0.33 meter behind the mirror, is virtual, and is 0.066 meter in height (erect).
8.
Calculate the magnification of the image (m) if an object whose height is 0.05 meter (ho) which has
an image has a height of 0.75 m (hi).
m = (hi / ho)
a. 0.38
b. 2.0
c. 9.5
d. 15
9.
Calculate the magnification of the image (f) if the distance of the image from the mirror is 0.25 m (di)
if an object is placed 0.75 meter (do) from a concave mirror.
m = (di / do)
a. 0.33
b. 1.0
c. 9.0
d. 18
SPHERICAL ABERRATION
Spherical mirrors are subject to a deficiency known as spherical aberration. A spherical concave mirror will
not focus the light exactly to one point. To correct this deficiency, a parabolic mirror must be used, as
illustrated below.
10.
Which statement is false?
a. Spherical mirrors are subject to a deficiency known as spherical aberration.
b. A spherical concave mirror will not focus the light exactly to one point. To correct the deficiency
spherical aberration, a parabolic mirror must be used.
c. A spherical concave mirror will always focus the light exactly to one point.
d. A parabolic mirror concave mirror can focus light exactly to one point.