20.2. Concave and Convex Mirrors http://www.ck12.org
b. What is the magnification?
Solution:
m=−
di
do=−
10. 91 cm
30. 0 cm=− 0. 3636 →− 0. 364
m=− 0. 364The image is inverted and reduced, sincemis negative and has absolute value smaller than 1.
c. What is the height of the image?
Solution:
m=
hi
ho→− 0. 3636 =
hi
10 cm
hi=− 3. 64 cmd. The object is now positioned 5.3 cm from the mirror. Find the image position.
Solution:
We are givenho= 10. 0 cm,do= 5. 3 cm and f= 8. 0 cm
Using the mirror equation:
1
do+
1
di=
1
f
1
5 .3 cm+
1
di=
1
8 .0 cm
1
di=
1
8 .0 cm−
1
5 .3 cm
1
di=
5. 3 − 8. 0
42 .4 cm
_.i=− 15 .7 cmNotice the result is negative. The image is, therefore, 15.7 cm behind the mirror, and is virtual.
An object placed between the mirror and the focal point, as in this case( 5. 3 cm< 8. 0 cm), will always produce a
virtual image.
e. What is the magnification of the object?
m=−
di
do=−
− 15. 7 cm
5. 3 cm= + 2. 96 →+ 3. 0
m= 3. 0Solution:The image is upright and magnified; its height is 30 cm. SeeFigure20.13.
Concave mirrors are often sold as “beauty aids” for applying makeup, since the image of an object placed close
to the mirror is magnified. If you hold one of these mirrors at arm’s length, you’ll see your image as inverted and
reduced. However, as you bring the mirror slowly toward your eyes, your image will grow larger, vanish, and then
reappear upright and magnified (and not always flattering, many people say!). Try doing this with a shiny spoon!