20.4. Thin Lenses http://www.ck12.org
b. What is the magnification?
Solution:
m=−di
do→−
− 10. 0 cm
35. 0 cm=
2
7
m=2
7
The object is upright and has a reduced image, sincemis positive and is smaller than 1. SeeFigure20.19.
c. What is the height of the image?
Solution:
m=
hi
ho
→hi=mho→hi=2
7
( 7. 0 cm) = 2. 0 cmThe link below helps explain how converging and diverging lens are used in eyeglasses.
http://demonstrations.wolfram.com/NearsightednessAndFarsightedness/