20.4. Thin Lenses http://www.ck12.org
m=−
di
do→
15. 0 cm
30. 0 cm=− 0. 500
m=−1
2
The image is inverted and reduced, sincemis negative and has absolute value smaller than 1.
c. What is the height of the image?
Solution:
The image is half the size of the object:
m=hi
ho
→hi=mho→hi=1
2
( 12. 0 cm) = 6. 0 cmd. The object is now positioned 7.0 cm from the lens. Find the image position.
Solution:
1
do+
1
di=
1
f→
1
7. 0
+
1
di=
1
10. 0
→
1
di=
1
10. 0
−
1
7. 0
=
7. 0 − 10. 0
70 cm→
di=− 23. 3 cmSince the image distance is negative, the image is formed on the same side of the lens as the object, and therefore is
virtual. SeeFigure20.19.
e. What is the magnification?
Solution:
m=−ddoi=−− 723. 0.^3 cmcm= 3. 33 → 3. 3
The image is upright and magnified.
The diverging lens
A lens that is thinner at its center than its edges is a diverging lens.
Constructing ray diagrams for diverging lenses
Three principal rays can be drawn to construct ray diagrams for diverging lenses.
Ray 1: A ray parallel to the principal axis refracts so that its extension (the dashed line) passes through the focal
pointFinFigure20.20.
Ray 2: A ray heading toward focal pointF′refracts parallel to the principal axis, as inFigure20.20.
Ray 3: A ray passes through the center of the lens with no apparent refraction.