Problem 3

A thin converging lens of focal length

f=20

cm stands in air. A small luminous object of height

H=2.0

cm is perpendicular to the principal axis,

u=30

cm in front of the lens. Use the convention that a real image has

v>0

, the lens equation

1/f=1/u+1/v

, and lateral magnification

m=-v/u

. (a) Find the image distance and height. (b) State whether the image is real or virtual and upright or inverted. (c) The object is moved farther away to

u=60

cm; find the new image and describe the size change.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

m=-v/u=-60/30=-2,\quad h'=mH=-4.0\ \mathrm{cm}

problems.proof.analysis

The magnitude $|m|=2$ means the image is twice as tall as the object. The negative sign indicates inversion, so its signed height is $-4.0$ cm.

problems.proof.pitfall. Magnification is signed:

m=-v/u

. Omitting the minus sign would incorrectly predict an upright image.