Problem 3

A small upright object of height

h

is placed perpendicular to the principal axis of a thin converging lens, at distance

u=3/2f

from its optical centre; the focal length is

f>0

and

u>f

. The media on both sides are the same. Use

u>0

, take a real image on the opposite side to have

v>0

, and define signed magnification by

m=-v/u

. (a) Find the image distance

v

in terms of

f

. (b) Find the magnification and image height, and state whether the image is real or virtual and upright or inverted. (c) If the object is moved a small distance farther from the lens, determine the image’s direction of motion from the dependence of

v

u

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

m=-\frac vu=-\frac{1}{3/2-1}

problems.proof.analysis

Signed lateral magnification is defined as $-v/u$ . Its negative sign means inversion; its magnitude tells whether the image is reduced or enlarged.

problems.proof.pitfall. Constraint condition: Do not discard the negative sign of

m

: it encodes inversion, while image size is

|m|h