Problem 3

On an optical bench, a thin converging lens of focal length

f=10.0\,\mathrm{cm}

is used with an object of height

h_o=2.00\,\mathrm{cm}

placed on the principal axis, initially at distance

u=30.0\,\mathrm{cm}

. Work in air and in the paraxial approximation. For a real object take

u>0

; a real image has

v>0

. Define transverse magnification by

M=h_i/h_o=-v/u

. (a) Find the image distance and magnification, and state whether the image is real or virtual and upright or inverted. (b) Find the image height. (c) Move the object to

u=15.0\,\mathrm{cm}

and repeat the position, magnification, and image-character determination, then compare the two placements.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

M=-\frac vu=-0.500

problems.proof.analysis

Using the distances in $M=-v/u$ gives $M=-0.500$ . The negative sign means inverted, and magnitude one-half means reduced by a factor of two.

problems.proof.pitfall. With the stated definition, an inverted image has negative

M

; do not confuse

|M|

with orientation.