Physic Labs

Problem 3

A small upright object of height hh is placed perpendicular to the principal axis of a thin converging lens, at distance u=3fu=3f from its optical centre; the focal length is f>0f>0 and u>fu>f. The media on both sides are the same. Use u>0u>0, take a real image on the opposite side to have v>0v>0, and define signed magnification by m=−v/um=-v/u. (a) Find the image distance vv in terms of ff. (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 vv on uu.
h′=mh=−h3−1;v>0h'=mh=-\frac{h}{3-1};\quad v>0
problems.proof.analysis

The signed image height is mhmh and its physical magnitude is ∣m∣h|m|h. Positive vv places the image on the far side, while negative mm makes it inverted.