Physic Labs

Problem 3

A transverse object of height HH is placed before a converging thin lens of focal length f>0f>0, perpendicular to its principal axis. Its distance from the lens is u=3fu=3f; use the convention that image distance v>0v>0 lies opposite the object and signed transverse magnification is M=−v/uM=-v/u. (a) Use the principal-ray construction to determine the image character and location. (b) Find the signed magnification and image height. (c) If a smaller object is placed at the same position, state what changes and what does not, explaining with the lens equation.
problems.proof.stepOf
problems.proof.prevStepproblems.proof.nextStep
M=−vu=−3f/23f=−12M=-\frac vu=-\frac{3f/2}{3f}=-\frac12
problems.proof.analysis

By convention M=−v/uM=-v/u. Since both distances are positive the image is inverted; magnitude 1/21/2 means half the object's size.