364 Image Formation(c)To find the area, first find qRand qL, along with the heights ′hRand ′hL, using the thin lensequation.111pqfRR+=becomes120 013 3..cmcmqRorqR=40 0. cm′==−FHGIKJ=−=−hhMhqpRR10 02 0020 0...cmcmafaf130 013 3cmcmqLorqL=24 0′−=−MLL10 00 8008 00cmcmafThus, the area of the image is:Area ′+−′ −′ =qqhRLL12224 cm2.P36.41(a)The image distance is:qdp.Thus,111pq fbecomespdp f+−=.This reduces to a quadratic equation:pdpfd20+ =which yields:pddfdfd=±−=±−224224.Since fd<4, both solutions are meaningful and the two solutions are not equal to eachother. Thus, there are two distinct lens positions that form an image on the screen.(b)The smaller solution for pgives a larger value for q, with a real, enlarged, inverted image .
This is the end of the preview. Sign up
access the rest of the document.