Answer to Question #296573 in Physics for Kitchie

Question #296573

2. What is the image description of a 3 cm tall pen light when placed at distance of 30.5 cm, from a converging lens having a focal length of 10.2 cm?


1
Expert's answer
2022-02-11T08:46:27-0500

Given:

do=30.5cmd_o=30.5\:\rm cm

f=10.2  cmf=10.2\;\rm cm

ho=3cmh_o=3\:\rm cm


The thin lens equation says

1do+1di=1f\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}

di=fdodof=30.510.230.510.2=15.3cmd_i=\frac{fd_o}{d_o-f}=\frac{30.5*10.2}{30.5-10.2}=15.3\:\rm cm

Image size

hi=hodido=315.330.5=1.51cmh_i=-h_o\frac{d_i}{d_o}=-3*\frac{15.3}{30.5}=-1.51\:\rm cm

Image: inverted, reduced, real.


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