Question #211030

An object 1.00cm high is placed 18.0cm from a converging lens forms a real image 2.00cm high. Calculate the focal length of the lens


1
Expert's answer
2021-06-28T17:14:05-0400

Given:

do=18.0cmd_o=18.0\:\rm cm

ho=1.00cmh_o=1.00\:\rm cm

hi=2.00cmh_i=-2.00\:\rm cm

The magnification of thin lenth


m=hiho=did0m=\frac{h_i}{h_o}=-\frac{d_i}{d_0}

Hence,


di=dohiho=18.0×2.001.00=36.0cmd_i=-d_o\frac{h_i}{h_o}=-18.0\times\frac{-2.00}{1.00}=36.0\:\rm cm

The thin lens equation says


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

Finally,


f=dodido+di=18.036.018.0+36.0=12.0  cmf=\frac{d_od_i}{d_o+d_i}=\frac{18.0*36.0}{18.0+36.0}=12.0\;\rm cm


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