Question

The figure shows a silvered lens. ${\mu }_A=1.6$ and ${\mu }_B=1.2$, $R_1=80$cm, $R_2=40$cm and $R_3=20$cm. An object is placed at a distance of $12$cm from this lens. Find the image position.

Answer 1

$r_1=-80cm$

$r_2=-40cm$

$r_3=20cm$

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

$=(1.6-1)[\frac{1}{-80}-\left(\frac{-1}{40}\right)]+2(1.2-1)[\frac{-1}{40}-\left(\frac{-1}{20}\right)]$

$=0.6[\frac{-1}{80}+\frac{1}{40}]+0.4[\frac{-1}{40}+\frac{1}{20}]$

$=0.6\left[\frac{-1+2}{80}\right]+0.4\left[\frac{-1+2}{40}\right]$

$=\frac{0.6}{80}+\frac{0.4}{40}=\frac{1}{100}(\frac{3}{40}+\frac{1}{10})=\frac{7}{4000}$

$t=\frac{4000}{7}cm$

$u=-12cm$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{7}{4000}-\frac{1}{12}$

$v=\frac{21-1000}{12000}$

$v=\frac{-12000}{979}cm$