The figure shows a silvered lens. μ_A = 1.6 and μ_B = 1.2, R_1 = 80 c…

Equivalent focal length of the two lenses

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

$=(1.6-1))(\frac{1}{-80}-\frac{1}{-40})+(1.2-1)(\frac{1}{-40}-\frac{1}{-20})=\frac{1}{80}$

∴ For this lens combination

u = -12 cm, f = +80 cm

$\frac{1}{v_1}-\frac{1}{-12}=\frac{1}{80}$

$⇒\frac{1}{v_1}=\frac{1}{80}-\frac{1}{12}=\frac{-68}{80×12}$

Now this image acts as an object for the mirror of $f=\frac{R_3}{2}=10cm$

$∴\frac{1}{v}+\frac{-68}{80×12}=\frac{1}{-10}$

$⇒\frac{1}{v_2}=\frac{68}{80×12}-\frac{1}{10}=\frac{68-96}{80×12}=\frac{-28}{80×12}$

This image formed by mirror again acts as an object for the lens system

$∴\frac{1}{v}-\frac{-28}{80×12}=\frac{-1}{80}$

(focal length is taken negative because rays are now coming from right and the principal focus of the lens system will be on left side)

$∴\frac{1}{v}=-\frac{1}{80}-\frac{28}{80×12}⇒v=-24cm$

∴ The final image is formed at a distance of 24 cm to the left of the silvered lens.

Alternate Method :

$\frac{1}{F_{eq}}=\frac{2}{f_1}+\frac{2}{f_2}+\frac{1}{f_m}=\frac{1}{8}$

∴ or the equivalent mirror

$\frac{1}{v}+\frac{1}{-12}=-\frac{1}{8}⇒v = -24 cm$