Question

Find the position of final image from 'O' for the
arrangement shown.
20cm
20cm
Object
f=40cm
f= 40cm
(1) -24 cm
(3) -40 cm
(2)-120 cm
(4) -20 cm​

Answer 1

$\bf \to \: \green{{ \underline{to \: find \div }}}$

$\sf \to \: the \: position \: of \: final \: image \: from \: o$

$\bf \to \: { \underline{solution \div }}$

$\sf \to \: for \: convex \: lens \\ \\ \sf \to \: distance \: of \: object \: = u \: = - 20cm \\ \\ \sf \to \: focal \: length \: = 40cm \\ \\ \sf \to \: from \: lens \: formula \: \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \\ \\ \sf \to \: \frac{1}{v} = \frac{1}{40} + \frac{1}{ - 20} \\ \\ \sf \to \: \frac{1}{v} = \frac{1 - 2}{40} \\ \\ \sf \to \: \frac{1}{v} = \frac{ - 1}{40} \\ \\ \sf \to \: v \: = - 40cm \\ \\ \sf \to \: distance \: from \: concave \: lens \: = -40 - 20 = - 60cm \\ \\ \sf \to \: by \: apply \: again \: lens \: formula \\ \\ \sf \to \: \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \\ \\ \sf \to \: \frac{1}{v} = \frac{1}{-40} + \frac{1}{ - 60} \\ \\ \sf \to \: \frac{1}{v} \: = \frac{-3-2}{120} \\ \\ \sf \to \: \frac{1}{v} = \frac{ - 5}{120} \\ \\ \sf \to \: v \: = - 24cm \\ \\ \sf \to \: \therefore \: position \: of \: final \: position \: from \: o = - 24cm$

Answer 2

Answer:

$\huge\bold\purple{A}\red{N}\green{S}\blue{W}\purple{E}\green{R}$

$\huge\bold\red{\boxed{-24}}$