Question

find the focal length of the convex lens which produces a real image at 60 cm from the lens when an object is placed at 40 in front of the lens​

Answer 1

Explanation:

$given \\ for \: lens \: 1$

$image \: distance \\ v = 60 \: cm$

$let \: object \: distance \: be \: u$

$\frac{1}{f1} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{f1} = \frac{1}{60} + \frac{1}{u}$

$\frac{1}{f1} = \frac{u + 60}{60u} .........(1)$

$when \: lenses \: are \: comined \: the \: image \: distance \: v2 = 60 - 40 = 20cm$

$\frac{1}{f} = \frac{1}{f1} + \frac{1}{f2} = \frac{1}{v2} - \frac{1}{u}$

$⤇\: \frac{1}{f1} + \frac{1}{f2} = \frac{1}{20} + \frac{1}{u}$

$⤇ \frac{u + 60}{60u} + \frac{1}{f2} = \frac{1}{20} + \frac{1}{u}$

$⤇\frac{1}{f2} = \frac{u + 20}{20u} - \frac{u + 60}{60u}$

$\frac{1}{f2} = \frac{60 {u }^{2 } + 1200u - 20 {u}^{2} - 1200u}{ {1200u}^{2} }$

$⤇\frac{40 {u}^{2} }{1200 {u}^{2} }$

$⤇f2 = 30cm$

$hope \: it \: is \: helpful \: to \: u$