Question

A convex lens of focal length 10 cm forms images having magnitude of magnification three for any two positions of object. The distance between two such positions of image will be

pls ans​

Answer 1

$\text{Focal Length of given lens}(f) = 10 \; cm$

$\text{We know that magnification of a lens}(m) = \dfrac{-v}{u}$

$\text{It is given that }|m| = 3$

$\implies m = \±3$

$\text{CASE 1: }m = +3$

$\implies 3 = \dfrac{-v}{u}$

$\implies -3u = v \text{ ------ (i)}$

$\text{We know that, for a lens }\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$\implies \dfrac{1}{v} - \dfrac{1}{u} =\dfrac{1}{10}$

$\text{From (i)}$

$\implies \dfrac{1}{-3u} - \dfrac{1}{u} =\dfrac{1}{10}$

$\implies \dfrac{-1-3}{3u} =\dfrac{1}{10}$

$\implies \dfrac{-4}{3u} =\dfrac{1}{10}$

$\implies \dfrac{-40}{3} \; cm = u$

$\text{In (i)}$

$\implies -3\times\dfrac{-40}{3} = v$

$\implies 40 \; cm = v \text{ ------ (ii)}$

$\text{CASE 2: }m = -3$

$\implies -3 = \dfrac{-v}{u}$

$\implies v = 3u \text{ ------ (iii)}$

$\implies \dfrac{1}{v} - \dfrac{1}{u} =\dfrac{1}{10}$

$\text{From (iii)}$

$\implies \dfrac{1}{3u} - \dfrac{1}{u} =\dfrac{1}{10}$

$\implies \dfrac{1-3}{3u} =\dfrac{1}{10}$

$\implies \dfrac{-2}{3u} =\dfrac{1}{10}$

$\implies \dfrac{-20}{3} \; cm = u$

$\text{In (iii)}$

$\implies 3\times\dfrac{-20}{3} = v$

$\implies -20 \; cm = v \text{ ------ (iv)}$

$\text{We have to find the distance between these to image positions}$

$\implies \boxed{\triangle{v} = |40 - (-20)| = |40 + 20| = 60 \; cm}$