Question

A convex lens has a focal length 15cm from a image 10 cm from the lens. how for is the object placed from the lens? Also find the magnification of the lens.​

Answer 1

Given :

▪ Focal length = 15cm

▪ Distance of image = 10cm

▪ Type of lens : Convex

To Find :

▪ Distance of object.

Concept :

↗ X-coordinates of centre of curvature and focus of convex lens are positive and those for concave lens are negative.

↗ Lens Formula :

$\bigstar\bf\:\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

↗ Lateral Magnification :

$\bigstar\bf\:m=\dfrac{v}{u}$

Calculation :

⚽ Position of Object :

$\implies\sf\:\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\ \\ \implies\sf\:\dfrac{1}{-10}-\dfrac{1}{u}=\dfrac{1}{15}\\ \\ \implies\sf\:\dfrac{1}{u}=\dfrac{1}{(-10)}-\dfrac{1}{15}\\ \\ \implies\sf\:\dfrac{1}{u}=\dfrac{(-25)}{150}\\ \\ \implies\underline{\underline{\bf{u=-6cm}}}$

⚽ Magnification :

$\dashrightarrow\sf\:m=\dfrac{v}{u}\\ \\ \dashrightarrow\sf\:m=\dfrac{(-10)}{(-6)}\\ \\ \dashrightarrow\underline{\underline{\bf{m=\dfrac{5}{3}}}}$

Answer 2

Given:-

Focal length = 15 cm

Distance of image = 10cm

Types of lens:- convex (+)

To Find:-

Distance of object from the lens.

Concept:-

Always remember, Focus of convex lens are positive and concave lens are negative.

Formula :-

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

lateral magnification:-

${\implies{\tt{m = \frac{v}{u}}}}$

By calculating, we find..

★Position of object

${\implies{\tt{\frac{1}{-10} - \frac{1}{u} = \frac{1}{15}}}}$

${\implies{\tt{\frac{1}{u} = \frac{1}{(-10)} = \frac{1}{15}}}}$

${\implies{\tt{\frac{1}{u} = \frac{-25}{150} }}}$

${\implies{\tt{u = - 6 cm}}}$

★Magnification of lens:-

${\implies{\tt{m = \frac{v}{u}}}}$

${\implies{\tt{m = \frac{-10}{-6}}}}$

${\implies{\tt{ m = \frac{5}{3}}}}$