Question

A concave lens has focal length of 15 cm . At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens . also find the magnification produced by the lens ? ​

Answer 1

Solution : Here, focal length of concave lens, `f = -15 cm`. <br> object distance, `u = ?`, image distance, `v = -10 cm`, magnification of lens, `m = ?` <br> As `(1)/(f) = (1)/(v) - 1/u, 1/u = (1)/(v) -(1)/(f) = 1/(-10) + 1/15 = (-1)/30 or u = -30 cm` <br> Thus the object should be placed at a distance of 30 cm on the left side of the concave lens. <br> Linear magnification, `m = v/u = (-10)/(-30) = 1/3`. <br> The positive sign of m shows that the image is virtual and erect, and its size is `(1//3)` of the size of the object.

Answer 2

Focal length f = -15cm

Image distance v = -10cm

By lens formula

$\sf{ \frac{1}{f} } = \frac{1}{v} - \frac{1}{u}$

$\sf{ \frac{1}{ - 15}} = \frac{1}{ - 10} - \frac{1}{u}$

$\sf{ \frac{1}{u} } = \frac{1}{ - 10 \: } - \frac{1}{ - 15}$

$\sf{ \frac{1}{u} } = - \frac{1}{30}$

$\sf{u = - 30 \: cm}$

$\sf{Magnification } \: m = \frac{v}{u}$

$\sf{m = \frac{ - 10}{ - 30}} = 0.33$