Question

size of the image of an object by a concave lens of focal length 20 cm is observed to be reduced 1/3 of its size find the distance of object from the lens​

Answer 1

answer : 40cm

explanation : focal length of concave lens, f = -20cm

a/c to question,

size of image = 1/3 × size of object.

so, magnification, m = 1/3 [ as lens is concave so, magnification always be positive ]

we know, magnification of lens, m = v/u

or, 1/3 = v/u => v = u/3 ......(1)

use lens maker formula,

1/v - 1/u = 1/f

or, 1/(u/3) - 1/u = 1/-20 [ from equation (1), ]

or, 3/u - 1/u = 1/-20

or, 2/u = 1/-20

or, u = -40cm

hence, distance of object from the lens is 40cm

Answer 2

$\bf\huge\textbf{\underline{\underline{According\:to\:the\:Question}}}$  

f = - 20cm

$\bf\huge{\implies Size\:of\:Image= \dfrac{1}{3}\:of\:Object}$          

$\bf\huge{\implies m = \dfrac{v}{u}}$          

$\bf\huge{\implies\dfrac{1}{3} = \dfrac{v}{u}}$          

$\bf\huge{\implies v = \dfrac{u}{3}}$          

$\bf\huge\bf\huge{\boxed{\bigstar{{Lens\:Formula}}}}$          

$\bf\huge{\implies\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}$          

$\bf\huge\bf\huge{\boxed{\bigstar{{ Substitute\: value\: of\: v }}}}$          

$\bf\huge{\implies\dfrac{1}{u/3} - \dfrac{1}{u} = \dfrac{1}{-20}}$          

$\bf\huge{\implies\dfrac{3}{u} - \dfrac{1}{u} = \dfrac{1}{-20}}$        

$\bf\huge{\implies\dfrac{2}{u} = \dfrac{1}{-20}}$          

u = - 40cm

$\bf\huge\bf\huge{\boxed{\bigstar{{Distance=\:40cm}}}}$