Question

Size of the image of an object by a concave lens of focal length 20cm is observed to be reduced to 1/3rd of it's size .find the distance of the object from the lens.

Answer 1

answer : object distance from the optical centre of the lens is 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

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

so, v/u = 1/3

so, v = u/3 ........(1)

now applying lens maker formula,

1/v - 1/u = 1/f

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

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

or, 2/u = 1/-20 cm

or, u = -40 cm

hence, object is placed 40cm from the pole of concave lens.

Answer 2

answer : object remove from the optical focal point of the focal point is 40cm.  

clarification : central length of sunken focal point , f = - 20cm  

a/c to address,  

size of picture = 1/3 × size of item.  

thus, amplification , m = 1/3  

we know, amplification of focal point , m = v/u  

thus, v/u = 1/3  

thus, v = u/3 ........(1)  

presently applying focal point creator equation,  

1/v - 1/u = 1/f  

or on the other hand, 1/(u/3) - 1/u = 1/ - 20  

or on the other hand, 3/u - 1/u = 1/20  

or on the other hand, 2/u = 1/ - 20 cm  

or on the other hand, u = - 40 cm  

henceforth, object is set 40cm from the post of sunken focal point.