Question

An object is first placed at a distance 24cm from the lens and then at a distance of 16 cm from the lens. The magnification of the image formed is same in both the cases. The focal length of the lens is?

Answer 1

case 1 :-
object is placed at a distance 24cm from the lens.
e.g., u = -24 cm
Let focal length of lens is f
then, 1/v - 1/u = 1/f
1/v - 1/-24 = 1/f
1/v = 1/-24 + 1/f
1/v = 1/f - 1/24 = (24 - f)/24f
v = 24f/(24 - f)
now magnification , m = v/u = 24f/(24 - f)/24
m = f/(24 - f) ........ (i)


case 2 :- an object is placed at a distance of 16cm from the lens.
e.g., u = -16 cm
so, 1/v - 1/-16 = 1/f
1/v = 1/-16 + 1/f
1/v = 1/f - 1/16
1/v = (16 - f)/16f
v = 16f/(16 - f)
now magnification, M = v/u = 16f/(16 - f)/16
M = f/(16 - f) ........(ii)

according to question,
m = M
but if we doesn't get answer { e.g., 16 ≠ 24}
it means if first case image is real then image formed in 2nd case must be virtual and vice -versa .

so, m = -M
then, f/(24 - f) = -f/(16 - f)
(24 - f) = (f - 16)
24 + 16 = f + f
f = 20cm

hence, focal length is 20cm