Question

an object is first placed at a distance 24 cm from the lens and then at a distance of 16 cm from the lens the magnification of the image is same in both the cases that the focal lenth of the lens is

Answer 1

when object is placed at two positions from lens it will have same magnification

this is possible when one of the magnification will be due to real image while other is due to virtual image.

So focal length must be between 16 cm and 24 cm

when object is placed at 16 cm then image must be virtual and hence we can say

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

here magnification = m

now when image is virtual and placed at 16 cm

$v = -m*16$

now when image is real and placed at 24 cm

$v = m*24$

now by lens formula

$\frac{1}{-16m} + \frac{1}{16} = \frac{1}{f}$

$\frac{1}{24m} + \frac{1}{24} = \frac{1}{f}$

now by above two equations

$\frac{1}{16}(\frac{1}{-m} + 1) = \frac{1}{24}(\frac{1}{m} + 1)$

$\frac{3}{-m} + 3 = \frac{2}{m} + 2$

$\frac{5}{m} = 1$

m = 5

now we have

$\frac{1}{-16m} + \frac{1}{16} = \frac{1}{f}$

now m = 5

$\frac{1}{-80} + \frac{1}{16} = \frac{1}{f}$

f = 20 cm

so focal length of the lens will be 20 cm

Answer 2

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)

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