Question

An object is first placed at a distance of 24 cm from the lensand then at a distance of 16 cm from the lens the magnification of the imageformed is same in both the cases . find the focal length of lens

Answer 1

Since the magnification is same in both cases so one of the image must be real image while other will be virtual image.

so we can say

for real image

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

here we have

$u = -24 cm$

$v = 24\alpha cm$

here $\alpha$ = magnification

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

now similarly for virtual image

$u = -16 cm$

$v = -16\alpha cm$

here $\alpha$ = magnification

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

now by above two equations

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

$-6 + 6\alpha = 4 + 4\alpha$

$\alpha = 5$

now we have

$\frac{1}{f} = \frac{1}{24*5}(1 + 5)$

$f = 20 cm$

so focal length 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