Question

An object placed in front of a convex lens, forms an image of same size on a screen. Moving the object 12 cm closer to the lens results in the formation of a real image which is three times the size of the object.

Calculate the focal length of the lens.

Answer 1

Let the initial object distance be u, the image distance be v, and the focal length be f.

Since the image is formed on a screen, it is real and inverted, and it is also of the same size as the object, which means the magnification produced by the lens is −1.

Magnification (m) of a lens is given by,

m = `v/u`

−1 = `v/u`

u = −V    ...(i)

From the lens formula,

`1/f = 1/v - 1/u`

`1/f = 1/v - 1/(- v)`

`1/f = 1/v + 1/v`

`1/f = 2/v`

f = `v/2`    ...(ii)

Let the final object distance be u', and the final image distance be v'.

u' = u + 12    ...(iii)

m' = −3    ...(iv)

Magnification (m') of a lens is given by,

m' = `(v')/(u')`

−3 = `(v')/(u')`

u' = `-(v')/3`    ...(v)

u + 12 = `-(v')/3`    ...[From equation (iii)]

−v + 12 = `-(v')/3`    ...[From equation (i)]

−3(v − 12) = −v'

3(v − 12) = v'

3v − 36 = v'    ...(vi)

From lens formula,

`1/f = 1/(v') - 1/(u')`

= `1/(v') - 3/(-v')`    ...[From equation (v)]

= `1/(v') + 3/(v')`

`1/f = 4/(v')`

f = `(v')/4`

= `(3 v)/4 - 36/4`    ...[From equation (vi)]

f = `(3f)/2 - 9`

`(3 f)/2 - (2 f)/2` = 9

`f/2` = 9

f = 2 × 9

= 18 cm

Hence, the focal length of the lens is 18 cm.