Question

A screen is placed 80 cm from an object. The image of the object on the screen is formed by a convex lens placed between them at two different locations separated by a distance 20 cm. determine the focal length of the lens.

Answer 1

Case 1:

Object distance (u) = x

image distance (v) = 80 − x

focal length = f

According to the lens formula:

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

`1/f = 1/(80 - x) + 1/x`    ...(1)

Case 2:

u = x + 20

v = 60 − x

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

`1/f = 1/(60 - x) + 1/(20 + x)`    ...(2)

On comparing equations (1) and (2), we get:

`1/f = 1/(80 - x) + 1/x`

= `1/(60 - x) + 1/(20 + x)`

`80/(x(80 - x)) = (80)/((60 - x)(20 + x)`

80 x − x² = 1200 + 40 x − x²

40 x = 1200

x = `1200/40`

x = 30 cm

Putting the value of x in equation (1),

`1/f = 1/(80 - 30) + 1/30`

`1/f = 1/50 + 1/30`

`1/f = 8/150`

f = (1 xx 150)/8`

= `150/8`

= 18.75 cm

Answer 2

For position 1:

u = −x

v = 80 − x

`1/-x - 1/(80 - x) = 1/f`    ...(i)

For position 2:

u = −(x + 20)

v = 80 − (x + 20)

`1/(-(x + 20)) - 1/(80 - (x + 20)) = 1/f`    ...(ii)

Comparing equations (i) and (ii),

`1/-x - 1/(80 - x) = 1/(-(x + 20)) - 1/(80 - (x + 20))`

⇒ `1/-x - 1/(80 - x) = 1/(-(x + 20)) - 1/(-x + 60)`

⇒ `(80 - x + x)/(x(80 - x)) = (-x + 60 + x + 20)/((x + 20)(-x + 60))`

⇒ x(80 − x) = (x + 20)(−x + 60)

⇒ 80x = 40x + 1200

⇒ 80x − 40x = 1200

⇒ 40x = 1200

⇒ x = `1200/40`

x = 30 cm

Putting the value of x in equation (i),

`1/-30 - 1/50 = 1/f`

⇒ `-8/150 = 1/f`

f = `-150/8`

= 18.75 cm