Question

A lens forms a real image 3 cm high of an object 1 cm high. If the separation of object and image is 15 cm, find the focal length of the lens.

Answer 1

Let 'u' be the object distance, 'v' be the image distance and 'f' be the focal length of the lens.

According to the question, the magnification is: m = -3 (negative sign comes as the image is real)

Therefore, using magnification formula: $m=\frac{v}{u}$

$\frac{v}{u}$ = - 3

$\Rightarrow$ v = - 3u ........eq(1)

Also, according to question separation of object and image is 15 cm, therfore

- u + v = 15 ( u will be on left side $\therefore$ -ve sign because of sign conventions)

$\Rightarrow$ -4u = 15 [using eq(1)]

$\Rightarrow$ u = $\frac{-15}{4}$

And v = -3 × $\frac{-15}{4}$ = $\frac{45}{4}$

Now,

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

$\Rightarrow$ $\frac{1}{f}$ = $\frac{4}{45}$ - $\frac{-4}{15}$

$\Rightarrow$ $\frac{1}{f}$ = $\frac{4}{45}$ + $\frac{4}{15}$

$\Rightarrow$ $\frac{1}{f}$ = $\frac{16}{45}$

$\Rightarrow$ f = 2.8 cm

So, the focal length is 2.8 cm.