Question

An object 4 cm tall is kept on the principal axis of a converging lens of focal

length 12 cm. Find the position, nature and size of the image formed if the object

is at 18cm from the lens. Also find the height of the image .​

Answer 1

Explanation:

Given = object height, h = 4cm

focal length = -12cm

object distance = -18 cm

Using lens formula,

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

$\boxed{ \frac{1}{ - 12} = \frac{1}{v} + \frac{1}{18} }$

$\boxed{ \frac{1}{ - 12} - \frac{1}{18} = \frac{1}{v} }$

$\boxed{ \frac{3 - 2}{ - 36} = \frac{1}{v} }$

$\boxed{v = - 36}$

We know that, Magnification = h'/h = v/u

$\boxed{h'/4 = \frac{ - 30}{ - 18} }$

$\boxed{h' = \frac{ - 120}{ - 18} }$

$\boxed{h' = 6.7 \: cm}$

Using the magnification formula,

$\boxed{ \frac{v}{u} = \frac{ - 30}{ - 18} = 1.4}$

Since the magnification is positive it means the image is virtual and erect and magnified.