Question

2. An object is placed perpendicular to the principal axis of convex lens
of focal length 18cm, at a distance of 12 cm from the lens. Determine the
position and the magnification of the image. Also illustrate this with the help of
a suitable ray diagram.

Answer 1

Explanation:

For Convex lens:-

• Focal length = +ve

• Object distance = -ve

Given, Focal length = 18cm

Object distance = -12cm

Using lens formula:-

$\sf \dashrightarrow\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\sf \dashrightarrow\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

$\sf \dashrightarrow\dfrac{1}{v} = \dfrac{1}{18} + \dfrac{1}{(-12)}$

$\sf \dashrightarrow\dfrac{1}{v} = \dfrac{1}{18} - \dfrac{1}{12}$

$\sf \dashrightarrow\dfrac{1}{v} = \dfrac{2 - 3}{36}$

$\sf \dashrightarrow\dfrac{1}{v} = \dfrac{-1}{36}$

$\sf v = -36cm$

$\sf Now,\: Magnification = \dfrac{-v}{u}$

$\sf \dashrightarrow m = \dfrac{-(-36)}{-12}$

$\sf \dashrightarrow m = \dfrac{36}{-12}$

$\sf \dashrightarrow m = -3$

Therefore:-

Image is formed 36cm infront of the lens .

And, Magnification of image = -3

Nature of image :- Virtual, erect and enlarged.

For Ray Diagram:- Refer the Attachment