Question

An object is kept at distance of 30cm from a diverging lens of focal length 15cm. At what distance the image is formed from the lens? find the magnification of the image.​

Answer 1

Given :

In diverging lens (concave lens),

Object distance = - 30 cm

Focal length = - 15 cm

To find :

The image distance and the Magnification of the image.

Solution :

using lens formula that is,

» The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\boxed{ \bf \dfrac{1}{v} - \dfrac{1}{u}= \dfrac{1}{f}}$

where,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

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

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{ - 30}= \dfrac{1}{ - 15}$

$\dashrightarrow \sf \dfrac{1}{v} + \dfrac{1}{ 30}= \dfrac{1}{ - 15}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{ - 15} - \dfrac{1}{ 30}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{ - 2- 1}{ 30}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{ - 3}{ 30}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{ - 1}{ 10}$

$\dashrightarrow \underline{\boxed{\sf v = - 10 \: cm}}$

Thus, the position of image is - 10cm.

Magnification of image :

» The Magnification produced by a lens is equal to the ratio of image distance to the object distance .i.e.,

$\leadsto\bf m = \dfrac{v}{u }$

$\leadsto\sf m = \dfrac{ - 10}{ - 30}$

$\leadsto\sf m = \dfrac{1}{3}$

$\leadsto \underline{\boxed{\sf m = 0.33}}$