Question

A student has focused on the screen the image of a 2cm high candle flame placed at a distance of 24cm from a convex lens of focal length 16cm. Use lens formula to calculate the distance of the screen from the lens and length of the image formed?​

Answer 1

Given:

• Height of candle flame, h = +2cm

• Distance of candle from lens,U =-24cm

• Focal length, F= +16cm

To find:

• Distance of image, V

• Height of image formed, h'

Solution:

• Applying lens formula

$\implies\sf \dfrac 1 F=\dfrac 1 V-\dfrac 1 U$

$\implies\sf \dfrac 1 {16cm}=\dfrac 1 V-\dfrac 1 {(-24cm)}$

$\implies\sf \dfrac 1 {16cm}=\dfrac 1 V+\dfrac 1 {24cm}$

$\implies\sf \dfrac 1 {16cm}-\dfrac 1 {24cm}=\dfrac 1 V$

$\implies\sf \dfrac {3-2}{48cm}=\dfrac 1 V$

$\implies\sf \dfrac {1}{48cm}=\dfrac 1 V$

$\boxed{\implies\sf V=48cm}$

So distance of screen from lens is 48cm.

$\rule{240}{1}$

• Now we have formula:

$\implies\sf \dfrac{ h'}h=\dfrac V U$

$\implies\sf \dfrac{ h'}{2cm}=\dfrac {48cm}{-24cm}$

$\implies\sf \dfrac{ h'}{2cm}=-2cm$

$\implies\sf { h'}=(-2cm)(2cm)$

$\boxed{\implies\sf { h'}=-4cm}$

Height of image is 4cm.

This negative sign shows that image is inverted.