Question

Find the position and size of the virtual image formed when an object 2cm tall is placed 20cm from (a) diverging lens of focal length 40cm (b) converging lens of focal length 40cm

Answer 1

Hope this helps
And if it helped please mark as brainliest

Answer 2

Given that,

Height of object = 2 cm

Object distance = 20 cm

Focal length of diverging lens = -40 cm

Focal length of converging lens = 40 cm

(I). For converging lens,

We need to calculate the image distance

Using formula of lens

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

Put the value into the formula

$\dfrac{1}{40}=\dfrac{1}{v}-\dfrac{1}{-20}$

$\dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{20}$

$\dfrac{1}{v}=-\dfrac{1}{40}$

$v=-40\ cm$

Negative sign shows that the image formed is virtual and erect.

We need to calculate the height of image

Using formula of magnification

$m=\dfrac{v}{u}=\dfrac{h'}{h}$

Where, v = image distance

u = object distance

h' = height of image

h = height of object

Put the value into the formula

$\dfrac{-40}{-20}=\dfrac{h'}{2}$

$h'=\dfrac{40\times2}{20}$

$h'=4\ cm$

The size of image is 4 cm.

(II). For converging lens

We need to calculate the image distance

Using formula of lens

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

Put the value into the formula

$\dfrac{1}{-40}=\dfrac{1}{v}-\dfrac{1}{-20}$

$\dfrac{1}{v}=\dfrac{1}{-40}-\dfrac{1}{20}$

$\dfrac{1}{v}=-\dfrac{3}{40}$

$v=-\dfrac{40}{3}\ cm$

$v=-13.33\ cm$

Negative sign shows that the image formed is virtual and erect.

We need to calculate the height of image

Using formula of magnification

$m=\dfrac{v}{u}=\dfrac{h'}{h}$

Where, v = image distance

u = object distance

h' = height of image

h = height of object

Put the value into the formula

$\dfrac{-40}{-20\times3}=\dfrac{h'}{2}$

$h'=\dfrac{40\times2}{20\times 3}$

$h'=1.33\ cm$

The size of image is 1.33 cm.

Hence, (I). The position of image is formed 40 cm distance from the converging lens in left side.

The size of image is 4 cm

(II). The position of image is formed -13.3 cm distance from the diverging lens in left side.

The size of image is 1.33 cm.