(a) Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from:(i) a perging lens of focal length 40 cm.(ii) a converging lens of focal length 40 cm.(b) Draw labelled ray diagrams to show the formation of images in case (i) and (ii) above (The diagrams may not be according to scale).

(a)(i) Given:

Object height, $h$ = $+$2 cm

Object distance, $u$ = $-$20 cm  (object distance is always taken negative)

Focal length, $f$ = $-$40 cm        (focal length of a diverging lens always taken negative)

To find: Position or image distance, $v$ and image size $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

$\frac {1}{v}-\frac {1}{(-20)}=\frac {1}{(-40)}$

$\frac {1}{v}+\frac {1}{20}=-\frac {1}{40}$

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

$\frac {1}{v}=\frac {-1-2}{40}$

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

$v=-\frac {40}{3}$

$v=-13.33cm$

Thus, the distance of the image $v$ is 13.33 cm from the lens. The negative $(-)$ sign implies that the image is formed in front of the lens (on the left side) and is virtual and erect. 

Now,

From the magnification formula, we know that-

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

$\frac {-13.33}{-20}=\frac {h'}{2}$

$\frac {1333}{20\times {100}}=\frac {h'}{2}$

(a)(i) Given:

Object height, $h$ = $+$2 cm

Object distance, $u$ = $-$20 cm  (object distance is always taken negative)

Focal length, $f$ = $+$40 cm        (focal length of a converging lens always taken negative)

To find: Position or image distance, $v$ and image size $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

$\frac {1}{v}-\frac {1}{(-20)}=\frac {1}{40}$

$\frac {1}{v}+\frac {1}{20}=-\frac {1}{40}$

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

$\frac {1}{v}=\frac {1-2}{40}$

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

$v=-40cm$

Thus, the distance of the image $v$ is 40 cm from the lens. The negative $(-)$ sign implies that the image is formed in front of the lens (on the left side) and is virtual and erect. 

Now,

From the magnification formula, we know that-

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

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

$\frac {40}{20}=\frac {h'}{2}$

$2=\frac {h'}{2}$