An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre ‘O’ of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical center‘ O’ and principal focus ‘F’ on the diagram. Also, find the approximate ratio of size of the image to the size of the object.

Given:

Object height, $h$ = $+$4.0 cm 

Object distance from the convex lens, $u$ = $-$30 cm    $(object\ distance\ is\ always\ taken\ negative)$

Focal length of the convex lens, $f$ = $+$20 cm       $(focal\ length\ of\ a\ convex\ lens\ is\ taken\ positive)$

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

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

$\frac {1}{v}-\frac {1}{(-30)}=\frac {1}{20}$

$\frac {1}{v}+\frac {1}{30}=\frac {1}{20}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{30}$

$\frac {1}{v}=\frac {3-2}{60}$

$\frac {1}{v}=\frac {1}{60}$

$v=+60cm$

Thus, the distance of the image $v$ is 60 cm from the lens. The positive sign implies that the image forms on the right side of the lens, therefore the image will be real in nature.

Now, 

From the magnification formula, we know that-

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

Substituting the given value, we get-

$\frac {60}{-30}=\frac {h'}{4}$

$-2=\frac {h'}{4}$

$h'=4\times {(-2)}$

$h'=-8cm$

Thus, the size or height of the image is 8 cm. The negative sign implies that the image is inverted.

A ray diagram showing the position and size of the image formed.

The approximate ratio of the size of the image to the size of the object is given as-

$\Rightarrow \frac {h'}{h}=\frac {8}{4}=2$.

Thus, the approximate ratio is 2.

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Updated on: 10-Oct-2022

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