(a) A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position, nature and size of the image formed.(b) Draw a labelled ray diagram showing object distance, image distance and focal length in the above case.

Given:

Type of lens - Convex Lens

Object distance, $u=-30cm$

Focal length, $f=+20cm$

Object height, $h=+5cm$

To find: Position, nature and size of the image.

Solution:

From lens formula we know that-

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

Substituting the required 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 position of the image is 60 cm from the lens, and the plus sign implies that it forms behind the lens (on the right side). Also, the nature of the image is real as it forms on the right side.

Now,

From the magnification formula of the lens, we know that-

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

Substituting the given values, we get-

$\frac {60}{-30}=\frac {h'}{5}$

$-2=\frac {h'}{5}$

$h'=-10cm$

Thus, the size of the image is 10 cm, and the minus sign implies that the image is inverted (below the principal axis).

Hence, the position of the image is 60 cm behind the lens, the nature of the image is real and inverted, and the size of the image is 10 cm.

(b) A labelled ray diagram is given below showing object distance, image distance and focal length in the above case.