A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object forms the lens is 15 cm. What is the nature, position and size of the image?

Nature, position and size of the image is virtual, erect and magnified and

h ’ = 16 cm

Given,
Distance of the object from the lens,

u = 15 cm

Focal length of the lens,

f = 20 cm

By using lens formula

\frac{1}{v}

\frac{1}{u}

\frac{1}{f}

where

u

is the object distance.

\Rightarrow \frac{1}{20} + \frac{1}{15} = \frac{1}{v}

\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{15}

\Rightarrow \frac{1}{v} = \frac{3 - 4}{60}

\Rightarrow v = - 60 cm

So, the image formed

60 cm

in front of the lens on the same side as the object. Negative sign shows that the image formed is Virtual.
The magnification of convex lens,

m = \frac{h'}{h}

\Rightarrow m = \frac{v'}{u}

\Rightarrow h ’ = 16 cm

As the height of image is positive and greater than height of object, So the image formed is erect and magnified.

A $4 cm$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $20 cm .$ The distance of the object forms the lens is $15 cm$ . What is the nature, position and size of the image?