A 10 cm tall object is placed perpendicular to the principle axis of a convex lens of focal length 12 cm. The distance from the object from the lens is 18 cm. Find the nature, position and size of the image by choosing the correct option?

Position:

36 cm

away from lens, nature: Real and inverted, size:

- 20 cm

. The size of the image is

20 cm

and negative sign denote the nature of image is inverted.
Given,
Object height,

h = + 10 cm

Focal length,

f = + 12 cm

Object distance,

u = - 18 cm

By applying the Lens formula, we get

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

\Rightarrow \frac{1}{v} - \frac{1}{- 18} = \frac{1}{12}

\Rightarrow \frac{1}{v} = \frac{1}{12} - \frac{1}{18}

\Rightarrow \frac{1}{v} = \frac{3 - 2}{36}

\Rightarrow \frac{1}{v} = \frac{1}{36}

\Rightarrow v = + 36 cm .

Hence, the position of the image is

36 cm

away from the lens, and the + sign denotes that the image is real. And the size of the image can be determined by

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

\Rightarrow \frac{h'}{10} = \frac{36}{- 18}

\Rightarrow h ’ = - 20 cm

A $10 cm$ tall object is placed perpendicular to the principle axis of a convex lens of focal length $12 cm$ . The distance from the object from the lens is $18 cm$ . Find the nature, position and size of the image by choosing the correct option?