An object 3 cm high is placed 20 cm from a convex lens of focal length 12 cm. Find the position, and height of the image.

The nature, position, and height of the image is

+ 30 cm

and

- 4.5 cm

.
Given data in the problem,
Distance of image from the lens,

v = ?

Distance of the object from the lens,

u = - 20 cm

The focal length of the lens,

f = + 12 cm

By using lens formula,

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

\Rightarrow \frac{1}{v} - \frac{1}{- 20} = \frac{1}{12}

\Rightarrow \frac{1}{v} + \frac{1}{20} = \frac{1}{12}

\Rightarrow \frac{1}{v} = \frac{1}{12} - \frac{1}{20}

\Rightarrow \frac{1}{v} = \frac{5 - 3}{60}

\Rightarrow \frac{1}{v} = \frac{2}{60}

\Rightarrow v = + 30 cm

So, the position of image is

+ 30 cm

.
Magnification,

m = \frac{v}{u}

\Rightarrow m = \frac{30}{- 20}

\Rightarrow m = - 1.5

m

is negative and greater than

1

, so the image will be inverted, real and larger than the object.

m = \frac{h'}{h}

\Rightarrow - 1.5 = \frac{h'}{3}

\Rightarrow h ’ = - 1.5 \times 3

\Rightarrow h ’ = - 4.5 cm

An object $3 cm$ high is placed $20 cm$ from a convex lens of focal length $12 cm$ . Find the position, and height of the image.