Find the power of a convex lens which forms a real, an inverted image of magnification -1, of the object placed at a distance of 20 cm from its optical centre.

The power of a convex lens which forms a real, an inverted image of magnification

- 1

, of the object placed at a distance of

20 cm

from its optical centre is

+ 10 D

.
Given, magnification,

m = - 1

Object distance,

u = - 20 cm

We know

m = \frac{v}{u}

Where

v

is the image distance

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

\Rightarrow v = + 20 cm

by lens formula,

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

\Rightarrow \frac{1}{f} = \frac{1}{+ 20} - \frac{1}{- 20}

\Rightarrow \frac{1}{f} = \frac{1 + 1}{20}

\Rightarrow \frac{1}{f} = \frac{2}{20}

\Rightarrow \frac{1}{f} = \frac{1}{10}

\Rightarrow f = + 10 cm

\Rightarrow f = 0.1 m

The power of the lens,

P = \frac{1}{f}

\Rightarrow P = \frac{1}{0.1}

\Rightarrow P = + 10 D

Find the power of a convex lens which forms a real, an inverted image of magnification $- 1$ , of the object placed at a distance of $20 cm$ from its optical centre.