An object is placed at a distance of 30 cm from a concave lens of focal lenth 15 cm. List four charcteristics (nature, positions, etc.) of the image formed by the lens.

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Solution

Given, object distance, u = - 30 cm and focal length, f = -15 cm
Let 'v' be the image distance.
We know that,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{(- 30)} = \frac{1}{(- 15)}$
$\frac{1}{v} = \frac{- 1}{15} - \frac{1}{30}$
$\frac{1}{v} = \frac{- 2 - 1}{30} = \frac{- 3}{30} = \frac{- 1}{10}$
$\Rightarrow v = - 10 c m$
Characteristics of image :
(i) The image is formed at a distance of 10 cm from the concave lens on the left side.
(ii) Image formed is virtual.
(iii) Image formed is erect.
(iv) The size of the image formed is diminished.

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