A divergent lens has a focal length of $30$ cm. At what distance should an object of height $5$ cm from the optical centre of the lens be placed so that its image is formed $15$ cm away from the lens? Find the size of the image also.

Question

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Answer 1

We know that for a divergent lens both u and f are -ive , from lens equation

$v = \frac{u f}{u + f}$ , as both u and f are -ive therefore v will be -ive ,

given $f = - 30 c m, v = - 15 c m$ ,

so $u = \frac{f v}{f - v} = \frac{- 3 0 \times - 1 5}{- 3 0 - ( - 1 5 )} = - 30 c m$ ,

now linear magnification $m = I / O = v / u$ ,

given height of object $O = 5 c m, I = ?$ ,

therefore $I / 5 = 15 / 30$ ,

or $I = 5 \times 15 / 30 = 2.5 c m$

Answer 2

We know that for a divergent lens both u and f are -ive , from lens equation

$v = \frac{u f}{u + f}$ , as both u and f are -ive therefore v will be -ive ,

given $f = - 30 c m, v = - 15 c m$ ,

so $u = \frac{f v}{f - v} = \frac{- 3 0 \times - 1 5}{- 3 0 - ( - 1 5 )} = - 30 c m$ ,

now linear magnification $m = I / O = v / u$ ,

given height of object $O = 5 c m, I = ?$ ,

therefore $I / 5 = 15 / 30$ ,

or $I = 5 \times 15 / 30 = 2.5 c m$