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.

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Solution

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 $u = \frac{f v}{f - v} = \frac{- 30 \times - 15}{- 30 - (- 15)} = - 30 c m$ ,

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

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

therefore $I / 5 = 15 / 30$ ,

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

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

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.