A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an erect and virtual image at 10 cm from the lens?

Foundation

Physics Foundation

Refraction and parallel slab

Question

15 cm

10 cm

30 cm

60 cm

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Given :

Focal length f =– 15cm

Image distance v = –10cm

Using lens formula,

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

$\Rightarrow \frac{1}{u} = - \frac{1}{10} - (\frac{1}{- 15})$

$= - \frac{1}{10} + \frac{1}{15} = \frac{- 3 + 2}{30}$

⇒ u =–30cm

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