An object is placed at a distance of f2 from a convex lens. The image will be

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An object is placed at a distance of $\frac{f}{2}$ from a convex lens. The image will be

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$At \frac{3 f}{2}, real and inverted$

At 2f, virtual and erect

At one of the foci, virtual and double its size

None of these

answer is A.

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Detailed Solution

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} (Given u = \frac{- f}{2}) \Rightarrow \frac{1}{f} = \frac{1}{v} + (\frac{1}{f / 2}) \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{2}{f} \Rightarrow \frac{1}{v} = \frac{- 1}{f} and m = \frac{v}{u} = \frac{f}{f / 2} = 2 So virtual at the focus and of double size$