Magnification for erect and inverted image is:

+ v e

and

- v e

respectively

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- v e

and

+ v e

respectively

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+ v e

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- v e

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Solution

The correct option is A $+ v e$ and $- v e$ respectively
Magnification is the ratio of height of image and object.
$m = \frac{h_{i}}{h_{o}}$
And by convention, height of image formed below principal axis is taken negative and above is taken positive.

Hence,
for erect image, $m = + v e$
for inverted image $m = - v e$ .

Answer is option (A).

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