In order to obtain a real image of magnificat | vedclass

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(D) For a converging lens,the magnification $m$ is given by $m = \frac{f}{u + f}$.Since the image is real,the magnification $m$ must be negative,so $m

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$-30$

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(D) For a converging lens,the magnification $m$ is given by $m = \frac{f}{u + f}$.Since the image is real,the magnification $m$ must be negative,so $m = -2$.The focal length $f$ of a converging lens is positive,so $f = 20 \ cm$.Substituting these values into the formula: $-2 = \frac{20}{u + 20}$.Multiplying both sides by $(u + 20)$,we get $-2(u + 20) = 20$.Dividing by $-2$,we get $u + 20 = -10$.Therefore,$u = -10 - 20 = -30 \ cm$.The object should be placed at a distance of $30 \ cm$ in front of the lens.

In order to obtain a real image of magnification $2$ using a converging lens of focal length $20 \ cm,$ where should an object be placed in $cm$?

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