A magnifying glass is to be used at a fixed o | vedclass

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(C) For a magnifying glass,the image formed is virtual and erect. Thus,the magnification $m = +5$.Using the magnification formula $m = \frac{v}{u}$,we

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$1.25$

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(C) For a magnifying glass,the image formed is virtual and erect. Thus,the magnification $m = +5$.Using the magnification formula $m = \frac{v}{u}$,we have $5 = \frac{v}{u}$,which implies $v = 5u$.Given the object distance $u = -1 	ext{ inch}$ (using sign convention).Therefore,the image distance $v = 5 	imes (-1) = -5 	ext{ inch}$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-5} - \frac{1}{-1} = -0.2 + 1 = 0.8$.Thus,$f = \frac{1}{0.8} = 1.25 	ext{ inch}$.

$A$ magnifying glass is to be used at a fixed object distance of $1 \text{ inch}$. If it is to produce an erect image magnified $5$ times,its focal length should be ........ $\text{inch}$.

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