A wire mesh consisting of very small squares | vedclass

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(A) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given,the object distance $u = -8 \, cm$ (using sign convention) and the fo

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$5$

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(A) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Given,the object distance $u = -8 \, cm$ (using sign convention) and the focal length $f = +10 \, cm$ for a converging lens.Substituting these values into the lens formula:$\frac{1}{v} - \frac{1}{-8} = \frac{1}{10}$$\frac{1}{v} + \frac{1}{8} = \frac{1}{10}$$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = -\frac{1}{40}$Thus,$v = -40 \, cm$.The magnification $m$ produced by the lens is given by $m = \frac{v}{u}$.$m = \frac{-40}{-8} = 5$.Therefore,the magnification produced by the lens is $5$.

$A$ wire mesh consisting of very small squares is viewed at a distance of $8 \, cm$ through a magnifying converging lens of focal length $10 \, cm$,kept close to the eye. The magnification produced by the lens is

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