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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Solution

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Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{8} = \frac{1}{10}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = - \frac{1}{40}$

$v = - 40 c m$

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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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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.581020

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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.
5
8
10
20

Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{8} = \frac{1}{10}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = - \frac{1}{40}$

$v = - 40 c m$