The near vision of an average person is $25 \, cm$. To view an object with an angular magnification of $10$,what should be the power of the microscope?

(40 D) For a healthy human eye,the near point is $D = 25 \, cm$. Therefore,the image distance is $v = D = 25 \, cm$.
Given magnification $m = 10$. For a simple microscope,the magnification is given by $m = \frac{D}{f}$.
Rearranging the formula to find the focal length $f$,we get $f = \frac{D}{m}$.
Substituting the values,$f = \frac{25 \, cm}{10} = 2.5 \, cm = 0.025 \, m$.
The power of the lens $P$ is given by $P = \frac{1}{f}$ (where $f$ is in meters).
$P = \frac{1}{0.025} = 40 \, D$ (dioptre).