91Ó°ÊÓ

The unit of measurement for the optical power of a lens or curved mirror is the dioptre of power of a lens, which is equal to the reciprocal of focal length.

Power of the normal eyes is $P_{Normal}=50.0d$.

The far point of man is,$d_{∘}=20.0cm\left(\frac{1m}{100cm}\right)=0.20m$

The lens to retina distance is$d_i=2.00cm\left(\frac{1m}{100cm}\right)=0.20m$

The power of the severely myopic eye P_Myopic can be expressed as,

$P_{Myopic}=\frac{1}{d_0}+\frac{1}{d_0}$........................(1)

And, the number of diopterscan be obtained using:

$P_{Corrected}=P_{Normal}-P_{Myopic}$...................(2)

Substitute the given data in equation (1), we will get,

$P_{Myopic}=\frac{1}{0.20m}+\frac{1}{0.02m}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(3)

Substituting data from equation (3) into equation (2) gives,

$$\begin{gathered} P_{Myopic}=50.0D-55.0D \\ =-5.00D \end{gathered}$$

Therefore, a myopic man's visual power should be decreased (corrected) by \( - {\rm{5}}{\rm{.00}}\;{\rm{D}}\).