Myopic person uses spectacles of power $-1.0$ diopter or concave lens of focal length $f=\frac{1}{P}=-100 \mathrm{~cm}$ in order to observe clearly object at infinity. For point of the person can be calculated $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}-\frac{1}{-\infty}=\frac{1}{-100}$ or $v=-100 \mathrm{~cm}$.
Similarly if the person uses spectacles of power $+2.0$ dioptre then he must be using convex lens of power $+$ $2.0$ dioptre.
Focal length and near point can be calculated.
$$
f=\frac{1}{P}=\frac{1}{+2}=0.5 \mathrm{~m}=50 \mathrm{~cm}
$$
For near point, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$$
\frac{1}{v}-\frac{1}{-25}=\frac{1}{50} \text { or } \frac{1}{v}-\frac{1}{50}=\frac{1}{25}
$$
$\frac{1}{v}=\frac{1-2}{50}$ or $v=-50 \mathrm{~cm}$
Thus the person also has the defect of hypermetropia and has a near point $50 \mathrm{~cm}$. So with both defects he need different lenses for distant vision and to see close object.