Myopic person uses spectacles of power – 1.0 dioptre or concave lens of focal length f = $\frac{1}{P}$ = - 100 cm in order to observe clearly object at infinity. Far point of the person can be calculated as $\frac{1}{v} - \frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v} - \frac{1}{-\infty}$ = $\frac{1}{-100}$ or v = - 100 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 as f = $\frac{1}{p}$ = $\frac{1}{+2}$ = 0.5 m = 50 cm For near point, $\frac{1}{v} - \frac{1}{u}$ = $\frac{1}{f}$ $\frac{1}{v} - \frac{1}{-25}$ = $\frac{1}{50}$ or $\frac{1}{v}$ = $\frac{1}{50} - \frac{1}{25}$ ⇒ $\frac{1}{v}$ = $\frac{1-2}{50}$ or v = - 50 cm Thus the person also has the defect of hypermetropia and has a near point 50 cm. So having both defects he needs different lenses for distant vision and to see closer objects.