Question

A presbyopic patient has near point as $30cm$ and far point as $40cm$. The dioptric power for the corrective lens for seeing distant objects is

• A. $40D$
• B. $4D$
• C. $-2.5D$
• D. $0.25D$

Answer 1

The correct option is C $-2.5D$

In this case, for seeing distant objects the far point is $40cm$.
Hence the required focal length is
$f=-d$ ( distance of a far point ) $=-40cm$
Power $P=\frac{100}{f}cm=\frac{100}{-40}=-2.5D$