The far point of a myopic person is 80 cm. The power of corrective lens will be

-1.25 D

The correct answer is Option (1) → -1.25 D

For a myopic person, the corrective lens should form the image of a distant object (at ∞) at the far point (80 cm).

Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Here, $u = ∞ \Rightarrow \frac{1}{u} = 0$, $v = -0.80 \, \text{m}$ (virtual image at far point)

$\frac{1}{f} = \frac{1}{-0.80} = -1.25 \, \text{D}$

Power of corrective lens = -1.25 D