Question

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct this defect of vision?

Answer 1

Answer:

The person is suffering from a eye defect called myopia . In this defect, image formed in the front of retina. Hence, a concave lens is used to correct this defect of vision.

Object distance, u = infinity = $\infty$

Image distance, v = -80 cm

Focal length = f

According to the lens formula -

$\\ \sf \: \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\ \\ \\ \implies \sf \: \dfrac{1}{ - 80} - \dfrac{1}{ \infty } = \dfrac{1}{f} \\ \\ \\ \implies \sf \: \dfrac{1}{f} = - \frac{1}{80} \\ \\ \\ \implies \sf \: f = - 80 \: \: cm = - 0.8 \: \: cm.$

We know ,

$\\ \sf \: Power,p = \frac{1}{f \: ( \: in \: metres \: ) \: } \\ \\ \\ \implies \sf \: P = \frac{1}{ - 0.8} = - 1.25 \: D$

A concave lens of power -1.25 D is required by the person to correct his defect.

Answer 2

Answer:

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The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem? The corrected lens should be such that it forms the image at the retina instead of, infront of the retina. Since, power of the lens is negative, remedial lens is a concave lens.