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 the problem?​

Answer 1

Answer:

The person is suffering from myopia and hence need a concave lens to correct the defect. The lens should be such that an object at infinity must form its image at the far point.

Nature of lens diverging.

f = −80 cm = −0.8 m

P = 1/f

P = 1/(−0.8)

P = −1.25 D

HOPE IT HELPS!

Answer 2

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.