The far point of a myopic person is 80 ,cm in | vedclass

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(C) The person is suffering from an eye defect called myopia. In this defect,the image is formed in front of the retina. Hence,a concave lens is used

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Concave lens ; $1.25 \,D$

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(C) The person is suffering from an eye defect called myopia. In this defect,the image is formed in front of the retina. Hence,a concave lens is used to correct this defect of vision.Object distance,$u = \infty$Image distance,$v = -80 \,cm = -0.8 \,m$Focal length = $f$According to the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$\frac{1}{-0.8} - \frac{1}{\infty} = \frac{1}{f}$$\frac{1}{f} = -1.25 \,m^{-1}$$f = -0.8 \,m$We know,Power,$P = \frac{1}{f(	ext{in metres})}$$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.

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?

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