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?

Asked by Vishal kumar | 1 year ago |  170

Solution :-

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

Object distance (u) = infinity = ∞

Image distance (v) = - 80 cm

Focal length = f

According to the lens formula,

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$

$$-\frac{1}{80}-\frac{1}{∞}=\frac{1}{f}$$

$$\frac{1}{f}=-\frac{1}{80}$$

$$f=-80\,cm=-0.8\,m$$

We know,

$$Power,P=\frac{1}{f(in \,meters)}$$

$$P=\frac{1}{-0.8}=-1.25\,D$$

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

Answered by Shivani Kumari | 1 year ago

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