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The far point of a myopic eye is 60 cm. What is the nature and power of the lens required to correct the problem

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The focal length required is

1/f = 1/v - 1/u

here u is taken as infinity and v is -60 cm because the person can see an object up to 60 cm. So a distant object is considered such that the image is formed at 60 cm


1/f=1/(-60) - 0

f=-60 cm=-0.6 m

Power = 1/f = 1/(- 0.6)= -1.667 Diopter.

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Safeer PP

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