Q19 of 25 Page 11

Make a diagram to show how hypermetropia is corrected and what are the causes of this problem. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm. ( CBSE 2008)

The eye defect called Hypermetropia is corrected by using a convex lens spectacles.

We will first calculate the focal length of the convex lens required in this case. For hypermetropic eye can see the nearby object kept at 25 cm (at near point if normal eye) clearly if the image of this object is formed at its own near point which is 1 meter here. So, in this case :

Object distance, u = -25 cm (Normal near point)

Image distance, v = -1 m (Near point of this defective eye) = -100 cm

Focal length, f= ? (To be calculated)

Putting these values in the lens formula,

By putting in values

We get : 


Or, 



Power of Lens needs to be Calculated ,
Given by P=
∴ We convert 33.3 cm into meter i.e 0.33m
P=
= +3.0 D
Thus, the power of convex lens required is +0.3 diopters.

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