Q20 of 70 Page 11

The near point of a hypermetropic eye is 1m. What is the nature and  power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.

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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A person is unable to see objects nearer than 50 cm. He wants to read a book placed at a distance of 25 cm. Find the nature, focal length and power of the lens, he requires for his spectacles.

 19

A student is unable to see clearly the words written on the blackboard placed at a distance of approximately 4 m from him. Name the defect of vision the boy is suffering from. Explain the method of correcting this defect Draw ray diagram for the:

i. defect of vision and also


ii. for its correction.

 21

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?

 22

An old man cannot see objects closer than 1 m from the eye clearly. Name the defect of vision he is suffering from. How can it be corrected? Draw ray diagram for the (i) defect of vision and also(ii) for its correction.