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CBSE Class 10 Science 2020 Delhi Set 3

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Question : 8
Total: 10
The near point of the eye of a person is 50‌cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25‌cm from the eye.
Solution:  
(a) It is the case of hypermetropia.
For a hypermetropic eye, u=−25‌cm and v=−50‌cm
Using lens formula
‌‌
1
f
=‌
1
v
−‌
1
u

‌‌
1
f
=−‌
1
50
+‌
1
25

‌‌
1
f
=‌
1
50

We get focal length is f=50‌cm=0.5m
From the formula, P=‌
1
f

P=‌
1
0.5
=2D
So the power of the lens is 2 dioptre and the nature of lens is a converging lens or a convex lens.
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