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Question

The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens?

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Solution

Given:
Near point of the child = 10 cm
Far point of the child = 100 cm
The retina is at a distance of 2 cm behind the eye lens.
Thus, we have:
v = 2 cm = 0.02 m
When the near point is 10 cm,
u = − 10 cm = − 0.1 m
v = 2 cm = 0.02 m
The lens formula is given by
1v-1u=1f
On putting the respective values, we get:
1f=10.02-1-0.1 =50+10=60 m
∴ Power of the lens, P = 1f = 60 D
Now, consider the near point 100 cm.
Here,
u = − 100 cm = − 1 m
v = 2 cm = 0.02 m
The lens formula is given by
1v-1u=1f
Putting the values, we get:
1f=10.02-1-1 =50 +1=51
∴ Power of the lens = 1f = 51 D
So, the range of the power of the eye lens is from + 60 D to + 51 D.

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