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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

`1/v -1/u = 1/f`

On putting the respective values, we get:

`1/f = 1/0.02 -1/((-0.1))`

=50+10=60 m

∴ Power of the lens,* P* = `1/f` = 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

`1/v-1/u=1/f`

Putting the values, we get:

`1/f =1/0.02 -1/((-1))`

=50+1=51

∴ Power of the lens = `1/f` = 51 D

So, the range of the power of the eye lens is from + 60 D to + 51 D

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