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?

Answer 1

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

Let us consider the near point 100 cm.

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.

Answer 2

HIIII BUDDY!!!!

Given,

Near point and far point is 10 cm and 100 cm

We know that the retina lies behind the eye lens

v= 2 cm (given)

= 0.02 cm

u= -10 cm (given)

=0.5 m

So,

$\frac{1}{ F \: near } = \frac{1}{v} - \frac{1}{u}$

$= \frac{1}{0.02} - \frac{1}{0.1}$

$= 50 + 10$

$= 60 \: \: D \:$

TO FIND NEAR POINT,

$\frac{1}{F \: far} = \frac{1}{v} - \frac{1}{u}$

$= \frac{1}{0.02} - \frac{1}{ - 1}$

$= 50 + 1$

$= 51 \: \: D \:$

THEREFORE,

RANGE IS +60 D TO + 51 D.