Question

A young boy can adjust the power of his lens between 50D to 60D. If his far point is infinity, then find his near point.​

Answer 1

Answer:

1.67 to 2 cm

Explanation:

D = 1m / f

f = 1m / D

case 1

f = 1m or 100 cm / 50

f = 2 cm

case 2

f = 100 / 60

f = 1.67 cm

so near point is between 1.67 to 2 cm

Answer 2

Answer:

The near point of the boy is 10cm, if his far point is infinity.

Explanation:

When the power of lens, $P=50D$

then focal length $f=\frac{1m}{P}=\frac{100cm}{50cm}=2cm$

When the power of lens, $P=60D$

Focal length, $f=\frac{1}{P}=\frac{1m}{60cm}=\frac{100cm}{60cm} =\frac{10}{6} =1.67cm$

If  the distance between $retina$ from eye-lens will $2cm$ then eye will be relaxed.

To find the near point, use lens maker formula:

$v=2cm, f=\frac{10}{6}$  and $u=?$

⇒  $\frac{1}{u} =\frac{1}{v} -\frac{1}{f}$

⇒  $\frac{1}{u}=\frac{1}{2}-\frac{6}{10}$

⇒  $\frac{1}{u}=\frac{-5+6}{10}$

⇒  $\frac{1}{u}=\frac{-1}{10}$

∴   $u=-10cm$