Question

An individual is nearsighted; his near point is 18.0 cm and his far point is 40.0 cm.

Answer 1

Answer

given,

near point = 18 cm

far point = 40 cm

a) The lens should form an upright, virtual image at far point from the distant object.

therefore, f = q  = -40 cm = -0.4 m

where f is the focal length.

the required power

$P =\dfrac{1}{f}$

$P =\dfrac{1}{-0.40}$

       P = -2.5 D

b) If the lens is used the Person's near point

The lens should form an upright, virtual image at near point from the distant object should be q = - 18 cm = -0.18 m

    $p = \dfrac{qf}{q-f}$

    $p = \dfrac{(-0.18)(-0.4)}{-0.18-(-0.4)}$

     p = 32.72 cm

The person's near point is 32.72 cm