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
[tex]P =\dfrac{1}{f}[/tex]
[tex]P =\dfrac{1}{-0.40}[/tex]
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
[tex]p = \dfrac{qf}{q-f}[/tex]
[tex]p = \dfrac{(-0.18)(-0.4)}{-0.18-(-0.4)}[/tex]
p = 32.72 cm
The person's near point is 32.72 cm
