Question

A person's right eye can see objects clearly only if they are between 25 cm and 87 cm away.

What power of contact lens is required so that objects far away are sharp?

What will be the near point with the lens in place?

Answer 1

a) In this case person is near sighted.To correct nearsightedness, contact lens ( diverging lens) is used.

For a distance object ( infinity far away) to be sharp, the virtual image formed by this object must be 80 cm from person's eye. so from lens equation:

1/V - 1/u = 1/f ( V = image distance = -87cm, u = object distance = infinity)

1/f = 1/ -87

f( focal length) = -87 cm

P(power) = 1/f

P = 1/ - 0.87m

P = -1.15D

b) When the object is at near point , the virtual image formed by lens will be 25cm in front of lens ( so that person can focus it)

So from lens equation:

1/V - 1/u = 1/f

1/u = 1/V - 1/f

1/u = 1/ (-25) - 1/( -87) ( here V= -25 cm, f= -87cm)

u = - 35.09 cm

near point will be at 35.09cm .

I hope help you !!