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The near point of a patient's eye is 50.0 cm. What po...

The near point of a patient's eye is 50.0 cm. What power (in diopters) must a corrective lens have to enable the eye to see clearly an object 25.0 cm away?

A.

2 diopters

B.

2.5 diopters

C.

0.5 diopters

D.

3 diopters

Correct answer is A

The patient cannot see clearly an object closer than 50 cm

Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens

So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens

u=25cm ;v=-50cm (virtual image); p=

1f=1u + 1v

1f =150

1f125-150

1f = f = 50cm =0.5m

1f = 2150

   p = 1f

   p =  10.5

   p =   2 diopters 
 
  ; The patient needs a converging lens with a power of 2 diopters