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?
2 diopters
2.5 diopters
0.5 diopters
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=
\(\frac{1}{f}\)=\(\frac{1}{u}\) + \(\frac{1}{v}\)
⇒\(\frac{1}{f}\) =\(\frac{1}{50}\)
⇒\(\frac{1}{f}\) = \(\frac{1}{25}\)-\(\frac{1}{50}\)
⇒\(\frac{1}{f}\) = f = 50cm =0.5m
⇒\(\frac{1}{f}\) = \(\frac{2-1}{50}\)
p = \(\frac{1}{f}\)
p = \(\frac{1}{0.5}\)
p = 2 diopters
; The patient needs a converging lens with a power of 2 diopters
