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?
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=
1f=1u + 1v
⇒1f =150
⇒1f = 125-150
⇒1f = f = 50cm =0.5m
⇒1f = 2−150
p = 1f
p = 10.5
p = 2 diopters
; The patient needs a converging lens with a power of 2 diopters
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