A woman can read the large print in a newspaper only when it is at a distance of
ID: 1263225 • Letter: A
Question
A woman can read the large print in a newspaper only when it is at a distance of 56 cm or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn 2.6 cmfrom the eyes), so she can read the newspaper at a distance of 28 cm from the eyes? A woman can read the large print in a newspaper only when it is at a distance of 56 cm or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn 2.6 cmfrom the eyes), so she can read the newspaper at a distance of 28 cm from the eyes?Explanation / Answer
(A) Because the age of the woman is not cleared in this question we can assume she is hyperopic because the near point is 25cm from the retina. Since her near point is further than that, she is unable to see objects closer than that clearly so there is farsighted (hyperopic) defect in her eyes. Hyperopic eye requires a positive power lens to move the hyperopic near and far points inward.
(B) Now for refractive power claculation-
We know that 1/v+1/u=1/f u-distance of object, v-distance of image and f-focal distance
here according to question distance of object will be u=(28 - 2.6)cm=25.4cm
and v=56 - 2.9=53.12cm
now 1/f= -1/53.12+1/25.4
1/f=0.0205
f=48.67cm
power P=100/f(cm)=100/48.67=2.05D. Ans.
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