(a) Calculate the power of eye lens of a normal eye when it is focused at (i) far point (ii) near point.Assume the distance of retina from the eye lens to be 2.5 cm.
(b) The far point of a myopic person is 150 cm. Calculate the focal length and the power of the lens required that enables him to see distant objects clearly.
Answers
the a) question is not complete. The correct question is
a) Calculate the power of the eye - lens of the normal eye when it is focused at its (a) far point, infinity and (b) near point, 25 cm from the eye. Assume the distance of the retina from the eye - lens to be 2.5 cm.
Answer:
a) Given:
As eye lens is convex lens so,
focal length ( f ) =2.5cm=.025m
power ( P )= 1 / f
Solution:
i) power ( P ) = 1 / f
= 1 / .025
= 40D
Hence, the power of normal eye will be 40 D.
ii) Near point ( u ) = 25 cm
and image at retina ( v ) = 2.5 cm
1 / f = 1 / v - 1 / u
= 1 / 2.5 - ( - 1 / 25 )
= 25 / 11 cm
f = 25 / 1100 cm
Thus, P = 1 / f
= 1 / 25/1100
= 44D
Hence, the power of normal eye will be 44 D.
b) Far point or image at retina ( v ) = 150 cm
Near point ( u ) = ∞
1 / f = 1 / v - 1 / u
= - 1 / 150 - 1 / ∞
= - 1 / 150
f = - 1.5 m
Thus, P = 1 / f
= 1 / 1.5
= - 0.66 D
Hence, the concave lens of power - 0.66 D will be used.
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