CBSE Class 10 Answered
The far point of a myopic eye is 1.5 m. To correct this defect of the eye, the power of lens is -
Asked by dhairya | 23 Aug, 2009, 10:23: AM
Expert Answer
u = - 1.5 m; v = -0.25 m;
Using
(1 / f) = (1/ v) - (1 / u)
1 / f = - 1 / 0.25 + 1 / 1.5 = (- 6 + 1 ) / 1.5 = - 3.33
P = - 3.33 D
Answered by | 24 Aug, 2009, 12:41: PM
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