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For a normal eye, the far point is at infinity and the near point of distanct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converaging power of the eye-lens) of a normal eye. |
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Answer» <P> SOLUTION :Total power = 40 + 20 = 60 dioptreu = - `infty ,f = (1)/(P) = (1)/(60) m, (1)/(V) - (1)/(u) = (1)/(f) "" therefore (1)/(v) - (1)/((-infty)) = 60 "" therefore v = (1)/(60) m = (100)/(60) ` cm i.e.,v = `(5)/(3) ` cm To focus the object at the near point , u = - 25 cm, v = `(5)/(3) ` cm `therefore (1)/(f) = (1)/(v) - (1)/(u) = (3)/(5) - (1)/((-25)) = (3)/(5) + (1)/(25)= (15 + 1)/(25) = (16)/(25) "" therefore " power" (1)/(f)= (16)/(25)xx 100 = 64 ` dioptres |
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