if the far point of eye lens is 1.1 m.find the power required to correct the defect.

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Solution

$H e r e f o c a l d i s \tan c e = f = - 1.1 m (n e g a t i v e a s c o n c a v e l e n s i s r e q u i r e d) S o p o w e r = p = 1 / f = - \frac{1}{1.1} = - 0.909 D$

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if the far point of eye lens is 1.1 m.find the power required to correct the defect.

Here focal distance=f= -1.1 m (negative as concave lens is required)So power= p=1/f =-11.1 =-0.909D

$H e r e f o c a l d i s \tan c e = f = - 1.1 m (n e g a t i v e a s c o n c a v e l e n s i s r e q u i r e d) S o p o w e r = p = 1 / f = - \frac{1}{1.1} = - 0.909 D$