A presbyopic patient has near point as 30cm and far point as 40cm. The dioptric power for the corrective lens for seeing distant objects is
A
40D
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B
4D
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C
−2.5D
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D
0.25D
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Solution
The correct option is C−2.5D In this case, for seeing distant objects the far point is 40cm. Hence the required focal length is f=−d ( distance of a far point ) =−40cm Power P=100fcm=100−40=−2.5D