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The near point of a hypermetropic eye is...

The near point of a hypermetropic eye is `50 cm`. Calculate the power of the lens to enable him to read a book at `40 cm` ?

Text Solution

Verified by Experts

The correct Answer is:
`0.5 D`
Here, `x' = 50 cm, d = 40 cm, P = ?`
As `f = (x' d)/(x' - d), f = (50 xx 40)/(50 - 40) = 200 cm`,
`P = (100)/(f) = (100)/(200) = 0.5 D`.
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Knowledge Check

  • The near point of hypermetropic eye is 40 cm . What is the power of the lens used to correct this defect ?

    A
    2.5 D
    B
    1.5 D
    C
    `-1.5 D `
    D
    0.5 D

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