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Question

The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is 25 cm)


A

+3 D

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B

- 3 D

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C

+4 D

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D

-4 D

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Solution

The correct option is A

+3 D


If the object is placed at 25 cm (N) , then the image has to be formed at 1 m (N) (because N is the far point for the hypermetropic eye) , which act as the object for eye lens, which forms the final image at the retina.

Let v be the image distance, u be the object distance and f be the focal length.
Using lens formula we have,
1v1u=1f
Now, applying cartesian convention,
v=1 m, u=25 cm=14 m
Therefore,
1141=1f
1f=+3 m
Thus power is +3 D


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