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The near point of a hypermetropic eye is 1 m. If the near point of the normal eye is 25 cm, then

what is the power of lens required to correct this defect?

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Final answer:

The power of lens required to correct the hypermetropic eye is 1 diopter.


Step-by-step explanation:

The near point of a hypermetropic eye is the closest point at which a person with hypermetropia (farsightedness) can see objects clearly without straining their eyes. In this case, the near point is given as 1 m. On the other hand, the near point of a normal eye is typically around 25 cm.

To correct the hypermetropic eye, a converging lens (convex lens) is needed. The power of the lens required can be calculated using the formula:

Power of lens = 1 / Near point (in meters)

Using the given near point of 1 m, the power of lens required to correct this defect would be:

Power of lens = 1 / 1 = 1 diopter


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