Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Answers (1)

correction for hypermetropia

Given the near point for the hypermetropic eye is 1m.

Assume the near point of the normal eye is 25 cm.

So,

The object distance will be, u = -25 cm (Normal near point).

The image distance, v = -1 m (Near point of this defective eye) or -100 cm.

Then the focal length can be found from the lens formula:

frac1v - frac1u = frac1f

Substituting the values in the equation, we obtain

Rightarrow frac1-100 - frac1-25 = frac1f

Rightarrow frac-1+4100 = frac1f

Rightarrow frac3100 = frac1f

Rightarrow f = frac1003 = 33.3 cm or 0.333 m

Hence the power of the lens will be:

P =frac1f(m) = frac1+0.333m = +3.0 D

Thus, the power of the convex lens required will be +0.3 D

 

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