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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.

Q.7.    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.

 

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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:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Substituting the values in the equation, we obtain

\Rightarrow \frac{1}{-100} - \frac{1}{-25} = \frac{1}{f}

\Rightarrow \frac{-1+4}{100} = \frac{1}{f}

\Rightarrow \frac{3}{100} = \frac{1}{f}

\Rightarrow f = \frac{100}{3} = 33.3\ cm\ or\ 0.333\ m

Hence the power of the lens will be:

P =\frac{1}{f(m)} = \frac{1}{+0.333m} = +3.0\ D

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

 

 

 

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