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25/09/2021 7:04 pm
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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.
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25/09/2021 7:10 pm
The required diagram is shown below:
Given: Distance of near point of defective eye, i.e.
v = -1 m = -100 cm
Distance of near point of normal eye, i.e.
u = -25 cm
Applying the formula \(\frac{1}{f}\) = \(\frac{1}{v}\) - \(\frac{1}{u}\), we get,
\(\frac{1}{f}\) = \(\frac{1}{-100}\) - \(\frac{1}{-25}\)
= \(\frac{-1}{100}\) + \(\frac{1}{25}\)
= \(\frac{-1 + 4}{100}\) = \(\frac{3}{100}\)
or f = \(\frac{100}{3}\) = 33.33 cm
∴ Power, P = \(\frac{100}{f}\) = \(\frac{100}{33.3}\)
= 3.0 D
Hence, the power of the required lens is 3.0 D.