  Question

The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).

Solution

A person suffering from hypermetropia can correct the defect by wearing spectacles with convex lenses. In order to find the power of the convex lens required, we have to first calculate its focal length. Given that the near point of the hypermetropic eye is 50 cm in front of the eye (the person can see an object kept at the normal near point of 25 cm from the eye if the image of the object is formed at the person's own near point of 50 cm from the eye). u = -25 cm (the distance of the object at the normal near point) v = -50 cm (the location of the near point of the defective eye) f = ? (focal length) The focal length can be calculated using the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. Substituting the values in the formula, we get $\frac{1}{-50}-\frac{1}{-25}=\frac{1}{f}$. $\frac{1}{-50}+\frac{1}{25}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}\frac{\left(-1+2\right)}{50}=\frac{1}{50}=\frac{1}{f}$  ∴ $f=50cm=0.5m$ Now that we know the focal length of the convex lens, the power can be calculated. Power $P=\frac{1}{f\left(inmetres\right)}$ ∴ Hence, the power of the convex lens required to rectify this defect is +2 dioptres.ScienceLakhmir Physics 2013Standard X

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