Question
A person is suffering from hypermetropia (long sightedness). It is a defect in which a human eye can see far off object clearly, but is unable to see nearby object distinctly. The near point of the person is $1.5 m$. Assume that the near point of the normal eye is $25 cm$.

(i) What type of lens should be used in his spectacles?
(ii) What should be the focal length of the lens he used?
(iii) What will be the potver of the lens?
or
(iv) Write one possible cause of this defect.

Answer

(i) Convex lens is to be used in his spectacles.
(ii) Here,
$\begin{aligned}d & =150 cm \\D & =25 cm \\v & =-150 cm \\u & =-25 cm\end{aligned}$
Using lens formula,
$\begin{aligned}\frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\& =\frac{1}{-150 cm }-\frac{1}{-25 cm } \\& =\frac{-1+6}{150 cm }=\frac{5}{150 cm } \\f & =30 cm\end{aligned}$
(iii) Power of the lens,
$P=\frac{100}{f} D =\frac{100}{30} D =3.3 D$
or
(iv) Increase in focal length of the eye lens, when the eye is fully relaxed is the cause of hypermetropia.

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