Question

The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect?
A. +1.0D
B. -1.0D
C. +3.0D
D. -3.0D

Answer 1

Hint: Proceed the solution of this question first understand what is given in question with clear understanding of near point and far point so that we can decide the value of u and v correctly then simply using Lens formula we can get our answer because we know that reciprocal of focal length will give the power of that particular lens.

Complete step-by-step solution -

In the question, it is given that the near point of defective or hypermetropic eye is 1 m but for a normal eye near point it should be 25 cm. That's why we can let u = −25 cm. The lens used by that person forms its virtual image at a near point of hypermetropic eye i.e., v = −1m =−100 cm. (Using 1m=100cm)
On putting the values of u and v in lens formula,
$ \Rightarrow \dfrac{1}{{\text{v}}} - \dfrac{1}{{\text{u}}} = \dfrac{1}{{\text{f}}}$
$ \Rightarrow \dfrac{1}{{ - 100}} - \dfrac{1}{{ - 25}} = \dfrac{1}{{\text{f}}}$
$ \Rightarrow \dfrac{1}{{ - 100}} + \dfrac{1}{{25}} = \dfrac{1}{{\text{f}}}$
On further solving
$ \Rightarrow \dfrac{{ - 1 + 4}}{{100}} = \dfrac{1}{{\text{f}}}$
$ \Rightarrow \dfrac{3}{{100}} = \dfrac{1}{{\text{f}}}$
On cross multiplication
$ \Rightarrow {\text{f = 33}}{\text{.33 cm or }}{\text{.33 m}}$

Power = $\dfrac{1}{{{\text{f (in meter)}}}} = + 3.0{\text{ Diopter or + 3}}{\text{.0D}}$

Hence C is the correct option.

Note- In this particular question we should know that for a hypermetropic eye convex lens is required to correct it. And convex lenses form virtual and erect images. And also the power of a convex lens is positive therefore the focal length and power of a hypermetropic lens is also positive.