Question

For a person near a point of vision is $100cm$. Then the power of lens he must wear so as to have normal vision, should be:
(A) $ + 1D$
(B) $ - 1D$
(C) $ + 3D$
(D) $ - 3D$

Answer 1

Hint We are given with the near point of vision and are asked about the power of the lens for correcting the vision of the person. Thus, we will use the lens formula for this problem.
Formulae Used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where,$f$ is the focal length of the required lens,$v$ is the image distance of the lens and$u$ is the required object distance.
$P = \dfrac{1}{f}$
Where,$P$ is the power of the lens.

Complete Step By Step Solution
Here,
For being able to see an object distinctively, the object should be placed at the minimum distance of distinct vision.
Thus,
$u = - 25cm$
Also,
Given that the near point of vision is$100cm$.
Thus,
$v = - 100cm$
Now,
Applying the lens formula, we get
$\dfrac{1}{f} = \dfrac{1}{{\left( { - 100} \right)}} - \dfrac{1}{{\left( { - 25} \right)}}$
Further, we get
$\dfrac{1}{f} = \dfrac{1}{{\left( { - 100} \right)}} + \dfrac{1}{{25}}$

Thus,
$\dfrac{1}{f} = \dfrac{1}{{25}} - \dfrac{1}{{100}}$
Further, we get
$\dfrac{1}{f} = \dfrac{{4 - 1}}{{100}}$
Then, we get
$\dfrac{1}{f} = \dfrac{3}{{100}}$
Then, we get
$f = \dfrac{{100}}{3}cm$
Now,
For calculating the power of the lens, we have to take the focal length in meters.
Thus,
We get
$f = \dfrac{{100}}{3} \times \dfrac{1}{{100}}m$
Further, we get
$f = \dfrac{1}{3}m$
Thus,
We will apply the formula for power
$P = \dfrac{1}{f}$
Then,
We will substitute the value for focal length here, we get
$P = \dfrac{1}{{\left( { + \dfrac{1}{3}} \right)}}$
Finally, we get
$P = + 3D$

Hence, the correct option is (C).

Additional Information Fundamentally, the power of the lens is the reciprocal of the focal length of the lens. But, in more depth we can define the power as the tangent of the angle formed by the line joining the focus of the lens with the principal axis.

Note The focal length and the power of the lens is positive. Thus, the lens to be used is a convex one.