Question

For a person with near point of vision is $100{\text{cm}}$then the power of lens he must wear so as to have a normal vision would be:
(A) $ + 1D$
(B) $ - 1D$
(C) $ + 3D$
(D) $ - 3D$

Answer 1

Hint
Normally the near vision of the human eye is taken to be $25cm$, in case of some defect, when this number becomes greater than that, it is desired to reduce this number back to $25cm$by using an appropriate lens. So first find the focal point of this lens and then it’s power.
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$(Lens formula)
Where, f= focal length of the lens
v= distance of the image formed
u= distance of the object
$\Rightarrow P = \dfrac{1}{f}$
Where P=Power of the lens

Complete step by step answer
Here, it is required to move the object’s position back from$100cm$to an image at a distance of 25 cm for comfortable viewing. So we take the values,
$\Rightarrow u = 100cm$
$\Rightarrow v = 25cm$
On applying the lens formula,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{25}} - \dfrac{1}{{100}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{{4 - 1}}{{100}} = \dfrac{3}{{100}}$
$\Rightarrow f = 33.33{\text{cm}}$
The power of a lens, is given by-
$\Rightarrow P = \dfrac{1}{f}$
Where f should be in meters so, on converting we get, $f = 0.33m$ $(\because 1cm = 0.01m)$
Then,
$\Rightarrow P = \dfrac{1}{{0.33}} = 3$
The power is $ + 3D$.
The positive sign tells us that the required lens for this job is a Convex lens or a converging lens.
Thus, option (C) is correct.

Additional Information
This medical condition is known as hypermetropia or long-sightedness. The cause of this condition is either too small eyeballs or when the focal length of the eye is too long. In both of the cases image is formed behind the retina, and by using a convex lens, the light rays are converged in such a way that they form in the retina.

Note
It should be taken care of that the calculation of power should always be done after converting the focal length to meters. Because the definition of Dioptre (D) itself is defined in meters, the unit $D$has a dimension of ${m^{ - 1}}$.