Question

The near point of a myopic eye is $15{\text{ }}cm$. What should be the power of the lens which will enable the eye to read a book placed at $30{\text{ }}cm$?
A. $3.33$ dioptre
B. $ - 3.33$ dioptre
C. $2.22$ dioptre
D. $ - 2.22$ dioptre

Answer 1

Hint:Myopia also known as short sightedness which refers to the defect in the eye when a person is not able to distinctly see objects at far distance. By using the lens formula we will find the focal length of the lens and then by inverting it we will find the power of the lens.

Complete step by step answer:
The Lens Formula is stated as-
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$.
Here, $v = $ the distance of the image, $u = $ the distance of the object and $f = $ focal length of the lens.

The near point of myopic eye refers to the distance till which an image is seen clearly. So, the image distance $v = 15{\text{ }}cm$ here. The object distance is given in the question as $u = 30{\text{ }}cm$.
$v = - 15{\text{ }}cm$ as the image is formed on the left side of the pole.
$u = - 30{\text{ }}cm$, object is placed in front of the lens.
Hence by substituting the values in the lens formula we get,
$ - \dfrac{1}{{15}} + \dfrac{1}{{30}} = \dfrac{1}{f}$
Simplifying the equation we get,
$ - \dfrac{1}{{30}} = \dfrac{1}{f}$
Thus we get, $f = - 30{\text{ }}cm = - 0.3{\text{ }}m$
So, the power of lens$ = \dfrac{1}{f} = - \dfrac{1}{{0.3}} = - 3.33{\text{ }}D$
So, the power of the lens is $ - 3.33$ Dioptre.

Hence, the correct answer is option A.

Note: It must be noted that in case of any lens, when the image is real then the sign is negative. For virtual images, sign convention is defined as positive. The power of a convex lens is negative which is used for the correction of myopia and the concave lens is used to correct hypermetropia which is positive in nature.