Answer

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**Hint:**The formulae such as, the lens formula, that relates the focal length, the object distance and the image distance, the power formula, that relates the focal length and the power of the lens should be used to solve this problem.

**Formula used:**

\[\begin{align}

& \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\

& P=\dfrac{1}{f} \\

\end{align}\]

**Complete step by step answer:**

From the given information, we have the data as follows.

A certain far-sighted person cannot see objects closer to the eye than 100 cm. The distance of 25 cm.

As the values of the image distance and the object distance are given, so, we will be using the formula that relates the focal length, the object distance and the image distance, the reason being, the inverse of the focal length gives the power of the lens.

Consider the lens formula.

\[\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\]

Substitute the values of the image distance and the object distance in the above equation.

\[\dfrac{1}{f}=\dfrac{1}{-100}-\dfrac{1}{-25}\]

Continue further computation.

\[\begin{align}

& \dfrac{1}{f}=\dfrac{1}{-100}+\dfrac{1}{25} \\

& \therefore \dfrac{1}{f}=\dfrac{3}{100}cm \\

\end{align}\]

Thus, the value of the focal length is,

\[\begin{align}

& \dfrac{1}{f}=\dfrac{3}{100} \\

& \Rightarrow \dfrac{1}{f}=\dfrac{3}{100}c{{m}^{-1}} \\

& \therefore f=\dfrac{1}{3}m \\

\end{align}\]

The power formula that relates the power of the lens and the focal length of the lens is given as follows.

Consider the lens formula.

\[P=\dfrac{1}{f}\]

Substitute the value of the focal length in the above equation.

\[P=\dfrac{1}{{}^{1}/{}_{3}}\]

Thus, the value of the power of the lens is,

\[P=3\,D\]

\[\therefore \] The power of the lens which will enable a certain far-sighted person to read at a distance of 25 cm will be 3 dioptres.

**So, the correct answer is “Option A”.**

**Note:**The lens formula gives the relation between the focal length, the image distance and the object distance, the mirror formula gives the relation between the focal length, the image distance and the object distance, but, the difference being the sign used (+/-).

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