A person with myopic vision see objects only up to a distance of $0.25m.$ To be able to read a book placed at the distance of $0.50m$ from the eyes, the spectacles needed by this person have power of
$
A.{\text{ - 1}}{\text{.0D}} \\
{\text{B}}{\text{. - 1}}{\text{.5D}} \\
{\text{C}}{\text{. - 2}}{\text{.0D}} \\
{\text{D}}{\text{. + 2}}{\text{.0D}} \\
$

Hint:The myopic vision is the common vision condition in which the person can see near objects clearly but far objects seem to be blurry and unclear. This condition is also known as near-sightedness. Here use the formula of lens and substitute the given values and simplify for the required answer.

Complete step by step answer:
Given that –
The Image distance, $v = - 0.25m$ (Since image formed here is virtual)
The object distance, $u = - 0.5m$ (since object is placed in front)
By using the Lens formula –
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Place the given values in the above equations –
$\dfrac{1}{{ - 0.25}} - \dfrac{1}{{ - 0.5}} = \dfrac{1}{f}$
Minus and minus makes plus and Simplify the above decimals first, convert decimals in to fractions –
$\dfrac{{ - 100}}{{25}} + \dfrac{{10}}{5} = \dfrac{1}{f}$ (Since, denominator’s denominator goes to the numerator)
Now, simplify and make the unknown “f” the subject –
\[\dfrac{1}{f} = \dfrac{{ - 100}}{{25}} + \dfrac{{10}}{5}\]
Take LCM and simplify
\[
\dfrac{1}{f} = \dfrac{{ - 100}}{{25}} + \dfrac{{50}}{{25}} \\
\Rightarrow\dfrac{1}{f} = \dfrac{{ - 100 + 50}}{{25}} \\
\Rightarrow\dfrac{1}{f} = \dfrac{{ - 50}}{{25}} \\
\Rightarrow\dfrac{1}{f} = - 2 \\
\Rightarrow f = \dfrac{{ - 1}}{2}m \\
\Rightarrow f = - 0.5m \\
\]
Now, Power of the lens is the ability to bend light. The power of lens is inversely proportional to the focal length of the lens and is measured in Dioptres (D).
$P = \dfrac{1}{{f{}_{meter}}}$
$\therefore P = - 2.0D$
Therefore, the spectacles needed by the person with myopic vision should have power of $ - 2.0D.$

Hence, from the given multiple choices- the option C is the correct answer.

Note: Remember the sign convention for both the convex and concave lenses to solve these types of word problems. It is similar to the convex and concave mirrors. The rest goes well once sign convention is done properly.