A newspaper placed nearer than $50\mathit{ cm}$ cannot be read by a person. What will be the power of the corrective lens required?

Here, u = - 25 cm, v = -50 cm

Using lens formula, 

      $\frac{1}{\mathrm{f}}= \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

           $= -\frac{1}{50}+\frac{1}{25}$

           $\Rightarrow \frac{1}{f}= \frac{1}{50}$
Focal length $f=50\mathit{ cm}$
It is known that the power of a lens is the reciprocal of the focal length of the lens. Mathematically, 
$P=\frac{1}{f}$
Where $f$ is the focal length of the lens in meters.($\therefore$f=50/100 m)
On substituting the given values, we get the power as
$\Rightarrow P=\frac{100}{50}$
$\Rightarrow P=2\mathit{ D}$
Hence, the power of the lens required is $2\mathit{ D}$.