The far point of a person suffering from long-sightedness is $75\mathit{ cm}$ in front of the eye. What will be the power of the corrective lens required?

Given,
Image distance, $v=75\mathit{ cm}$ and object distance, $u=\mathit{–\infty }$
Since the person is suffering from long-sightedness or hypermetropia, the corrective lens used will be a convex lens; hence, the focal length should be positive.
From the lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Where $f$, $v$, and $u$ are the focal length, image distance, and object distance, respectively.
On substituting the given values, we get the focal length as
$\Rightarrow \frac{1}{f}=\frac{1}{75}-\frac{1}{-\infty }$
$\Rightarrow \frac{1}{f}=\frac{1}{75}$
$\Rightarrow \mathit{ f}= 75\mathit{ cm}$
Hence, the focal length is $75\mathit{ cm}$.
Now, it is known that the power of a lens is the reciprocal of the focal length of the lens. Mathematically, 
$P=\frac{100}{f}$
Where $f$ is the focal length of the lens in centimetres.
On substituting the given values, we get the power as
$\Rightarrow P=\frac{100}{74}$
$\Rightarrow P=1.33\mathit{ D}$
Hence, the power of the lens required is $1.33\mathit{ D}$.