The near point of the person is defected.

Near point of the defected eye, d = 50 cm.

He wants to correct it (up to 25 cm)

If image of the object (u = –25cm) is at 

v = –50 cm, then the defected eye can see it easily.

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

 $\frac{1}{\mathrm{f}}=\frac{-1}{50}+\frac{1}{25}=\frac{-1+2}{50}$ 

or $\frac{1}{\mathrm{f}}=\frac{1}{50}$              ⇒  f = + 50 cm

He will use a convex lens of focal length 50 cm.