For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

For a normal eye, near point of distinct vision = 25 cm.
i.e.,                u = –25 cm
Converging power = +40 D 

To see objects at infinity, the eye uses its least converging power = 40 + 20 = 60 dioptres. This gives the rough idea of the distance between the retina and cornea eye-lens. 
$\therefore$ to focus an object at near point,
$F\mathrm{ocal} \mathrm{length} \mathrm{of} \mathrm{eye}-\mathrm{lens} = \frac{100}{\mathrm{P}}=\frac{10}{6}=\frac{5}{3}\mathrm{cm}$
$\therefore$                              $\mathrm{v} = -\frac{5}{3}\mathrm{cm}$

Now, using the formula, 
                          $\frac{1}{\mathrm{f}} =\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$ 

                               $= \frac{3}{5}+\frac{1}{25}= \frac{16}{25}$ 

i.e., $\mathrm{f} = \frac{25}{16},$ corresponding to a converging power given by, 

                 $\mathrm{P} = \frac{{\displaystyle \frac{100}{25}}}{16} = 64 \mathrm{dioptre}.$ 

$\therefore$ the power of the eye lens = 64 - 60 = 24 dioptre.

Hence, we can say that the range of accommodation of the eye-lens is roughly 20 to 24 dioptre.