Hint: \(p=\frac{1}{f}\)

Step 1: Find the focal length of the cornea-lens combination.
Least distance of distinct vision, d = 25 cm
The far point of a normal eye, d’ = ∞
Converging power of the cornea, ${\mathrm{P}}_{\mathrm{c}}=40<mspace\; width="1em"></mspace>\mathrm{D}$
Least converging power of the eye-lens, ${\mathrm{P}}_{\mathrm{e}}=20<mspace\; width="1em"></mspace>\mathrm{D}$

To see the objects at infinity the eye uses its least converging power of eye-lens.

Power of the cornea-lens combination, $\mathrm{P}={\mathrm{P}}_{\mathrm{c}}+{\mathrm{P}}_{\mathrm{e}}=40+20=60<mspace\; width="1em"></mspace>\mathrm{D}$

The focal length of the cornea-lens combination is given by:

$\mathrm{f}=\frac{1}{\mathrm{P}}=\frac{1}{60\mathrm{D}}=\frac{100}{60}=\frac{5}{3}<mspace\; width="1em"></mspace>\mathrm{cm}<mspace\; width="1em"></mspace>$

Step 2: Find the power of the eye-lens.
For an object at the near point,

The object distance (u)=−d=−25 cm
Image distance= the distance between the cornea and the retina = the focal length of the cornea-lens combination
Hence, image distance, $v=53cm$
Using the lens formula,

$\mathrm{If}<mspace\; width="1em"></mspace>\mathrm{f}\text{'}<mspace\; width="1em"></mspace>\mathrm{is}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{focal}<mspace\; width="1em"></mspace>\mathrm{length}<mspace\; width="1em"></mspace>\mathrm{of}<mspace\; width="1em"></mspace>\mathrm{the}<mspace\; width="1em"></mspace>\mathrm{cornea}-\mathrm{lens}<mspace\; width="1em"></mspace>\mathrm{combination}<mspace\; width="1em"></mspace>\mathrm{for}<mspace\; width="1em"></mspace>\mathrm{near}<mspace\; width="1em"></mspace>\mathrm{point}:$
$\frac{1}{\mathrm{f}\text{'}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$
$\frac{1}{\mathrm{f}\text{'}}=\frac{3}{5}+\frac{1}{25}=\frac{15+1}{25}=\frac{16}{25}{\mathrm{cm}}^{-1}$
$\mathrm{Power},<mspace\; width="1em"></mspace>\mathrm{P}\text{'}<mspace\; width="1em"></mspace>=\frac{1}{\mathrm{f}\text{'}}\times 100=\frac{16}{25}\times 100=64<mspace\; width="1em"></mspace>\mathrm{D}$

Therefore power of the eye-lens = 64 − 40 = 24 D

Hence, 20 D to 24 D is the range of accommodation of the eye-lens.