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**Hint:**At first find out the net power of the eye, then calculate the distance between the eye lens and the retina. Now write the mirror formula and calculate the focal length then from the focal length find out the power of the lens of the eye. Then to calculate the accommodating power of the eye find out the net power.

**Formula used:**

Net power of the eye = Maximum power + least power.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

f = 1 / P.

**Complete step by step answer:**

From the question, we can say that the far point is at infinity and the near point of vision is at 25cm.

The cornea is providing a converging power of about 40 D.

The least converging power of the eye lens is behind the cornea about 20D.

So, we can say that the net power of the eye is,

Net power of the eye = Maximum power + least power.

Net power of the eye = (60 + 20) D = 60 D.

Therefore we can find the distance between the eye lens and the retina by,

\[\dfrac{100}{P}\],

$\Rightarrow f=\dfrac{100}{60}$

\[\Rightarrow f=\dfrac{5}{3}cm\]

So, we know that,

v = 5 / 3 cm, u = -25 cm

Now, by the lens formula, we know that,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\Rightarrow \dfrac{1}{f}=\dfrac{3}{5}+\dfrac{1}{25}$,

On solving the above equation we will get,

$\Rightarrow f=\dfrac{25}{16}cm$

Now, power of eye lens is = 1 / f

= $\dfrac{100}{25}\times 16$

= 64 cm.

Therefore net power = Power of eye lens – actual power of the lens

= $(64-40)$cm

= 24 cm

**So the correct option is option B.**

**Note:**The distance between the lens and the retina is actually the focal length of the lens. While calculating the net power be careful as the least power of the lens is also given and mistakes can happen.

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