# For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye.

**Question:**

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm 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.

**Solution:**

Least distance of distinct vision, *d* = 25 cm

Far point of a normal eye, $d^{\prime}=\infty$

Converging power of the cornea, $P_{\mathrm{c}}=40 \mathrm{D}$

Least converging power of the eye-lens, $P_{\mathrm{e}}=20 \mathrm{D}$

To see the objects at infinity, the eye uses its least converging power.

Power of the eye-lens, *P* = *P*c + *P*e = 40 + 20 = 60 D

Power of the eye-lens is given as:

$P=\frac{1}{\text { Focal length of the eye lens }(f)}$

$f=\frac{1}{P}$

$=\frac{1}{60 \mathrm{D}}$

$=\frac{100}{60}=\frac{5}{3} \mathrm{~cm}$

To focus an object at the near point, object distance (*u*) = −*d* = −25 cm

Focal length of the eye-lens = Distance between the cornea and the retina

= Image distance

Hence, image distance, $v=\frac{5}{3} \mathrm{~cm}$

According to the lens formula, we can write:

$\frac{1}{f^{\prime}}=\frac{1}{v}-\frac{1}{u}$

Where,

$f^{\prime}=$ Focal length

$\frac{1}{f^{\prime}}=\frac{3}{5}+\frac{1}{25}=\frac{15+1}{25}=\frac{16}{25} \mathrm{~cm}^{-1}$

Power, $P^{\prime}=\frac{1}{f^{\prime}} \times 100$

$=\frac{16}{25} \times 100=64 \mathrm{D}$

∴Power of the eye-lens = 64 − 40 = 24 D

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