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.

Solution

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.
$∴$ to focus an object at near point,
$F ocal length of eye - lens = \frac{100}{P} = \frac{10}{6} = \frac{5}{3} cm$
$∴$ $v = - \frac{5}{3} cm$

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

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

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

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

$∴$ 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.

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