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Question

For a normal eye, the far point is at infinity and the near point ofdistinct vision is about 25cm in front of the eye. The cornea of theeye provides a converging power of about 40 dioptres, and the leastconverging 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.

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Solution

Given: The least distance of distinct vision is 25cm, far point of a normal eye is infinite, the converging power of the cornea is 40D and the least converging power of the eye-lens is 20D.

Power of the eye-lens is given as,

P= P c + P e

Where, the power of the eye-lens is P, the converging power of the cornea is P c and the least converging power of the eye-lens is P e .

By substituting the given values in the above expression, we get

P=40+20 =60D

Power of the eye-lens is given as,

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

f e =D =v

Where, the focal length of the eye-lens is f e ( 100 60 cm ), the distance between the corona and the retina is D and the image distance is v.

By substituting the given values in the above expression, we get

v= 100 60 = 5 3 cm

The lens formula is given as,

1 f = 1 ν 1 u

Where f is the focal length of the lens.

By substituting the given values in the above expression, we get

1 f = 3 5 + 1 25 = 15+1 25 = 16 25 cm 1

Power is denoted as P'.

P = 1 f ×100

By substituting the given values in the above expression, we get

P = 16 25 ×100 =64D

The power of the lens is given as,

P L = P P

By substituting the given values in the above expression, we get

Where, the power of the lens is P L .

P L =64-40 =24D

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


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