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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' =oo`
Converging power of the cornea, `P_c =40D`
Least converging power of the eye-lens, `P_c = 20D`
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P = Pc + Pe = 40 + 20 = 60 D
Power of the eye-lens is given as: `P = 1/"Focal length of the eye lens(f)"`
`f = 1/P`
= `1/60D`
`= 100/60 = 5/3 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 = 5/3 cm`
According to the lens formula, we can write:
`1/f' = 1/v - 1/u`
Where,
f' = Focal length
`1/f' = 3/2 + 1/25 = (15+1)/25 = 16/25 cm`
Power, P' = `1/"f'" xx 100`
`= 16/25 xx 100 = 64D`
∴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.
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