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.
Answer:
Least distance of distance vision (d) = 25 cm
Far point a normal eye (d1) = ∞
Converging power of the cornea (Pc) = 40 D
Least Converging power of the eye lens (Pe) 20 D
So, power of the eye lens (P) = Pc + Pe = 40 + 20 = 60D
Power of the eye lens (P) = 1/ focal length of the eye lens (f)
f = 1/P = 1/60D
or, f = 100/60 cm = 5/3 cm
Object distance (u) = -d = -25 cm
Focal length of the eye lens = Distance between cornea and the retina = Image distance
So, image distance (v) = 5/3 cm
According to lens formula,
1/f = 1/v - 1/u
1/f = 3/5 + 1/25 = 15 +1/25 = 16/25 cm-1
So, power (P) = 1x 100/f = 16x100/25 64 D
Therefore, power of the lens = 64-40 =24 D
Hence, we can say that eye lens range is from 20 D to 24 D.
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