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:

We have –
Least distance of distinct vision is d = 25 cm
Far point of a normal eye is d’ = ∞
Converging power of the cornea is $P_{c}=40D$
Least converging power of the given eye-lens is $P_{e}=20D$
it is a fact that the human eyes use the least converging power to see the objects which are at infinity.
Power of the eye-lens then becomes

 P = $P_{c} + P_{e} = 40 + 20 = 60 D$

Power of the eye-lens is given as –

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

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

$\frac{1}{60D}$

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

To focus an object at the near point, object distance is (u) = −d = −25 cm
Focal length of the eye-lens is equal to the distance between the cornea and the retina which is equal to the Image distance
Hence, image distance is $v=\frac{5}{3}$
According to the lens formula, we can write,

$\frac{1}{f’}=\frac{1}{v}+\frac{1}{u}$
Where, f’=focal length

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

Power $P’=\frac{1}{f’}\times100$

$P’=\frac{16}{25}\times100=64D$

Therefore, Power of the eye-lens is

 = 64 − 40 = 24 D
Hence, 20D to 24D is the range of accommodation of the said eye-lens.