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.

• A. $10$ to $14D$
• B. $20$ to $24D$
• C. $28$ to $32D$
• D. $14$ to $18D$

Answer 1

Answer: (A)

Step 1: Given that:

The far point of the normal eye = infinity

Near point of normal eye = 25cm

The converging power of the cornea of the eye = $+40D$

The least converging power of the eye lens behind the cornea = $+20D$

Step 2: Calculation of range of accommodation of normal eye:

The net power of two lenses having individual powers; $P_1$ and $P_2$ is the sum of their individual powers.

That is $P_{net}=P_1+P_2$

Therefore;

Net power of the eye lens =$\text{40D+20D}$ = $60D$

Thus, the focal length of the eye lens will be;

$f\left(inm\right)=\frac{1}{P}$

Now;

$f=\frac{1}{60}\times 100cm$

$f=\frac{5}{3}cm$

Now,

To focus on an object at a near point;

Object distance(u) = -25cm

Since the focal length of the eye lens is the distance between the cornea and the retina

Image distance(v) = $\frac{5}{3}cm$

Using lens formula; $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Putting the values, we get

$\frac{1}{\frac{5}{3}cm}-\frac{1}{-25cm}=\frac{1}{f}$

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

$\frac{1}{f}=\frac{15+1}{25}$

$f=\frac{25}{16}cm$

Now, power is given as;

$P=\frac{100}{f\left(incm\right)}$

$P=\frac{100}{25}\times 16$

$P=64D$

The range of accommodation;

maximum power = 64D- 40D= 24D

minimum value of power = 20D

Thus,

The range of accommodation of normal eye is 20D to 24D.

Hence,

Option B) 20D to 24D is the correct option.