Question

The diameter of the eye ball of a normal eye is about $2.5cm$. The power of the eye lens varies from

• A. $9D$ to $8D$
• B. $40D$ to $32D$
• C. $44D$ to $40D$
• D. None of these

Answer 1

Answer: (C)

$44D$ to $40D$

An eye see distant objects with full relaxation.
So, $\frac{1}{2.5\times {10}^{-2}}-\frac{1}{-\infty }=\frac{1}{f}$
or $P=\frac{1}{f}=\frac{1}{2.5\times {10}^{-2}}=40D$
An eye see an object at $25cm$ with strain
$=\frac{1}{2.5\times {10}^{-2}}-\frac{1}{25\times {10}^{-2}}=\frac{1}{f}$
$P=\frac{1}{f}=40+4=44D$.