Question

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

• A. 2 D to 10 D
• B. 40 D to 32 D
• C. 9 D to 8 D
• D. 44 D to 40 D

Answer 1

The correct option is D 44 D to 40 D

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