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

Answer: (D)

44 D to 40 D

Answer 2

An eye sees distant objects with full relaxation so $\frac { 1 } { 2.5 \times 10 ^ { - 2 } } - \frac { 1 } { - \infty } = \frac { 1 } { f }$ or

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