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
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B
40 D to 32 D
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C
9 D to 8 D
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D
44 D to 40 D
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Solution
The correct option is D 44 D to 40 D An eye sees distant objects with full relaxation so 12.5×10−2−1−∞=1f or P=1f=125×10−2=40D
An eye sees an object at 25 cm with strain so 12.5×10−2−1−25×10−2=1forP=1f=40+4=44D