Q6. A normal eye has maximum accommodation of 5 dioptres and a far point at infinity. Its near point is:
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Question

Q6. A normal eye has maximum accommodation of 5 dioptres and a far point at infinity. Its near point is:

G1 Physics · Accepted Answer

GIVEN: Maximum accommodation A = 5 D, far point F = ∞. Step 1: Accommodation is the difference between eye powers for near and far objects: A = (1/N) − (1/F), with distances in metres. Step 2: Since F = ∞, the second term is zero: 5 = 1/N. Step 3: N = 1/5 m = 0.20 m = 20 cm. FINAL ANSWER: Near point = 20 cm. Option C.