Q4. The distance between the crystalline lens and retina is 2.5 cm. When the eye focuses from infinity to an object 25 cm away, the increase in optical power is:
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Question

Q4. The distance between the crystalline lens and retina is 2.5 cm. When the eye focuses from infinity to an object 25 cm away, the increase in optical power is:

G1 Physics · Accepted Answer

GIVEN: Image distance (eye lens to retina) v = 2.5 cm = 0.025 m, unchanged during accommodation. Step 1: With a distant object u = −∞, lens power P∞ = 1/v − 1/u = 1/0.025 = 40 D. Step 2: For an object at 25 cm, u = −0.25 m. Required power P₂ = 1/0.025 − (−1/0.25) = 40 + 4 = 44 D. Step 3: Change in power ΔP = P₂ − P∞ = 44 − 40 = 4 D. FINAL ANSWER: Increase in optical power = 4 D. Option B.