A compound microscope has an objective of focal length 1.0 cm and an eyepiece of focal length 5.0 cm. With the final image at the least distance of distinct vision (D = 25 cm) and the tube length L ≈ 15 cm, its magnification is approximately
A{'text': '15 × (25/5) = 75', 'label': 'A'}
B{'text': '(15/1) × (1 + 25/5) = 90', 'label': 'B'}
C{'text': '(15/1) × (25/5) = 75', 'label': 'C'}
D{'text': '15 × 5 = 75', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': '(15/1) × (1 + 25/5) = 90', 'label': 'B'}
1. Compound microscope, image at D: m = (L / f_o) × (1 + D / f_e).
2. m = (15 / 1.0) × (1 + 25 / 5) = 15 × 6 = 90.
3. (A), (C), (D) use only the normal-adjustment form (image at ∞), which gives 75.
_Source: NCERT Class 12 Physics Part II, Ch 9 §9.7.1_
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