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The maximum and minimum intensities of two sources is 4:1. The ratio of amplitude is
Options
(a) 3 : 1
(b) 1 : 3
(c) 1 : √3
(d) √3 : 1
Correct Answer:
3 : 1
Explanation:
The intensity is proportional to the square of the amplitude.
Therefore if the intensities-maximum of one source and minimum of the other is taken, then the answer is different. But from the answer we presume that the light from two sources of different amplitudes are superposed. In that case,
Iₘₐₓ / Iₘᵢₙ = [A₁+A₂ / A₁-A₂]²
⇒ 4/1 = [A₁+A₂ / A₁-A₂]²
2/1 = [A₁+A₂ / A₁-A₂]
Using componendo and dividendo
2+1 / 2-1 = A₁/A₂ ⇒ A₁/A₂ = 3/1 = 3 : 1
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Topics: Waves
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- The threshold wavelength for photoelectric emission from a material is 4800 Å.
- An elevator car whose floor to ceiling distance is equal to 2.7m starts
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- 24 cells of emf 1.5 V each having internal resistance of 1 ohm are connected
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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