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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
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- One way in which the operation of a n-p-n transistor differs from that of a p-n-p
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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