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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
- Graph of force per unit length between two long parallel current carrying
- Complete the reaction: ₀n¹+ ₉₂U²³⁵→₅₆Ba¹⁴⁴+……..+3n
- A lead bullet of 10 gm travelling at 300 m/s strikes against a block
- The current in a conductor varies with time t as I=2t+3t² Where I is in amperes
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Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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