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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
Related Questions: - To observed diffraction, the size of the obstacle
- A tuning fork of frequency x produces 4 beats with a source of 256 Hz and 8 beats
- A neutron is moving with a velocity u.It collides head on and elastically with an atom
- A stationary particle explodes into two particles of masses m₁ and m₂ which move
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Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- To observed diffraction, the size of the obstacle
- A tuning fork of frequency x produces 4 beats with a source of 256 Hz and 8 beats
- A neutron is moving with a velocity u.It collides head on and elastically with an atom
- A stationary particle explodes into two particles of masses m₁ and m₂ which move
- The intensity at the maximum in a Young’s double slit experiment is I₀.
Topics: Waves (80)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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