⇦ | ![]() | ⇨ |
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:
- The product of pressure and volume hve thesame units as the product of
- If λ₁ and λ₂ are the wavelengths of the first members of the Lyman and Paschen
- Which of the following figures represent the variation of particle momentum
- A body has 80 microcoulomb of charge. Number of additional electrons in it will be
- In a n-type semiconductor, which of the following statement is true?
Topics: Waves
(80)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply