⇦ | ![]() | ⇨ |
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 order of radius of the nucleus of an atom is
- In Young’s double silt experiment, 12 fringes are observed to be formed
- In Young’s double slit experiment, the phase difference between the two waves
- In the diagram shown in figure, match the following about weight
- A bullet is shot from a riffle. As a result the rifle recoils.The kinetic energy of riffle
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