Interference and Intensity Ratio – Rankers Physics

Principle of Superposition, Interference and Beats: Practice Problem & Solution

Two waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to:
10 : 8
9 : 1
4 : 1
2 : 1

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Principle of Superposition, Interference and Beats. Understanding the underlying formula is key to arriving at the correct answer below:

The ratio of maximum to minimum intensity is given by \(frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{\sqrt{I_1/I_2} + 1}{\sqrt{I_1/I_2} - 1}\right)^2\). Substituting \(\frac{I_1}{I_2} = 9\) yields \(\left(\frac{3 + 1}{3 - 1}\right\)^2 = 4\), which is \(4:1\).

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The \(4^{\text{th}}\) overtone of a closed organ pipe is same as that of \(3^{\text{rd}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
2. Assertion (A): When two tuning fork of frequency \(256 \text{ Hz}\) and \(324 \text{ Hz}\) are vibrating together. Beats will not be heard. Reason (R): Superposition of sound waves is possible for all frequencies of sound.  
3. Two sources of sound placed close to each other are emitting progressive waves given by $y_1 = 4 \sin 600 \pi t$ and $y_2 = 5 \sin 608 \pi t$. An observer located near these two sources of sound will hear: (2012 Pre)
4. Assertion (A): Beats are not observed in case of light waves from two independent sources. Reason (R): The phase difference between two light sources changes randomly.  
5. Wave has simple harmonic motion whose period is 4 seconds while another wave which also possesses simple harmonic motion has its period 3 sec. If both are combined, then the resultant wave will have the period equal to: (1993)