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10 Questions around this concept.
In interference, intensities of two interfering waves are 4 units and 1 unit. Find the ratio of maximum to minimum possible intensities.
Interference of sound waves
We have studied the principle of superposition, this principle of superposition is valid for sound waves also. If two or more waves pass through the same region of a medium, so the resultant disturbance is equal to the sum of the disturbances produced by individual waves. Based on the phase difference, the waves can interfere constructively or destructively leading to a corresponding increase or decrease in the resultant intensity. Here the waves are expressed in terms of pressure change. The resultant change in pressure is the algebraic sum of the changes in pressure due to the individual waves. So, there is no need for displacement vectors so as to obtain the resultant displacement wave.
Let us take two tuning forks S1 and S2 placed side by side. which vibrate with equal frequency and equal magnitude. The point P is situated at a distance x from S1 and x + $\Delta$x from S2 .
The forks may be set into vibration with a phase difference $\delta_\circ$. In case of tuning forks, the phase difference $\delta_\circ$ remains constant in time. Suppose the two forks are vibrating in phase so that $\delta_\circ$ = 0. Also, let p01 and p02 be the amplitudes of the waves from S1 and S2 respectively. Let us examine the resultant change in pressure at a point P. The pressure-change at A due to the two waves are described by
$$
\begin{aligned}
p_1 & =p_{01} \sin (k x-\omega t) \\
p_2 & =p_{02} \sin [k(x+\Delta x)-\omega t] \\
& =p_{02} \sin [(k x-\omega t)+\delta]
\end{aligned}
$$
where $\delta=k \Delta x=\frac{2 \pi \Delta x}{\lambda} \ldots$
Here, $\delta$ is the phase difference between the two waves reaching P . So, the resultant wave at P is given by -
$$
\begin{array}{ll}
& p=p_0 \sin [(k x-\omega t)+\varepsilon] \\
\text { where } & p_0^2=p_{01}^2+p_{02}^2+2 p_{01} p_{02} \cos \delta \\
\text { and } \quad & \tan \varepsilon=\frac{p_{02} \sin \delta}{p_{01}+p_0 \cos \delta}
\end{array}
$$
The resultant amplitude is maximum when $=2 \pi n$ and is minimum when $\mathrm{S}=(2 n+1) \pi$, where n is an integer. These are correspondingly the conditions for constructive and destructive interference. A similar condition in terms of path difference can be written as -
$$
\begin{array}{ll}
\Delta x=n \lambda & \text { (constructive) } \\
\Delta x=(n+1 / 2) \lambda & \text { (destructive) }
\end{array}
$$
The above equation is obtained with the help of the (1) equation.
At constructive interference,
$$
P_0=P_{01}+P_{02}
$$
At destructive interfernece -
$$
P_0=\left|P_{01}-P_{02}\right|
$$
Constructive interference | Destructive interference |
1. When the waves meet with the same phase, it forms constructive interference | 1. When the waves meet with opposite phase, it forms destructive interference |
2. Phase difference at the point of observation. $\delta=0^{\circ}$ or $2 n \pi$ |
2. Here, phase difference = $180^{\circ}$ or $(2 n-1) \pi$ where $n=1,2,3 \ldots$ |
3. Path difference = $n \lambda$ | 3. Path difference = $(2 n-1) \frac{\lambda}{2}$ |
4. Resultant amplitude = $A_{\max }=a_1+a_2$ |
4. Resultant amplitude = $A_{\min }=a_1-a_2$ |
5. Resultant intensity will be maximum = $I_{\text {max }}=I_1+I_2+2 \sqrt{I_1 I_2}$ | 5. Resultant intensity will be minimum = $I_{\text {max }}=I_1+I_2-2 \sqrt{I_1 I_2}$ |
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