Question

By the principle of interference, condition for constructive interference in terms of the path difference?
A. ${n^{ - 1}}\lambda$
B. $n\lambda$
C. $n{\lambda ^2}$
D. $n{\lambda ^4}$

Answer 1

Consider the equations of the two waves as ${y_1} = {A_1}\sin \left( {kx - \omega t} \right)$ and ${y_2} = {A_2}\sin \left( {kx - \omega t + \delta } \right)$. Add the two equations to get the resultant of the interference. Apply conditions to get the maximum amplitude. Because for constructive interference, the amplitude will be maximum.

Complete step by step answer:
The sources identical to each other produce waves having equations as mentioned above. The waves produced will have the same angular frequency $\omega$ and same velocity. As $k = \dfrac{\omega }{v}$, the wave number $k$ will also be the same for both the waves.
Now, if one wave starts a few moments later or the distance travelled by the two waves is different, there will be a phase different when they will reach the desired point for interference. Let the phase difference be $\delta$. Therefore, the equations of the wave are as follows:
${y_1} = {A_1}\sin \left( {kx - \omega t} \right)$
$\Rightarrow {y_2} = {A_2}\sin \left( {kx - \omega t + \delta } \right)$
Applying the principle of superposition, the resultant of these two waves will be
$y = {y_1} + {y_2}$

$$y = {A_1}\sin \left( {kx - \omega t} \right) + {A_2}\sin \left( {kx - \omega t + \delta } \right) \\\ \Rightarrow y = {A_1}\sin \left( {kx - \omega t} \right) + {A_2}\sin \left( {kx - \omega t} \right)\cos \delta + {A_2}\cos \left( {kx - \omega t} \right)\sin \delta \\\ \Rightarrow y = \left( {{A_1} + {A_2}\cos \delta } \right)\sin \left( {kx - \omega t} \right) + \left( {{A_2}\sin \delta } \right)\cos \left( {kx - \omega t} \right) \\\$$

Let us substitute the amplitudes of the above two waves with ${A_3}$ and ${A_4}$.
$y = {A_3}\sin \left( {kx - \omega t} \right) + {A_4}\cos \left( {kx - \omega t} \right)$
Now we multiply and divide the right-hand side by $\sqrt {{A_3}^2 + {A_4}^2}$.
$y = \sqrt {{A_3}^2 + {A_4}^2} \left( {\dfrac{{{A_3}}}{{\sqrt {{A_3}^2 + {A_4}^2} }}\sin \left( {kx - \omega t} \right) + \dfrac{{{A_4}}}{{\sqrt {{A_3}^2 + {A_4}^2} }}\cos \left( {kx - \omega t} \right)} \right)$

Now, since $\dfrac{{{A_3}}}{{\sqrt {{A_3}^2 + {A_4}^2} }}$ and$\dfrac{{{A_4}}}{{\sqrt {{A_3}^2 + {A_4}^2} }}$ are both less than $1$, these we can write these as sine and cosine of some angle between $0$ and $2\pi$.
So, we write $\sin \phi = \dfrac{{{A_4}}}{{\sqrt {{A_3}^2 + {A_4}^2} }}$ and $\cos \phi = \dfrac{{{A_3}}}{{\sqrt {{A_3}^2 + {A_4}^2} }}$
Therefore,
$y = \sqrt {{A_3}^2 + {A_4}^2} \left( {\sin \phi \cos \left( {kx - \omega t} \right) + \cos \phi \sin \left( {kx - \omega t} \right)} \right) \\\ \Rightarrow y = \sqrt {{A_3}^2 + {A_4}^2} \left( {\sin \left( {kx - \omega t + \phi } \right)} \right) \\\$
Here, we have the result of the interference. The amplitude is given as,
$A = \sqrt {{A_3}^2 + {A_4}^2}$
$\Rightarrow A = \sqrt {{{\left( {{A_1} + {A_2}\cos \delta } \right)}^2} + {{\left( {{A_2}\sin \delta } \right)}^2}} \\\ \Rightarrow y = \sqrt {{A_1}^2 + {A_2}^2{{\cos }^2}\delta + 2{A_1}{A_2}\cos \delta + {A_{}}^2{{\sin }^2}\delta } \\\ \Rightarrow y = \sqrt {{A_1}^2 + {A_2}^2\left( {{{\cos }^2}\delta + {{\sin }^2}\delta } \right) + 2{A_1}{A_2}\cos \delta } \\\ \Rightarrow y = \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \delta } \\\$
Now, to get the maximum amplitude, we should have $\cos \delta = 1$
General solution of $\cos \delta = 1$ is $\delta = 2n\pi$.
Here, $\delta$ is given as $\delta = \dfrac{\omega }{v}\Delta x$
$\delta = k\Delta x \\\ \Rightarrow \delta = \dfrac{{2\pi }}{\lambda }\Delta x \\\$
Substituting this value of phase difference in the obtained
$\dfrac{{2\pi }}{c}\Delta x = 2n\pi \\\ \therefore \Delta x = n\lambda \\\$
Therefore, by the principle of interference, the condition for constructive interference in terms of the path difference is $n\lambda$. Hence,option B is correct.

Note: The phase difference $\delta$ is given as $\delta = k\Delta x$, here $k$ is the wave number and $\Delta x$ is the path difference. The phase difference is generally due to time lag by any one wave, or if the distance travelled by one any one of the waves is less or greater than the other.