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The path difference between two waves for constructive interference, must be?A. $$\left( {2n\, + \,1} \right)\dfrac{\lambda }{2}$$ B. $$\left( {2n\, + \,1} \right)\lambda$$ C. $n\dfrac{\lambda }{2}$ D. $n\lambda$

Last updated date: 23rd Apr 2024
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Hint: For constructive interference, the two interfering waves must meet in the same phase ( that is, the phase difference between them is 0, $2\pi$, $$4\pi$$…... ).
Then only the resultant intensity or amplitude will be maximum.

For interference between any two waves, the resultant intensity determines whether the interference has been constructive or destructive. For constructive interference, the resultant intensity is maximum while for destructive interference, the resultant intensity is minimum.
The resultant intensity at any point depends upon the phase difference between two waves as:-
$${\text{I}}\,{\text{ = }}\,{{\text{I}}_1}\, + \,\,{{\text{I}}_{2\,}}\, + \,\,2\sqrt {{{\text{I}}_1}\,{{\text{I}}_{2\,}}} \cos \phi$$ …………..(1)
Here, ${{\text{I}}_1}$ is the intensity of first wave, ${{\text{I}}_2}$ is the intensity of second wave and $$\phi$$ is the phase difference between these two waves and $I$ is the resultant frequency.
And the relation between phase difference and path difference is given by:-
$$x\, = \,\,\dfrac{\lambda }{{2\pi }}\phi$$ …………...(2)
where, $\phi$ is the phase difference, $x$ is the path difference and $$\lambda$$ is the wavelength.
Now, for the constructive interference, the resultant intensity given in equation (1) must be maximum. For that,
$\cos \phi \, = \,1$ that is, Angle of Cos equals to 1 when,
$$\phi \, = \,\,2n\pi$$ ………......(3)
Therefore, using equation (3) in equation (2), we get path difference as:-
\eqalign{ & x\, = \,\,\dfrac{\lambda }{{2\pi }}\left( {2n\pi } \right) \cr & \cr}
On simplifying we get, $$x\, = \,\,n\lambda$$
Therefore, for the constructive interference the required path difference comes out to be $n\lambda$ .
Hence, for the above question (D) option is correct.

Note: For the constructive interference, the resultant intensity of two waves comes out to be:-
${{\text{I}}_{{\text{max}}}}\, = \,\,{{\text{I}}_{1\,}}\, + \,\,{{\text{I}}_2}\, + \,\,2\sqrt {{{\text{I}}_1}\,{{\text{I}}_2}} \, = \,\,{\left( {\sqrt {{{\text{I}}_1}} \, + \,\,\sqrt {{{\text{I}}_2}} } \right)^2}$
Here, ${{\text{I}}_1}$ is the intensity of first wave, ${{\text{I}}_2}$ is the intensity of second wave and $$\phi$$ is the phase difference between these two waves and $I$ is the resultant frequency.
Thus, the constructive interference occurs if the path difference between the interfering waves is zero or an integral multiple of the wavelength $\lambda$ .