S_{1} and S_{2} are coherent point sources of | vedclass

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Question

(B) For constructive interference, the path difference $\Delta x$ between the two waves must be an integral multiple of the wavelength, i.e.,$\Delta x

Vedclass · Accepted Answer

$(4\lambda, 0)$

Vedclass · Answer

(B) For constructive interference, the path difference $\Delta x$ between the two waves must be an integral multiple of the wavelength, i.e.,$\Delta x = n\lambda$, where $n = 0, 1, 2, \dots$.Let the point be $P(x, 0)$ on the $x$-axis.The distance from $S_{1}(0,0)$ to $P(x,0)$ is $r_{1} = \sqrt{(x-0)^{2} + (0-0)^{2}} = x$.The distance from $S_{2}(0,3\lambda)$ to $P(x,0)$ is $r_{2} = \sqrt{(x-0)^{2} + (0-3\lambda)^{2}} = \sqrt{x^{2} + 9\lambda^{2}}$.The path difference is $\Delta x = |r_{2} - r_{1}| = |\sqrt{x^{2} + 9\lambda^{2}} - x|$.Testing option $(B)$ $(4\lambda, 0)$:$r_{1} = 4\lambda$.$r_{2} = \sqrt{(4\lambda)^{2} + (3\lambda)^{2}} = \sqrt{16\lambda^{2} + 9\lambda^{2}} = \sqrt{25\lambda^{2}} = 5\lambda$.Path difference $\Delta x = |5\lambda - 4\lambda| = \lambda$.Since $\Delta x = 1\lambda$, which is an integral multiple of $\lambda$, constructive interference occurs at $(4\lambda, 0)$.

$S_{1}$ and $S_{2}$ are coherent point sources of light located in the $xy$-plane at points $(0,0)$ and $(0,3\lambda)$ respectively. Here $\lambda$ is the wavelength of light. At which one of the following points (given as coordinates) will the intensity of interference be maximum?

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