Four monochromatic and coherent sources of li | vedclass

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Question

(B) The sources are at $x = 0, d, 2d, 3d$. For a point $P$ far away on the $+x$ axis,the path difference between consecutive sources is $\Delta x = d$

Vedclass · Accepted Answer

If $d = \lambda /6$,the intensity at $P$ is $3I_0$.

Vedclass · Answer

(B) The sources are at $x = 0, d, 2d, 3d$. For a point $P$ far away on the $+x$ axis,the path difference between consecutive sources is $\Delta x = d$. The phase difference is $\phi = (2\pi / \lambda) \Delta x = (2\pi d) / \lambda$.Let the amplitude of each source be $A$,such that $I_0 = kA^2$. The resultant amplitude $A_R$ is the sum of four phasors each of magnitude $A$ with phase difference $\phi$: $A_R = A(1 + e^{i\phi} + e^{i2\phi} + e^{i3\phi}) = A \frac{1 - e^{i4\phi}}{1 - e^{i\phi}}$.The resultant intensity $I = |A_R|^2 = A^2 \left| \frac{\sin(2\phi)}{\sin(\phi/2)} \right|^2 = I_0 \left( \frac{\sin(2\phi)}{\sin(\phi/2)} \right)^2$.Case $A$: If $d = \lambda / 4$,then $\phi = (2\pi / \lambda) * (\lambda / 4) = \pi / 2$. $I = I_0 (\sin(\pi) / \sin(\pi / 4))^2 = 0$.Case $B$: If $d = \lambda / 6$,then $\phi = (2\pi / \lambda) * (\lambda / 6) = \pi / 3$. $I = I_0 (\sin(2\pi / 3) / \sin(\pi / 6))^2 = I_0 ((\sqrt{3}/2) / (1/2))^2 = 3I_0$.Case $C$: If $d = \lambda / 2$,then $\phi = (2\pi / \lambda) * (\lambda / 2) = \pi$. $I = I_0 (\sin(2\pi) / \sin(\pi / 2))^2 = 0$.Thus,option $B$ is correct.

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