Four monochromatic and coherent sources of li | vedclass

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Question

(A) For points at a large distance $(|x| >> d)$,the path difference between waves from adjacent sources is $\Delta x = d \cos 	heta$. For a point on

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points having $|x| >> d$ appear dark if $d = \lambda /4$

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(A) For points at a large distance $(|x| >> d)$,the path difference between waves from adjacent sources is $\Delta x = d \cos 	heta$. For a point on the $x$-axis at a large distance,$	heta = 0$,so $\Delta x = d$.The total amplitude at a distant point is the sum of the four waves: $A = A_0(1 + e^{ikd} + e^{i2kd} + e^{i3kd})$,where $k = 2\pi / \lambda$.This is a geometric series with $r = e^{ikd}$. The sum is $A = A_0 \frac{1 - e^{i4kd}}{1 - e^{ikd}}$.The intensity $I$ is proportional to $|A|^2 = A_0^2 \left| \frac{1 - e^{i4kd}}{1 - e^{ikd}} ight|^2 = A_0^2 \frac{\sin^2(2kd)}{\sin^2(kd/2)}$.For dark points (intensity = $0$),we need $\sin(2kd) = 0$ but $\sin(kd/2) 
eq 0$.$2kd = n\pi$ (where $n$ is an integer). For $n=1$,$2(2\pi/\lambda)d = \pi \implies d = \lambda / 4$.Thus,if $d = \lambda / 4$,the intensity is zero,and the points appear dark.

Four monochromatic and coherent sources of light,emitting waves in phase of wavelength $\lambda$,are placed at the points $x = 0, d, 2d$,and $3d$ on the $x$-axis. Then:

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