Two coherent sources S_1 and S_2 are separate | vedclass

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(B) Let the distance between the sources be $d = 4\lambda$. The detector is at a point $D$ on the $x$-axis at a distance $x$ from the origin (which is

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$3$

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(B) Let the distance between the sources be $d = 4\lambda$. The detector is at a point $D$ on the $x$-axis at a distance $x$ from the origin (which is $S_2$).The path difference $\Delta x$ between the waves reaching $D$ from $S_1$ and $S_2$ is given by $\Delta x = S_1D - S_2D$.In the right-angled triangle $\Delta S_1S_2D$,$S_1D = \sqrt{d^2 + x^2} = \sqrt{(4\lambda)^2 + x^2}$ and $S_2D = x$.For maxima,the path difference must be an integer multiple of the wavelength: $\Delta x = n\lambda$,where $n$ is an integer.So,$\sqrt{16\lambda^2 + x^2} - x = n\lambda$.Rearranging gives $\sqrt{16\lambda^2 + x^2} = n\lambda + x$.Squaring both sides: $16\lambda^2 + x^2 = n^2\lambda^2 + 2nx\lambda + x^2$.$16\lambda^2 - n^2\lambda^2 = 2nx\lambda$.$x = \frac{(16 - n^2)\lambda}{2n}$.For $x > 0$,we must have $16 - n^2 > 0$,which means $n^2 If $n=1$,$x = \frac{15\lambda}{2} = 7.5\lambda$.If $n=2$,$x = \frac{12\lambda}{4} = 3\lambda$.If $n=3$,$x = \frac{7\lambda}{6} \approx 1.17\lambda$.Thus,there are $3$ points where maxima are observed.

Two coherent sources $S_1$ and $S_2$ are separated by a distance four times the wavelength $\lambda$ of the source. The sources lie along the $y$-axis,whereas a detector moves along the $+x$-axis. Leaving the origin and far-off points,the number of points where maxima are observed is

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