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(A) The path difference $\Delta x$ at a point $P$ on the circle is given by $\Delta x = x \cos 	heta$,where $	heta$ is the angle with the diameter c

Vedclass · Accepted Answer

$20$

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(A) The path difference $\Delta x$ at a point $P$ on the circle is given by $\Delta x = x \cos 	heta$,where $	heta$ is the angle with the diameter containing the sources.For maximum intensity,the condition is $\Delta x = n \lambda$,where $n$ is an integer.Thus,$x \cos 	heta = n \lambda$,which implies $\cos 	heta = \frac{n \lambda}{x}$.Since $|\cos 	heta| \le 1$,we have $|\frac{n \lambda}{x}| \le 1$,so $|n| \le \frac{x}{\lambda}$.Given $x = 5 \lambda$,we have $|n| \le 5$.The possible values for $n$ are $0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$.For $n = 0$,there are $2$ points (where $\cos 	heta = 0$,i.e.,$	heta = 90^\circ, 270^\circ$).For each $n \in \{1, 2, 3, 4\}$,there are $4$ points (two in each semicircle,symmetric about the diameter).For $n = 5$,there are $2$ points (where $\cos 	heta = \pm 1$,i.e.,$	heta = 0^\circ, 180^\circ$).Total number of points $= 2 + (4 	imes 4) + 2 = 2 + 16 + 2 = 20$.

Two identical coherent sources are placed on a diameter of a circle of radius $R$ at a separation $x (x << R)$,symmetrically about the center of the circle. The sources emit waves of identical wavelength $\lambda$. Find the number of points on the circle with maximum intensity,given $x = 5 \lambda$.

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