Two monochromatic (wavelength lambda = a/5) a | vedclass

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(A) The distance between the two sources is $d = 2a - (-a) = 3a$.The path difference $\Delta x$ at any point on the circle is given by $\Delta x = d \

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) The distance between the two sources is $d = 2a - (-a) = 3a$.The path difference $\Delta x$ at any point on the circle is given by $\Delta x = d \cos 	heta$,where $	heta$ is the angle with the $x$-axis.The maximum possible path difference is at the points on the $x$-axis,where $\Delta x = \pm d = \pm 3a$.Given $\lambda = a/5$,the maximum path difference in terms of wavelength is $\Delta x_{max} = 3a / (a/5) = 15\lambda$.For constructive interference (maxima),the path difference must be $n\lambda$,where $n$ is an integer.On the positive $x$-axis (at $P$),$\Delta x = +15\lambda$ $(n = 15)$.On the negative $x$-axis (at $Q$),$\Delta x = -15\lambda$ $(n = -15)$.As the detector moves in a full circle,the path difference varies from $+15\lambda$ to $-15\lambda$ and back to $+15\lambda$.The values of $n$ for maxima are $n = \pm 15, \pm 14, \dots, \pm 1, 0$.For each value of $n$ from $1$ to $14$,there are two points on the circle (one in the upper semicircle and one in the lower semicircle). This gives $14 	imes 2 = 28$ maxima.For $n = -1$ to $-14$,there are also $14 	imes 2 = 28$ maxima.For $n = 0$,there are two points (where $\Delta x = 0$,i.e.,the $y$-axis).For $n = 15$ and $n = -15$,there is only one point each (the poles $P$ and $Q$).Total number of maximas $= 28 + 28 + 2 + 1 + 1 = 60$.

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