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Question

(B) The path difference $\Delta x$ at a point $P$ on the circle is given by the projection of the separation $d$ onto the line connecting the point $P

Vedclass · Accepted Answer

${\cos ^{ - 1}}\frac{{4\lambda }}{d}$

Vedclass · Answer

(B) The path difference $\Delta x$ at a point $P$ on the circle is given by the projection of the separation $d$ onto the line connecting the point $P$ to the sources. From the geometry,$\Delta x = d \cos 	heta$,where $	heta$ is the angle with the line joining the sources.For constructive interference (maxima) at point $P$,the condition is $\Delta x = n\lambda$.Equating the two expressions for path difference:$d \cos 	heta = n\lambda$$\cos 	heta = \frac{n\lambda}{d}$$	heta = \cos^{-1} \left( \frac{n\lambda}{d} ight)$Given $n = 4$,the angular position is $	heta = \cos^{-1} \left( \frac{4\lambda}{d} ight)$.

Two coherent sources separated by a distance $d$ are radiating in phase with a wavelength $\lambda$. $A$ detector moves in a large circle around the two sources in the plane of the two sources. The angular position of the $n = 4$ interference maxima is given as

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