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(B) Let the intensity of each source be $I_0$. The maximum intensity is $I_{max} = 4I_0$.We are looking for points where the intensity $I = \frac{I_{m

Vedclass · Accepted Answer

$\frac{\lambda}{2d}$

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(B) Let the intensity of each source be $I_0$. The maximum intensity is $I_{max} = 4I_0$.We are looking for points where the intensity $I = \frac{I_{max}}{2} = 2I_0$.The resultant intensity is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi)$.Setting $I = 2I_0$,we get $2I_0 = 2I_0(1 + \cos \phi)$,which implies $1 + \cos \phi = 1$,so $\cos \phi = 0$.This gives the phase difference $\phi = \frac{\pi}{2}$.The path difference $\Delta x$ is related to the phase difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.Thus,$\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x$,which gives $\Delta x = \frac{\lambda}{4}$.For small angles,the path difference is $\Delta x = d \sin 	heta \approx d 	heta$.Therefore,$d 	heta = \frac{\lambda}{4}$,which gives $	heta = \frac{\lambda}{4d}$.The angular separation between the points on either side of the central maximum is $\Delta 	heta = 	heta - (-	heta) = 2	heta = 2 	imes \frac{\lambda}{4d} = \frac{\lambda}{2d}$.

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