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(C) Let the distance between the sources be $d = 10 \mu m$. The distance of each source from the center $O$ is $d/2 = 5 \mu m$. The radius of the circ

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points $A$ and $C$ are dark and points $B$ and $D$ are bright

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(C) Let the distance between the sources be $d = 10 \mu m$. The distance of each source from the center $O$ is $d/2 = 5 \mu m$. The radius of the circle is $R = 40 \mu m$.At points $B$ and $D$,the path difference is $\Delta x = S_1P - S_2P = 0$ because these points lie on the perpendicular bisector of the line joining $S_1$ and $S_2$.Since $\Delta x = 0$,the phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = 0$. Thus,points $B$ and $D$ are bright.At points $A$ and $C$,the path difference is maximum. For point $A$,the distance $S_1A = R - d/2 = 40 - 5 = 35 \mu m$ and $S_2A = R + d/2 = 40 + 5 = 45 \mu m$. The path difference is $\Delta x_A = |S_2A - S_1A| = 10 \mu m$.Given $\lambda = 4 \mu m$,the path difference in terms of wavelength is $\Delta x_A = 10/4 \lambda = 2.5 \lambda$.Since the path difference is an odd multiple of $\lambda/2$ (i.e.,$5\lambda/2$),the interference is destructive,and points $A$ and $C$ are dark.

Two point sources $S_1$ and $S_2$ separated by a distance $10 \mu m$ emit light waves of wavelength $4 \mu m$ in phase. $A$ circular wire of radius $40 \mu m$ is placed around the sources as shown in the figure,where $O$ is the centre of the circle and $OS_1 = OS_2$. Then:

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