Two coherent monochromatic point sources S_1 | vedclass

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(D) The path difference at a point $P$ on the circumference,where the line joining the center to $P$ makes an angle $	heta$ with the line joining the

Vedclass · Accepted Answer

$B, C$

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(D) The path difference at a point $P$ on the circumference,where the line joining the center to $P$ makes an angle $	heta$ with the line joining the sources,is given by $\Delta x = d \cos 	heta$.At point $P_1$,$	heta = 90^{\circ}$,so $\Delta x = d \cos 90^{\circ} = 0$. This corresponds to the central maximum.At point $P_2$,$	heta = 0^{\circ}$,so $\Delta x = d \cos 0^{\circ} = d = 1.8 \ mm$.The order of the fringe $n$ at $P_2$ is $n = \frac{d}{\lambda} = \frac{1.8 	imes 10^{-3} \ m}{600 	imes 10^{-9} \ m} = 3000$.Since $n = 3000$ is an integer,a bright spot (maxima) is formed at $P_2$. Thus,option $[A]$ is incorrect and option $[B]$ is correct.The number of fringes between $P_1$ $(	heta = 90^{\circ}, n=0)$ and $P_2$ $(	heta = 0^{\circ}, n=3000)$ is $3000$. Thus,option $[C]$ is correct.For maxima,$d \cos 	heta = n \lambda$. Differentiating,$-d \sin 	heta \ d	heta = \lambda \ dn$,so the angular fringe width is $\Delta 	heta \approx |d	heta| = \frac{\lambda}{d \sin 	heta}$.As we move from $P_1$ $(	heta = 90^{\circ})$ to $P_2$ $(	heta = 0^{\circ})$,$\sin 	heta$ decreases,so $\Delta 	heta$ increases. Thus,option $[D]$ is incorrect.Therefore,the correct options are $[B]$ and $[C]$.

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