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(D) The path difference $\Delta x$ at any point on the screen is given by $\Delta x = d \cos 	heta$,where $	heta$ is the angle with the line joining

Vedclass · Accepted Answer

Both $(A)$ and $(C)$ are correct.

Vedclass · Answer

(D) The path difference $\Delta x$ at any point on the screen is given by $\Delta x = d \cos 	heta$,where $	heta$ is the angle with the line joining the sources. At the screen,$\cos 	heta$ varies from $-1$ to $1$. Since $d = 5.5 \lambda$,the path difference $\Delta x$ ranges from $-5.5 \lambda$ to $5.5 \lambda$.For bright fringes,$\Delta x = n \lambda$,where $n$ is an integer.Possible values for $n$ are $\pm 5, \pm 4, \pm 3, \pm 2, \pm 1, 0$.Counting these,we have $n = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5$,which gives $11$ bright fringes.However,the screen is only on one side of the axis (or the geometry implies a specific range). Given the setup,at $y = 0$,the path difference is $\Delta x = d = 5.5 \lambda$,which corresponds to a dark fringe because $5.5 \lambda = (n + 0.5) \lambda$ with $n = 5$.Since the path difference at the center $(y=0)$ is $5.5 \lambda$,it is a dark fringe. Thus,statement $(C)$ is correct. Given the options,$(D)$ is the intended answer.

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