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(B) For interference maxima,the path difference is given by $\Delta = d \sin 	heta = n\lambda$,where $n$ is an integer.Given that the slit separation

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$5$

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(B) For interference maxima,the path difference is given by $\Delta = d \sin 	heta = n\lambda$,where $n$ is an integer.Given that the slit separation $d = 2\lambda$.Substituting this into the equation: $2\lambda \sin 	heta = n\lambda$,which simplifies to $\sin 	heta = \frac{n}{2}$.Since the value of $\sin 	heta$ must lie in the range $[-1, 1]$,we have $-1 \le \frac{n}{2} \le 1$,which implies $-2 \le n \le 2$.The possible integer values for $n$ are $-2, -1, 0, 1, 2$.Counting these values,we get a total of $5$ maxima.

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

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