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(B) In a Young's double slit experiment,the condition for constructive interference (maxima) is given by $d \sin 	heta = n \lambda$,where $d$ is the

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$3$

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(B) In a Young's double slit experiment,the condition for constructive interference (maxima) is given by $d \sin 	heta = n \lambda$,where $d$ is the slit separation,$\lambda$ is the wavelength,and $n$ is the order of the maxima.Given $d = 1.8 \lambda$,the condition becomes $1.8 \lambda \sin 	heta = n \lambda$,which simplifies to $n = 1.8 \sin 	heta$.Since the maximum value of $\sin 	heta$ is $1$,the maximum value of $n$ is $1.8 	imes 1 = 1.8$.Thus,the possible integer values for $n$ are $0, \pm 1$.Therefore,the total number of maxima is $1$ (for $n=0$) $+ 2$ (for $n=1$ and $n=-1$),which equals $3$.

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

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