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

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$5$

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(B) In Young's double-slit experiment,the condition for constructive interference (bright fringes) is given by $d \sin 	heta = n \lambda$,where $n = 0, \pm 1, \pm 2, \dots$ and $d$ is the slit separation.Given that the distance between the slits $d = 2 \lambda$.Substituting this into the condition: $(2 \lambda) \sin 	heta = n \lambda$.This simplifies to $\sin 	heta = n / 2$.Since the maximum value of $\sin 	heta$ is $1$,we have $n / 2 \le 1$,which means $n \le 2$.Thus,the possible values for $n$ are $0, \pm 1, \pm 2$.For $n = 0$,we have the central bright fringe.For $n = 1$ and $n = -1$,we have two bright fringes.For $n = 2$ and $n = -2$,we have two bright fringes.Total number of bright fringes = $1 + 2 + 2 = 5$.

In a Young's double-slit experiment,the distance between the two slits is twice the wavelength of the light used. Find the total number of bright fringes formed on the screen.

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