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(B) The width of a certain region on the screen is $L$. The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wav

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$12$

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(B) The width of a certain region on the screen is $L$. The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance between the slits and the screen, and $d$ is the distance between the slits.The number of fringes $n$ in a region of width $L$ is given by $n = \frac{L}{\beta} = \frac{Ld}{\lambda D}$.Since $L, d,$ and $D$ are constant, we have $n \propto \frac{1}{\lambda}$, which implies $n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 8$, $\lambda_1 = 6600 	ext{ Å}$, and $\lambda_2 = 4400 	ext{ Å}$.Substituting the values: $8 	imes 6600 = n_2 	imes 4400$.$n_2 = \frac{8 	imes 6600}{4400} = \frac{8 	imes 66}{44} = \frac{8 	imes 3}{2} = 12$.Therefore, the number of fringes observed is $12$.

In a certain region of the screen, the number of fringes formed is observed to be $8$ in Young's double slit experiment when light of wavelength $6600 \text{ Å}$ is used. If light of wavelength $4400 \text{ Å}$ is used, the number of fringes observed in the same region of the screen will be:

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