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(D) $1$. The distance between the two virtual sources $d$ is given by $d = 2a(\mu - 1)\alpha$, where $a = 1 \, m$ is the distance of the source from t

Vedclass · Accepted Answer

$6$

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(D) $1$. The distance between the two virtual sources $d$ is given by $d = 2a(\mu - 1)\alpha$, where $a = 1 \, m$ is the distance of the source from the biprism, $\mu = 1.5$, and $\alpha = 2 	imes 10^{-3} \, 	ext{rad}$.

$2$. Substituting the values: $d = 2 	imes 1 	imes (1.5 - 1) 	imes 2 	imes 10^{-3} = 2 	imes 0.5 	imes 2 	imes 10^{-3} = 2 	imes 10^{-3} \, m$.

$3$. The fringe width $\beta$ is given by $\beta = \frac{D\lambda}{d}$, where $D$ is the total distance from source to screen, $D = 1 + 4 = 5 \, m$, and $\lambda = 6000 	imes 10^{-10} \, m$.

$4$. $\beta = \frac{5 	imes 6000 	imes 10^{-10}}{2 	imes 10^{-3}} = 1.5 	imes 10^{-3} \, m = 1.5 \, mm$.

$5$. The width of the interference pattern on the screen is limited by the geometry of the biprism. The width of the overlapping region on the screen is $W = \frac{2a(\mu - 1)\alpha(a+b)}{a} = 2(\mu - 1)\alpha(a+b) = 2(0.5)(2 	imes 10^{-3})(5) = 10 	imes 10^{-3} \, m = 10 \, mm$.

$6$. The number of fringes $N = \frac{W}{\beta} = \frac{10 \, mm}{1.5 \, mm} \approx 6.66$. Since the question asks for the number of fringes seen, we consider the integer part or the closest physical count, which is $6$.

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