In a Fresnel biprism experiment,the eyepiece | vedclass

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Question

(A) Given: Distance $D = 120 \, cm = 1.2 \, m$. Distance between virtual sources $d = 0.075 \, cm = 7.5 	imes 10^{-4} \, m$. Number of fringes $n = 2

Vedclass · Accepted Answer

$6000$

Vedclass · Answer

(A) Given: Distance $D = 120 \, cm = 1.2 \, m$. Distance between virtual sources $d = 0.075 \, cm = 7.5 	imes 10^{-4} \, m$. Number of fringes $n = 20$. Shift in eyepiece $\Delta x = 1.92 \, cm = 1.92 	imes 10^{-2} \, m$. The fringe width $\beta$ is given by $\beta = \frac{\Delta x}{n} = \frac{1.92 	imes 10^{-2}}{20} = 0.096 	imes 10^{-2} \, m = 9.6 	imes 10^{-4} \, m$. Using the formula $\beta = \frac{\lambda D}{d}$,we get $\lambda = \frac{\beta d}{D}$. Substituting the values: $\lambda = \frac{(9.6 	imes 10^{-4}) 	imes (7.5 	imes 10^{-4})}{1.2} = 8 	imes 7.5 	imes 10^{-7} \, m = 60 	imes 10^{-8} \, m = 6000 	imes 10^{-10} \, m$. Therefore,$\lambda = 6000 \, \mathop A\limits^o$.

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