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If prism angle $\alpha=1^{\circ}, \quad \mu=1.54$, distance between screen and prism $(D)=0.7 \mathrm{~m}$, distance between prism and source $a=0.3 \mathrm{~m}$, $\lambda=180 \pi \mathrm{nm}$, then in Fresnal biprism find the value of $\beta$ (fringe width).
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The correct answer is:
$10^{-3} \mathrm{~mm}$
$a=0.3 \quad \alpha=1^{\circ}, \quad \mu=1.54, \quad$ we know that $d=2 a(\mu-1) \alpha$
$\begin{aligned} & =2 \times 0.3(1.54-1) \times 1 \\ & =2 \times 0.3 \times 0.54=0.324\end{aligned}$
Fringe width $F=\frac{D \lambda}{d}$
$=\frac{0.7 \times 180 \pi \times 10^{-9}}{0.324}$
$=\frac{7 \times 180 \pi \times 10^{-6}}{324}=10^{-3} \mathrm{~mm}$
$\begin{aligned} & =2 \times 0.3(1.54-1) \times 1 \\ & =2 \times 0.3 \times 0.54=0.324\end{aligned}$
Fringe width $F=\frac{D \lambda}{d}$
$=\frac{0.7 \times 180 \pi \times 10^{-9}}{0.324}$
$=\frac{7 \times 180 \pi \times 10^{-6}}{324}=10^{-3} \mathrm{~mm}$
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