Q101 / 160 · JIPMER 2014 · Physics

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).

  • A
    $10^{-4} \mathrm{~m}$
  • Correct answer
    B
    $10^{-3} \mathrm{~mm}$
  • C
    $10^{-4} \times \pi \mathrm{m}$
  • D
    $\pi \times 10^{-3} \mathrm{~m}$

Solution

$a=0.3 \quad \alpha=1^{\circ}, \quad \mu=1.54, \quad$ 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 $F=\frac{D \lambda}{d}$ $=\frac{0.7 \times 180 \pi \times 10^{-9}}{0.324}$ $=\frac{7 \times 180 \pi \times 10^{-6}}{324}=10^{-3} \mathrm{~mm}$

Hence, the correct answer is option: 'B'.

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