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If fringe width is $0.4 \mathrm{~mm}$, the distance between fifth bright and third dark band on same side is
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Verified Answer
The correct answer is:
$1 \mathrm{~mm}$
Position of $n$ th bright fringe from central
$$
\begin{array}{l}
\text { maxima } x_{r_{1}}=\frac{n_{1} \lambda D}{d} \text { here } n_{1}=5 \\
\therefore \quad x_{n_{4}}=\frac{5 \lambda D}{d}
\end{array}
$$
Position of $n$ th dark fringe from central maxima
$$
\begin{aligned}
x_{n} &=\frac{(2 n-1) \lambda D}{2 d}, \text { Here } n=3 \\
x_{n} &=\frac{5}{2} \frac{\lambda D}{d} \\
x_{r_{4}}-x_{n} &=\frac{2.5 \lambda D}{d}=2.5 \beta
\end{aligned}
$$
Given $\quad \beta=0.4 \mathrm{~mm}$
$$
\Rightarrow \quad x_{n_{1}}-x_{n}=1 \mathrm{~mm}
$$
$$
\begin{array}{l}
\text { maxima } x_{r_{1}}=\frac{n_{1} \lambda D}{d} \text { here } n_{1}=5 \\
\therefore \quad x_{n_{4}}=\frac{5 \lambda D}{d}
\end{array}
$$
Position of $n$ th dark fringe from central maxima
$$
\begin{aligned}
x_{n} &=\frac{(2 n-1) \lambda D}{2 d}, \text { Here } n=3 \\
x_{n} &=\frac{5}{2} \frac{\lambda D}{d} \\
x_{r_{4}}-x_{n} &=\frac{2.5 \lambda D}{d}=2.5 \beta
\end{aligned}
$$
Given $\quad \beta=0.4 \mathrm{~mm}$
$$
\Rightarrow \quad x_{n_{1}}-x_{n}=1 \mathrm{~mm}
$$
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