In Young's double slit experiment, two wavelengths \( \lambda_{1}=780 \mathrm{~nm} \) and \( \lambda_{2}=520 \mathrm{~nm} \) are used to
obtain interferance fringes. If the \( n^{\text {th }} \) bright band due to \( \lambda_{1} \) concides with \( (n+1)^{\text {th }} \) bright band
due to \( \lambda_{2} \) then the value of \( n \) is

Given,
\( n^{\text {th }} \) bright band due to \( \lambda_{1}=(n+1)^{\text {th }} \) bright band due to \( \lambda_{2} \)
\[
\begin{array}{l}
\Rightarrow \frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \\
\Rightarrow n \lambda_{1}=(n+1) \lambda_{2} \\
\Rightarrow \frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}
\end{array}
\]
Given, \( \lambda_{1}=780 \mathrm{~nm} \) and \( \lambda_{2}=520 \mathrm{~nm} \), therefore
\[
\frac{n+1}{n}=\frac{780}{520}
\]
On rearranging, we get
\[
\begin{array}{l}
52(n+1)=78 n \\
\Rightarrow 52 n+52=78 n \\
\Rightarrow 78 n-52 n=52 \\
\Rightarrow 26 n=52 \\
n=\frac{52}{26}=2
\end{array}
\]
Thus, value of \( n \) is \( 2 \).