a
Let \(n_{1}\) bright fringe of \(\lambda_{1}\) coincides with \(n_{2}\) bright fringe of \(\lambda_{2}\). Then

\(\frac{{{n_1}{\lambda _1}D}}{d} = \frac{{{n_2}{\lambda _2}D}}{d}\)  or  \({n_1}{\lambda _1} = {n_2}{\lambda _2}\)

\(\frac{{{n_1}}}{{{n_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{10000}}{{12000}} = \frac{5}{6}\)

Let \(x\) be given distance.

\(\therefore \quad x=\frac{n_{1} \lambda_{1} D}{d}\)

Here, \({n_1} = 5,\) \(D = 2\,{\text{m}},\) \(d = 2\,{\text{mm}}\) \( = 2 \times {10^{ - 3}}{\text{m}}\)

\({\lambda _1} = 12000\,\mathop {\text{A}}\limits^o \) \( = 12000 \times {10^{ - 10}}{\text{m}}\)  \( = 12 \times {10^{ - 7}}{\text{m}}\)

\(x = \frac{{5 \times 12 \times {{10}^{ - 7}}{\text{m}} \times 2{\text{m}}}}{{2 \times {{10}^{ - 3}}{\text{m}}}}\) \( = 6 \times {10^{ - 3}}{\text{m}} = 6\,\,{\text{mm}}\)