\(\mu=1.6\)
\(\mathrm{n}=2(\text { from figure })\)
Applying formula \((\mu-1) \mathrm{t}=\mathrm{n} \lambda\)
\((1.6-1) \times 1.8 \times 10^{-6}=2 \lambda\)
or, \({\lambda = \frac{{1.8 \times {{10}^{ - 6}} \times 0.6}}{2}}\)
\(=\,540 \,\mathrm{nm}\)