આથી \( n_1 = 4, n_2 \)= \(\infty\) માટે,
આથી સમીકરણ \(\,\frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\)
\(\frac{1}{\lambda } = R\left[ {\frac{1}{{{4^2}}} - \frac{1}{\infty }} \right]\,\,\,\)
\(\Rightarrow \,\,\,\,\frac{1}{\lambda } = \frac{R}{{16}}\,\)
\(\therefore\) \(\lambda = \frac{{16}}{R}\)
\(R\,\, = \,\,10.97\,\, \times \,\,{10^6}\) મુક્તા,
\(\,\lambda = \frac{{16}}{{10.97 \times {{10}^6}}} = 1.46\,\,\mu\, m\)