Explanation:
We know that
For principle maxima in grating spectra,
\(\dfrac{\sin \theta}{N}=n \lambda\)
where, \({n=1,2,3, \ldots}\) order of principle maxima
\(\,\,\,\,\,\,\,\,\,\,\,\theta = \) angle of diffraction
In this case \({\theta=90^{\circ}}\)
\({\therefore \quad \dfrac{1}{N}=n \lambda \quad \Rightarrow \quad n=\dfrac{1}{\lambda N}}\)
\({\Rightarrow n=\dfrac{1}{6000 \times 10^{-10} \times 5000 \text { line } / {cm}}}\)
\({\Rightarrow n=\dfrac{1}{6000 \times 10^{-10} \times 5 \times 10^{5} \text { line } / {m}}}\)
\({\Rightarrow n=\dfrac{1}{30} \times 10^{2} \Rightarrow n=\dfrac{10}{3} \simeq 3}\).
So correct option is (3)