c
(c) \(I = {I_0}{\left[ {\frac{{\sin \alpha }}{\alpha }} \right]^2},\) where \(\alpha = \frac{\phi }{2}\)
For \({n^{th}}\) secondary maxima \(d\sin \theta = \left( {\frac{{2n + 1}}{2}} \right)\lambda \)
\( \Rightarrow \alpha = \frac{\phi }{2} = \frac{\pi }{\lambda }\left[ {d\sin \theta } \right] = \left( {\frac{{2n + 1}}{2}} \right)\pi \)
\(\therefore \,\,\,I = {I_0}{\left[ {\frac{{\sin \left( {\frac{{2n + 1}}{2}} \right)\pi }}{{\left( {\frac{{2n + 1}}{n}} \right)\pi }}} \right]^2} = \frac{{{I_0}}}{{{{\left\{ {\frac{{(2n + 1)}}{2}\pi } \right\}}^2}}}\)
So \({I_0}:{I_1}:{I_2} = {I_0}:\frac{4}{{9{\pi ^2}}}{I_0}:\frac{4}{{25{\pi ^2}}}{I_0}\)
\( = 1:\frac{4}{{9{\pi ^2}}}:\frac{4}{{25{\pi ^2}}}\)