Explanation

(c) $I = I_{0}\left\lbrack \frac{\sin\alpha}{\alpha} \right\rbrack^{2}$, where $\alpha = \frac{\phi}{2}$
For $n^{th}$ secondary maxima $d\sin\theta = \left( \frac{2n + 1}{2} \right)\lambda$
$\Rightarrow \alpha = \frac{\phi}{2} = \frac{\pi}{\lambda}\left\lbrack d\sin\theta \right\rbrack = \left( \frac{2n + 1}{2} \right)\pi$
$\therefore I = I_{0}\left\lbrack \frac{\sin{\left( \frac{2n + 1}{2} \right)\pi}}{\left( \frac{2n + 1}{2} \right)\pi} \right\rbrack^{2} = \frac{I_{0}}{\left\{ \left( \frac{2n + 1}{2} \right)\pi \right\}^{2}}$
So $I_{0}:I_{1}:I_{2} = I_{0}:\frac{4}{9\pi^{2}}I_{0}:\frac{4}{25\pi^{2}}I_{0}$
$= 1:\frac{4}{9\pi^{2}}:\frac{4}{25\pi^{2}}$