The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

1: $\frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}$

$I=I_0\left[\frac{\sin \alpha}{\alpha}\right]^2$, where $\alpha=\frac{\varphi}{2}$

For n^th secondary maxima $d \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda$

$\Rightarrow \alpha=\frac{\varphi}{2}=\frac{\pi}{\lambda}[d \sin \theta]=\left(\frac{2 n+1}{2}\right) \pi$

∴ $I=I_0\left[\frac{\sin \left(\frac{2 n+1}{2}\right) \pi}{\left(\frac{2 n+1}{n}\right) \pi}\right]^2=\frac{I_0}{\left\{\frac{(2 n+1)}{2} \pi\right\}^2}$

So $I_0: I_1: I_2=I_0: \frac{4}{9 \pi^2} I_0: \frac{4}{25 \pi^2} I_0=1: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}$