In a diffraction pattern due to a single slit of width $\text{'}a\text{'}$, the first minimum is observed at an angle $30°$ when light of wavelength $500 \mathrm{nm}$ is incident on the slit. The first secondary maximum is observed at an angle of

Given that for slit of width $a$the first minimum form at $\theta =30°,$the wavelength of light is $500 \mathrm{nm}$

Using Bragg's law of diffraction, $n\lambda =a\sin \theta ,$here $n$is the whole number and for the first minimum $n=1$

$$\begin{gathered} \Rightarrow \left(1\right)\lambda =a\sin 30° \\ \Rightarrow a=2\lambda \end{gathered}$$

For the first secondary maximum, $\left(n+\frac{1}{2}\right)\lambda =a\sin \theta \text{'}$where $n=1$

$$\begin{gathered} \Rightarrow \frac{3}{2}\lambda =a\sin \theta \text{'} \\ \Rightarrow \frac{3}{2}\lambda =\left(2\lambda \right)\sin \theta \text{'} \\ \Rightarrow \theta \text{'}={\sin }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$