A slit of width a is illuminated by a light of wavelength 700nm. what will be the value of slit width a when

a) first minimum faces at an angle of diffraction 30 degree.

b) first maximum faces at an angle of diffraction 30 degree. pls.. explain

Given
Wavelength, $\lambda$=700 nm
Angle of diffraction, $\theta ={30}^0$
Width of slit = a
We know that
(i) For destructive interference (minima) in single slit
$$\begin{gathered} a \sin \theta =n\lambda , for 1st minima n=1 \\ \Rightarrow a \sin \theta =\lambda \\ \Rightarrow a=\frac{\lambda }{\sin \theta }=\frac{700 nm}{\sin {30}^0}=\frac{700 nm}{1/2}=1400 nm \end{gathered}$$
(ii) For constructive interference (maxima) in single slit,
$a \sin \theta =\left(n+\frac{1}{2}\right)\lambda$
In single slit, first maxima occurs at the perpendicular bisector of the slit (n = 0 for central maxima),since in this case occurring at diffraction angle $\theta$ so it must be of other order , let n = 1 (say)
$$\begin{gathered} \Rightarrow a \sin \theta =\frac{3\lambda }{2} \\ \Rightarrow a=\frac{3\lambda }{2 \sin \theta }=\frac{3\times 700 nm}{2\times \sin {30}^0}=2100 nm \end{gathered}$$