In a diffraction pattern due to a single slit of width $$a,$$ the first minimum is observed at an angle $${30^ \circ }$$ when light of wavelength $$5000\,\mathop {\text{A}}\limits^ \circ $$  is incident on the slit. The first secondary maximum is observed at an angle of

Solution :
As the first minimum is observed at an angle of $${30^ \circ }$$ in a diffraction pattern due to a single slit of width $$a.$$
i.e. $$n = 1,\,\theta = {30^ \circ }$$
$$\because $$ According to Bragg’s law of diffraction,
$$\eqalign{ & a\sin \theta = n\lambda \Rightarrow a\sin {30^ \circ } = \left( 1 \right)\lambda \,\,\left( {n = 1} \right) \cr & \Rightarrow a = 2\lambda \,.......\left( {\text{i}} \right)\,\left\{ {\because \sin {{30}^ \circ } = \frac{1}{2}} \right\} \cr} $$
For 1^st secondary maxima
$$ \Rightarrow a\sin {\theta _1} = \frac{{3\lambda }}{2} \Rightarrow \sin {\theta _1} = \frac{{3\lambda }}{{2a}}\,\,........\left( {{\text{ii}}} \right)$$
Substitute value of a from Eq. (i) to Eq. (ii), we get
$$\sin {\theta _1} = \frac{{3\lambda }}{{4\lambda }} \Rightarrow \sin {\theta _1} = \frac{3}{4} \Rightarrow {\theta _1} = {\sin ^{ - 1}}\left( {\frac{3}{4}} \right)$$