In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle $3 0^{\circ}$ when light of wavelength $5000$ Å is incident on the slit. The first secondary maximum is observed at an angle of
${sin (30)^{\circ} = cos (60)^{\circ} = \frac{1}{2}}$

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The condition for the first minimum in a single slit diffraction pattern is given by $a sin θ = λ$ .

Substituting $θ = 3 0^{\circ}$ , we get $a sin 3 0^{\circ} = λ$ .

Since $sin 3 0^{\circ} = \frac{1}{2}$ , we have $a (\frac{1}{2}) = λ$ , which gives $a = 2 λ$ .

The condition for the first secondary maximum is given by $a sin θ^{'} = \frac{3 λ}{2}$ .

Substituting $a = 2 λ$ , we get $2 λ sin θ^{'} = \frac{3 λ}{2}$ .

Solving for $sin θ^{'}$ , we get $sin θ^{'} = \frac{3}{4}$ .

Therefore, $θ^{'} = sin^{- 1} (\frac{3}{4})$ .

Answer: $sin^{- 1} (\frac{3}{4})$

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