a
(a)In a single slit diffraction experiment, position of minima is given by \(d\sin \theta = n\lambda \)
So for first minima of red \(\sin \theta = 1 \times \left( {\frac{{{\lambda _R}}}{d}} \right)\)
and as first maxima is midway between first and second minima, for wavelength \(\lambda '\),
its position will be
\(d\sin \theta ' = \frac{{\lambda ' + 2\lambda '}}{2} \Rightarrow \sin \theta ' = \frac{{3\lambda '}}{{2d}}\)
According to given condition \(\sin \theta = \sin \theta '\)
\( \Rightarrow \lambda ' = \frac{2}{3}{\lambda _R}\) so \(\lambda ' = \frac{2}{3} \times 6600 = 440\,nm\)\( = 4400\,{Å}\)