In a single slit diffraction experiment first minimum for ${\lambda }_1=660nm$  coincides with first maxima for wavelength  ${\lambda }_2$ . Then ${\lambda }_2$  is

Position of minima in diffraction patter is given by;  $dsin\theta =n\lambda$
For first minima of  ${\lambda }_1,$ we have
$dsin{\theta }_1=\left(1\right){\lambda }_1$  or  $sin{\theta }_1=\frac{{\lambda }_1}{d}$             …(i)
The first maxima approximately lies between first and second minima. For wavelength ${\lambda }_2$  its positive will be, $dsin{\theta }_2=\frac{3}{2}{\lambda }_2 \therefore sin{\theta }_2=\frac{3{\lambda }_2}{2d}$             …….(ii)
The two will coincider if, ${\theta }_1={\theta }_2$  are  $\sin {\theta }_1=\sin {\theta }_2\therefore \frac{{\lambda }_1}{d}=\frac{3{\lambda }_2}{2d}or{\lambda }_2=\frac{3}{2}{\lambda }_1=\frac{3}{2}\times 660nm=440nm.$