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

Position of minima in diffraction pattern is given by;  $d\sin \theta =n\lambda$
For first minima of  ${\lambda }_1$, we have 
$d\sin {\theta }_1=\left(1\right){\lambda }_1\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sin {\theta }_1=\frac{{\lambda }_1}{d}$   ….. (i)   T h e
First maxima approximately lies between first and second maxima. For wavelength ${\lambda }_2$  its position will be. 
$d\sin {\theta }_2=\frac{3}{2}{\lambda }_2\therefore \sin {\theta }_2=\frac{3{\lambda }_2}{2d}\mathrm{...}\left(ii\right)$

The two will coincide if,    

${\theta }_1={\theta }_{2}or\sin {\theta }_1=\sin {\theta }_2\therefore \frac{{\lambda }_1}{d}=\frac{3{\lambda }_2}{2d}or$

${\lambda }_2=\frac{2}{3}{\lambda }_1=\frac{2}{3}\times 660nm=440nm$