In Fraunhoffer diffraction pattern width of central bright fringe is $\beta$. If wavelength of light is doubled and slit width is halved, distance of second dark fringe from the centre of central bright fringe is 

If ‘a’ be the width of the slit and $\lambda$ be the wavelength of light used. Then $\mathrm{\beta }=\raisebox{1ex}{$2\mathrm{\lambda }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right.$. 

In the second case width of central maxima is 

${\mathrm{\beta }}^1=\frac{2{\mathrm{\lambda }}^1}{{\mathrm{a}}^1}=\frac{2\left(2\mathrm{\lambda }\right)}{\left(\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}=4\left(\frac{2\mathrm{\lambda }}{\mathrm{a}}\right)=4\mathrm{\beta }$

$\therefore$ Width of first secondary maxima $=\frac{{\mathrm{\beta }}^1}{2}=2\mathrm{\beta }$

$\therefore$ Distance of second dark fringe 

= Half of the width of central bright fringe + width of first secondary bright fringe

$=\frac{{\mathrm{\beta }}^1}{2}+\frac{{\mathrm{\beta }}^1}{2}={\mathrm{\beta }}^1=4\mathrm{\beta }$