In a Young’s double slit experiment, the ratio of the slit’s width is $4:1$ . The ratio of the intensity of maxima to minima, close to the central fringe on the screen will be

$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{I_1}+\sqrt{I_2{)}^2}}{(\sqrt{I_1}-\sqrt{I_2{)}^2}}$

Intensity coming out from width means amount of energy coming out from slit.

Therefore, we can say intensity is proportional to width of slit.

$\Rightarrow \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=\frac{{\mathrm{w}}_1}{{ \mathrm{w}}_2}=\frac{4}{1}$

$\frac{{\mathrm{I}}_{\max }}{{\mathrm{I}}_{\min }}=\frac{{(\sqrt{{\mathrm{w}}_1}+\sqrt{{\mathrm{w}}_2})}^2}{{(\sqrt{{\mathrm{w}}_1}-\sqrt{{\mathrm{w}}_2})}^2}={\left(\frac{2+1}{2-1}\right)}^2=\frac{9}{1}$