In a Young's double slit experiment, the width of the one of the...

$⇒ {A}_{2} = 3{A}_{1}$

\dfrac{{I}_\max}{{I}_\min} = {\left(\dfrac{\sqrt{{I}_{1}} + \sqrt{{I}_{2}}}{\left|\sqrt{{I}_{1}} - \sqrt{{I}_{2}}\right|}\right)}^{2}

$∵$ Intensity $I ∝ {A}^{2}$

⇒ \dfrac{{I}_\max}{{I}_\min} ={\left(\dfrac{{A}_{1} + {A}_{2}}{\left|{A}_{1} - {A}_{2}\right|}\right)}^{2}

$= {\left(\dfrac{{A}_{1} + 3{A}_{1}}{\left|{A}_{1} - 3{A}_{1}\right|}\right)}^{2}$

$= {\left(\dfrac{4{A}_{1}}{2{A}_{1}}\right)}^{2} = 4: 1$