Given, The ratio of slits width = ${1 \over {25}}$

$\therefore$ ${{{I_1}} \over {{I_2}}} = {{25} \over 1}$

I $\propto$ A²

$\Rightarrow$ ${{{I_1}} \over {{I_2}}} = {{A_1^2} \over {A_2^2}}$ = ${{25} \over 1}$

$\Rightarrow$ ${{{A_1}} \over {{A_2}}} = {5 \over 1}$

${{{A_{\max }}} \over {{A_{\min }}}} = {{{A_1} + {A_2}} \over {{A_1} - {A_2}}}$ = ${{5 + 1} \over {5 - 1}}$ = ${6 \over 4}$ = ${3 \over 2}$

$\therefore$ ${{{I_{\max }}} \over {{I_{\min }}}} = {{A_{\max }^2} \over {A_{\min }^2}}$ = ${\left( {{3 \over 2}} \right)^2}$ = ${9 \over 4}$