Explanation:
\(\frac{{{I_{\max }}}}{{{I_{{\text{min }}}}}} = {\left( {\frac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}\)
\(\therefore \quad \dfrac{16}{1}=\left(\dfrac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}\)
\(\therefore 4\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)=\sqrt{I_{1}}+\sqrt{I_{2}}\)
\(\therefore 5 \sqrt{I_{2}}=3 \sqrt{I_{1}}\)
\(\therefore \dfrac{{I}_{1}}{{I}_{2}}=\dfrac{25}{9}\)
\(\therefore \frac{{{A_1}^2}}{{\;{A_2}^2}} = \frac{{25}}{9}\)\(\,\,\,\,\,\,\,\,\,\,\,\,\,\)\(\left( {\because \,\,1 \propto {A^2}} \right)\)
\(\therefore \frac{{{A_1}}}{{\;{A_2}}} = \frac{5}{3} = 1.67\)