Explanation

(d) Here, $\frac{I_{1}}{I_{2}} = \frac{16}{9}$
Since, intensity $\propto (amplitude)^{2}$
$\therefore\frac{I_{1}}{I_{2}} = \left( \frac{A_{1}}{A_{2}} \right)^{2} = \frac{16}{9}$
Or $\frac{A_{1}}{A_{2}} = \sqrt{\frac{16}{9}} = \frac{4}{3}$
$\frac{I_{\max}}{I_{\min}} = \frac{\left( A_{1} + A_{2} \right)^{2}}{\left( A_{1} - A_{2} \right)^{2}} = \frac{\left( \frac{A_{1}}{A_{2}} + 1 \right)^{2}}{\left( \frac{A_{1}}{A_{2}} - 1 \right)^{2}}$
$= \frac{\left( \frac{4}{3} + 1 \right)^{2}}{\left( \frac{4}{3} - 1 \right)^{2}} = \frac{49}{1}$