Explanation

(a) The intensity of light reflected from upper surface is
$I_{1} = I_{0} \times 25\%$
$= I_{0} \times \frac{25}{100}$
$= \frac{I_{0}}{4}$
The intensity of transmitted light from upper surface is
$I = I_{0} - \frac{I_{0}}{4} = \frac{3I_{0}}{4}$
$\therefore$ The intensity of reflected light from upper surface is
$I_{2} = \frac{3I_{0}}{4} \times \frac{50}{100} = \frac{3I_{0}}{8}\ $
$\therefore\frac{\ I_{\max}}{\ I_{\min}} = \frac{\left( \sqrt{I_{1}} + \sqrt{I_{2}} \right)^{2}}{\left( \sqrt{I_{1}} - \sqrt{I_{2}} \right)^{2}}$
$\therefore\ $ $\frac{\ I_{\max}}{\ I_{\min}} = \frac{\left( \sqrt{I_{\frac{0}{4}}} + \sqrt{\frac{3I_{0}}{8}} \right)^{2}}{\left( \sqrt{\frac{I_{0}}{4}} - \sqrt{\frac{{3I}_{0}}{8}} \right)^{2}}$
$= \frac{\left( \frac{1}{2} + \sqrt{\frac{3}{8}} \right)^{2}}{\left( \frac{1}{2} - \sqrt{\frac{3}{8}} \right)^{2}}$