$\large\frac{I_{max}} {I_{min}}$$ = \bigg(\large\frac{\sqrt {I_1}+ \sqrt {I_2}}{\sqrt {I_1}- \sqrt {I_2}}\bigg)^2$

$\qquad= \bigg[\large \frac{\sqrt {\frac {I_1}{I_2}} +1}{\sqrt {\frac{I_1}{I_2}}-1}\bigg]^2$

$\qquad=9:1$

Hence d is the correct answer.