A) 2 : 1
B) 5 : 3
C) 4 : 1
D) 16 : 1
Correct Answer: D
Solution :
\[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}=\frac{{{(5+3)}^{2}}}{{{(5-3)}^{2}}}=\frac{16}{1}\]
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