A) 10 : 8
B) 9 : 1
C) 4 : 1
D) 2 : 1
Correct Answer: C
Solution :
\[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{9}{1}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\frac{9}{1}}\] \[\Rightarrow \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{1}\] Then, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[=\frac{{{(3+1)}^{2}}}{{{(3-1)}^{2}}}\] \[=\frac{{{(4)}^{2}}}{{{(2)}^{2}}}=\frac{16}{4}\] \[=\frac{4}{1}\] Thus, \[{{I}_{\max }}:{{I}_{\min }}=4:1\]
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