A) \[\frac{\text{1+ }\!\!\beta\!\!\text{ }}{\sqrt{\text{ }\!\!\beta\!\!\text{ }}}\]
B) \[\sqrt{\left( \frac{\text{1+ }\!\!\beta\!\!\text{ }}{\text{ }\!\!\beta\!\!\text{ }} \right)}\]
C) \[\frac{\text{1+ }\!\!\beta\!\!\text{ }}{\text{2}\sqrt{\text{ }\!\!\beta\!\!\text{ }}}\]
D) \[\frac{\text{2}\sqrt{\text{ }\!\!\beta\!\!\text{ }}}{\text{1+ }\!\!\beta\!\!\text{ }}\]
Correct Answer: B
Solution :
We know \[I\propto {{a}^{2}}\]or \[a\propto \sqrt{I}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\left( \frac{{{I}_{1}}}{{{I}_{2}}} \right)}\] So, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}=\frac{({{a}_{1}}+{{a}_{2}})}{{{({{a}_{2}}-{{a}_{1}})}^{2}}}=\frac{{{(1+\sqrt{\beta })}^{2}}}{{{(1-\sqrt{\beta })}^{2}}}\] Applying componendo and dividendo \[\frac{{{I}_{\max }}+{{I}_{\min }}}{{{I}_{\max }}-{{I}_{\min }}}=\frac{{{(1+\sqrt{\beta })}^{2}}+{{(1-\sqrt{\beta })}^{2}}}{{{(1+\sqrt{\beta })}^{2}}-{{(1-\sqrt{\beta })}^{2}}}\] or \[=\frac{2+2\beta }{4\sqrt{\beta }}\] or \[\frac{{{I}_{\max }}-{{I}_{\min }}}{{{I}_{\max }}+{{I}_{\min }}}=\frac{2\sqrt{\beta }}{1+\beta }\]
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