A) \[\frac{1+\beta }{\sqrt{\beta }}\]
B) \[\sqrt{\left( \frac{1+\beta }{\beta } \right)}\]
C) \[\frac{1+\beta }{2\sqrt{\beta }}\]
D) \[\frac{2\sqrt{\beta }}{1+\beta }\]
Correct Answer: D
Solution :
We know that, \[I\propto {{a}^{2}}\] \[a\propto \sqrt{l}\] \[\frac{{{l}_{\max }}}{{{l}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}=\frac{{{(1+\sqrt{\beta })}^{2}}}{{{(1-\sqrt{\beta })}^{2}}}\] \[\Rightarrow \,\,\,\,\frac{{{l}_{\max }}+{{l}_{\min }}}{{{l}_{\max }}-{{l}_{\min }}}=\,\,\frac{{{({{a}_{1}}+\sqrt{\beta })}^{2}}+{{(1-\sqrt{\beta })}^{2}}}{{{({{a}_{1}}+\sqrt{\beta })}^{2}}-{{(1-\sqrt{\beta })}^{2}}}=\frac{2(1+\beta )}{4\sqrt{\beta }}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{{{l}_{\max }}-{{l}_{\min }}}{{{l}_{\max }}+{{l}_{\min }}}=\frac{2\sqrt{\beta }}{1+\beta }\]
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