A) \[\frac{1+\beta }{\sqrt{\beta }}\]
B) \[\sqrt{\frac{1+\beta }{\beta }}\]
C) \[\frac{1+\beta }{2\sqrt{\beta }}\]
D) \[\frac{2\sqrt{\beta }}{1+\beta }\]
Correct Answer: D
Solution :
Given,\[\frac{{{I}_{1}}}{{{I}_{2}}}=\beta =\frac{{{a}^{2}}}{{{b}^{2}}}\] \[\Rightarrow \] \[\frac{a}{b}=\sqrt{\beta }\] \[{{I}_{\max }}={{(a+b)}^{2}}\] and \[{{I}_{\min }}={{(a-b)}^{2}}\] \[\therefore \] \[\frac{{{I}_{\max }}-{{I}_{\min }}}{{{I}_{\max }}+{{I}_{\min }}}=\frac{{{(a+b)}^{2}}-{{(a-b)}^{2}}}{{{(a+b)}^{2}}+{{(a-b)}^{2}}}\] \[=\frac{4ab}{2({{a}^{2}}+{{b}^{2}})}=\frac{2ab}{({{a}^{2}}+{{b}^{2}})}\] \[=\frac{2b\cdot b\sqrt{\beta }}{{{b}^{2}}+\beta {{b}^{2}}}=\frac{2\sqrt{\beta }}{1+\beta }\]
You need to login to perform this action.
You will be redirected in
3 sec
