A) \[{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}\]
B) \[{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}\]
C) \[2\left( {{I}_{1}}-{{I}_{2}} \right)\]
D) \[{{I}_{1}}\,\,+\,\,{{I}_{2}}\]
Correct Answer: C
Solution :
As, intensity, \[I \propto \,\,{{A}^{2}}\,(A\,\,is amplitude)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,I=k{{A}^{2}}\,\,\,\,\,\Rightarrow \,\,\,A=\sqrt{\frac{I}{A}}\] On superposition, \[{{A}_{\max }}={{A}_{1}}+{{A}_{2}};\,\,{{A}_{\min }}={{A}_{1}}-{{A}_{2}}\] \[\therefore \,\,\,\,\,\,\,A_{\max }^{2}\,\,=\,\,A_{1}^{2}+A_{2}^{2}\,\,+\,\,2{{A}_{1}}{{A}_{2}}\] \[\frac{{{\operatorname{I}}_{max}}}{k}=\frac{{{I}_{1}}}{k}+\frac{{{I}_{2}}}{k}+\frac{2\sqrt{{{I}_{1}}{{I}_{2}}}}{k}\] \[{{\operatorname{I}}_{max}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\]` ?. (i) Now, \[A_{\min }^{2}=A_{1}^{2}+A_{2}^{2}-2{{A}_{1}}{{A}_{2}}\] \[\frac{I_{\min }^{{}}}{k}=\,\,\frac{{{I}_{1}}}{k}+\frac{{{I}_{2}}}{k}-2\frac{\sqrt{{{I}_{1}}{{I}_{2}}}}{k}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\operatorname{I}}_{min}}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}\] ?. (ii) From, Eqs. (i) and (ii) \[{{\operatorname{I}}_{\max }}\,\,+\,\,{{\operatorname{I}}_{min}}\,=\,\,2({{I}_{1}}+{{I}_{2}})\]
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