A) \[{{I}_{1}}+{{I}_{2}}\]
B) \[{{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\]
C) \[{{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\]
D) \[2(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})\]
Correct Answer: D
Solution :
Resultant intensity of two periodic waves is given by\[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \delta \]where \[\delta \] is the phase difference between the waves. For maximum intensity, \[\delta =2n\pi ;n=0,1,2,\] ...etc. Therefore, for zero order maxima, \[\delta =1\] \[{{\operatorname{I}}_{max}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] For minimum intensity, \[\delta =(2n-1)\pi ;\] n = 1, 2,...etc Therefore, for Ist order minima, \[\cos \delta =-1\] \[{{\operatorname{I}}_{min}}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}\] \[={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] Therefore, \[{{\operatorname{I}}_{max}}+{{\operatorname{I}}_{min}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}+{{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] \[=2({{I}_{1}}+{{I}_{2}})\]
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