A) \[{{I}_{1}}+{{I}_{2}}\]
B) \[{{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\]
C) \[{{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\]
D) \[2({{I}_{1}}+{{I}_{2}})\]
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, \[\delta =2\,n\pi ,n=0,1,2....\]etc. Therefore, for zero order maxima, \[\cos \delta =1\] \[{{I}_{\max }}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}+{{I}_{2}}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] For minimum intensity, \[\delta =(2N-1)\pi ,\] \[n=1,2,...\]etc. Therefore, for Ist order minima \[\cos \delta =-1\] \[{{I}_{\min }}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}\] \[={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] Therefore, \[{{I}_{\max }}+{{I}_{\min }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}+{{({{I}_{1}}-\sqrt{{{I}_{2}}})}^{2}}\] \[=2({{I}_{1}}+{{I}_{2}})\]
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