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, \[\delta =2n\pi ;n=0,1,2,\] ...etc. |
| Therefore, for zero order maxima, \[\delta =1\] |
| \[{{I}_{\max }}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] |
| For minimum intensity, \[\delta =(2n-1)\pi ;\] |
| \[n=1,\text{ }2,\] ...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, |
| \[{{I}_{\max }}+{{I}_{\min }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}+{{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] |
| \[=2({{I}_{1}}+{{I}_{2}})\] |
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