A) \[\frac{I}{2}\]
B) \[I\]
C) \[2I\]
D) \[4I\]
Correct Answer: D
Solution :
Key Idea: When two interfering waves meet in the same phase, resultant intensity is maximum. The resultant intensity due to two waves\[{{I}_{1}}\]and\[{{I}_{2}}\]and having phase difference\[\phi \]between them is given by \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] For maximum intensity, \[\phi =0\] \[\therefore \] \[{{I}_{\max }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] Given, \[{{I}_{1}}={{I}_{2}}=I\] \[\therefore \] \[{{I}_{\max }}={{(\sqrt{I}+\sqrt{I})}^{2}}=4I\] Note: Maximum intensity is also known as constructive interference.You need to login to perform this action.
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