A) \[\frac{1}{3}\]
B) \[\frac{1}{2}\]
C) \[\frac{3}{2}\]
D) \[\frac{4}{5}\]
Correct Answer: C
Solution :
[c] \[{{\left( \frac{{{I}_{P}}}{{{I}_{\max }}} \right)}_{{{\lambda }_{1}}}}={{\cos }^{2}}\frac{\phi }{2}=0.75\] [\[\phi =\]phase difference between two waves arriving at P] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( \frac{\phi }{2} \right)=\frac{\sqrt{3}}{2}\] \[\phi =\frac{\pi }{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2\pi }{{{\lambda }_{1}}}{{(\Delta x)}_{1}}=\frac{\pi }{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,{{(\Delta x)}_{1}}=\frac{{{\lambda }_{1}}}{6}\] ?. (i) Similarly \[{{\left( \frac{{{I}_{P}}}{{{I}_{\max }}} \right)}_{{{\lambda }_{2}}}}={{\cos }^{2}}\frac{\phi '}{2}=0.5\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\cos \frac{\phi '}{2}=\frac{1}{\sqrt{2}}\] \[\phi '=\frac{\pi }{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{(\Delta x)}_{2}}=\frac{{{\lambda }_{2}}}{4}\] ? (ii) Because \[{{(\Delta x)}_{1}}={{(\Delta x)}_{2}}\] \[\frac{{{\lambda }_{1}}}{6}=\frac{{{\lambda }_{2}}}{4}\Rightarrow \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{3}{2}\]You need to login to perform this action.
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