A) 1 : 16
B) 1 : 4
C) 1 : 2
D) 1 : 1
Correct Answer: A
Solution :
Intensity of the wave is given by \[l=2{{\pi }^{2}}{{n}^{2}}{{A}^{2}}\rho \upsilon \] Here from the given equation \[{{A}_{1}}=5,{{A}_{2}}=10,{{\omega }_{1}}=10{{\omega }_{2}}=20\] \[{{k}_{1}}=0.1\]and\[{{k}_{2}}=0.2\] so\[{{\upsilon }_{1}}=\frac{{{\omega }_{1}}}{{{k}_{1}}}=\frac{10}{01}=100,{{\upsilon }_{2}}=\frac{{{\omega }_{2}}}{{{k}_{2}}}=\frac{20}{0.2}=100\] As \[{{\upsilon }_{1}}={{\upsilon }_{2}}\] and medium is same. Thus \[I\propto {{A}^{2}}{{n}^{2}}\] As \[n=\frac{\omega }{2\pi }\] so \[I\propto {{A}^{2}}{{\omega }^{2}}\] \[\frac{{{i}_{1}}}{{{I}_{2}}}=\frac{A_{1}^{2}\omega _{1}^{2}}{A_{2}^{2}\omega _{2}^{2}}={{\left( \frac{A_{1}^{{}}\omega _{1}^{{}}}{A_{2}^{{}}\omega _{2}^{{}}} \right)}^{2}}\] \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{5}{10}\times \frac{10}{20} \right)}^{2}}=\frac{1}{16}\] Hence \[{{I}_{1}}:{{I}_{2}}=1:16.\]
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