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NEET Sample Paper NEET Sample Test Paper-20

  • question_answer
    In the figure the intensity of waves arriving at D from two coherent sources\[{{s}_{1}}\]and \[{{s}_{2}}\] is \[{{I}_{0}}.\]The wavelength of the wave is \[\lambda =4\,m.\]Resultant intensity at D will be

    A) \[4{{I}_{0}}\]                        

    B) \[{{I}_{0}}\]              

    C)      \[2{{I}_{0}}\]            

    D)       0                                

    Correct Answer: C

    Solution :

    Path difference \[=\Delta x={{s}_{2}}D-{{s}_{1}}D=5-4=1\,m\] \[\therefore \]The corresponding phase difference will be \[\phi =\frac{2\pi }{\lambda }.\Delta x=\left( \frac{2\pi }{4} \right)(1)=\frac{\pi }{2}\] Using \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] \[I={{I}_{0}}+{{I}_{0}}+2\sqrt{{{I}_{0}}{{I}_{0}}}\cos \frac{\pi }{2}=2{{I}_{0}}\] Hence, the correction option is (c).


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