JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Waves

  • question_answer
    In a standing wave formed as a result of reflection from a surface, the ratio of the amplitude at an antinode to that at node is x. The fraction of energy that is reflected is

    A)  \[{{\left[ \frac{x-1}{x} \right]}^{2}}\]

    B)  \[{{\left[ \frac{x}{x+1} \right]}^{2}}\]

    C) \[{{\left[ \frac{x-1}{x+1} \right]}^{2}}\]

    D) \[{{\left[ \frac{1}{x} \right]}^{2}}\]

    Correct Answer: C

    Solution :

    [c] \[\frac{{{A}_{1}}+{{A}_{2}}}{{{A}_{1}}-{{A}_{2}}}=x;\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{x-1}{x+1};\,Energy\propto {{A}^{2}}\] \[\Rightarrow {{\left( \frac{x-1}{x+1} \right)}^{2}}\]


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