JCECE Medical JCECE Medical Solved Paper-2003

  • question_answer
    The ratio of intensities of two waves is 9:16. If they interfere, the ratio of maximum to minimum intensity will be:

    A)  4 : 1          

    B)  1 : 25

    C)  1 : 3         

    D)  49 : 1

    Correct Answer: D

    Solution :

     Key Idea: Intensity of wave is proportional to square of amplitude of the wave. \[\text{Intensity}\,\propto {{\text{(Amplitude)}}^{\text{2}}}\] or \[I\propto {{a}^{2}}\] or \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{a}_{1}}}{{{a}_{2}}} \right)}^{2}}\] Given, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{9}{16}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}=}\sqrt{\frac{9}{16}}=\frac{3}{4}\] Maximum intensity, \[{{I}_{\max }}={{({{a}_{1}}+{{a}_{2}})}^{2}}\] \[={{(3a+4a)}^{2}}\] \[={{(7a)}^{2}}\] \[{{I}_{\min }}={{({{a}_{1}}-{{a}_{2}})}^{2}}\] \[={{(3a-4a)}^{2}}\] \[={{(-a)}^{2}}\] Hence, the, ratio, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(7a)}^{2}}}{{{(-a)}^{2}}}\] \[=\frac{49}{1}\]


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