JCECE Engineering JCECE Engineering Solved Paper-2010

  • question_answer
    Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is\[25\]. The intensities of the sources are in the ratio

    A) \[25:1\]                               

    B) \[5:1\]

    C) \[9:4\]                                  

    D)  \[25:16\]

    Correct Answer: C

    Solution :

    Let \[{{a}_{1}}\] and \[{{a}_{2}}\] be amplitudes of the two waves. For maximum intensity                 \[{{I}_{\max }}={{({{a}_{2}}+{{a}_{1}})}^{2}}\] For minimum intensity                 \[{{I}_{\min }}={{({{a}_{1}}-{{a}_{2}})}^{2}}\] Given,   \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{25}{1}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[\Rightarrow \]               \[\frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}}=\frac{5}{1}\Rightarrow \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2}\]                     (law of componendo and dividendo) Also, Intensity\[\propto {{(amplitude)}^{2}}\] \[\therefore \]  \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{{{a}_{1}}}{{{a}_{2}}} \right)}^{2}}=\frac{9}{4}\]


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