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JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer
    The two coherent sources of equal intensity produce maximum intensity of \[100\] units at a point. If the intensity of one of the sources is reduced by \[36%\] \[\text{by}\] reducing its width then the intensity of light at the same point will be

    A) \[67\]                                   

    B) \[81\]

    C)  \[89\]                                  

    D)  \[90\]

    Correct Answer: B

    Solution :

    Intensity of each source,\[I=\frac{100}{4}=25\]unit If the intensity of one source is reduced by\[36%\], then \[{{I}_{1}}=25\] unit and                 \[{{I}_{2}}=25-\frac{25\times 36}{100}=16\,\,unit\] Hence, resultant intensity at the same point will be                 \[I'={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\]                     \[=25+16+\sqrt{25\times 16}=81\,\,unit\]


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