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


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