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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
    A ray of light of intensity I is incident on a parallel glass slab at point A as shown India gram. It undergoes partial reflection and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A 'B' undergo interference. The ratio of \[{{\operatorname{I}}_{\max }}\] and\[{{\operatorname{I}}_{\min }}\] is :              JEE Main Online Paper (Held On 09 April 2013)     

    A) 49 : 1                                     

    B) 7 : 1                       

    C)  4 : 1                       

    D) 8 : 1                

    Correct Answer: C

    Solution :

    \[{{l}_{1}}=\frac{l}{4}\]                 \[{{l}_{2}}=\frac{1}{4}({{l}_{1}})\]                 \[=\frac{1}{16}l\] \[\Rightarrow \]               \[\frac{{{l}_{\max }}}{{{l}_{\min }}}=\frac{{{({{A}_{1}}+{{A}_{2}})}^{2}}}{{{({{A}_{1}}-{{A}_{2}})}^{2}}}=\frac{{{(\sqrt{{{l}_{1}}}+\sqrt{{{l}_{2}}})}^{2}}}{{{(\sqrt{{{l}_{1}}}-\sqrt{{{l}_{2}}})}^{2}}}\]                 \[=\frac{(\sqrt{l}/2+\sqrt{l}/4}{{{(\sqrt{l}/2-\sqrt{l}/4)}^{2}}}\]                 \[=\frac{\sqrt{l}/2\,{{\left( 1+\frac{1}{2} \right)}^{2}}}{\sqrt{l}/2\left( 1-\frac{1}{2} \right)}\]                 \[=\frac{{{(3/2)}^{2}}}{{{(1/2)}^{2}}}=\frac{9}{1}\]                


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