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JEE Main & Advanced AIEEE Solved Paper-2011

  • question_answer
    Let the \[x-z\] plane be the boundary between two transparent media. Medium 1 in \[z\ge 0\] has a refractive index of 2 and medium 2 with \[z>0\] has a refractive index of 3. A ray of light in medium 1 given by the vector \[\vec{A}=6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k}\] is incident on the plane of separation. The angle of refraction in medium 2 is   AIEEE  Solved  Paper-2011

    A) \[{{75}^{o}}\]                                       

    B) \[{{30}^{o}}\]

    C) \[{{45}^{o}}\]                                       

    D) \[{{60}^{o}}\]   

    Correct Answer: C

    Solution :

                              The data is inconsistent Taking boundary as \[x-y\] plane instead of \[x-z\] plane, the angle of incidence is given as \[\cos \theta =\frac{(-\hat{k}).(6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k})}{\sqrt{108+192+100}}\]    \[=\frac{1}{2}\Rightarrow \theta ={{60}^{o}}\] Now, \[\sqrt{2}\sin {{60}^{o}}=\sqrt{3}\sin r\] \[\Rightarrow \,\,\sin r=\frac{1}{\sqrt{2}}\]  \[\Rightarrow \,\,r={{45}^{o}}\]


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