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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    A ray of light passes through an equilateral prism such that an angle of incidence is equal to the angle of emergence and the latter is equal to\[\frac{3}{4}\]th the angle of prism. The angle of deviation is:

    A)  \[45{}^\circ \]                  

    B)         \[39{}^\circ \]

    C)  \[20{}^\circ \]                  

    D)         \[30{}^\circ \]

    E)  \[90{}^\circ \]

    Correct Answer: D

    Solution :

    Angle of incidence = angle of emergence, i.e.,        \[i=i\] Also,\[i=\frac{3}{4}\times \]angle of equilateral prism \[=\frac{3}{4}\times 60{}^\circ =45{}^\circ \] Thus, angle of deviation \[=i+i-A\] \[=(45{}^\circ +45{}^\circ -60{}^\circ )=30{}^\circ \]


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