WB JEE Medical WB JEE Medical Solved Paper-2007

  • 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 \[\frac{3}{4}\text{th}\] the angle of prism. The angle of deviation is

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

    B)  \[{{39}^{o}}\]

    C)  \[\text{2}{{\text{0}}^{\text{o}}}\]                                          

    D)  \[\text{3}{{\text{0}}^{\text{o}}}\]

    Correct Answer: D

    Solution :

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


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