CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    A ray of light is incident normally on one of the falls of a prism of angle \[{{30}^{o}}\] and refractive index \[\sqrt{2}.\] The angle of deviation will be:

    A) \[{{26}^{o}}\]                                    

    B) \[{{0}^{o}}\]

    C) \[{{23}^{o}}\]                                    

    D) \[{{13}^{o}}\]

    Correct Answer: D

    Solution :

    Given: Angle of the prism \[={{30}^{o}}\] Refractive index\[\mu =\sqrt{2}\] The relation for refractive index is given by \[\mu =\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{{{30}^{o}}+\delta }{2}}{\sin \frac{\sin {{30}^{o}}}{2}}\] \[\sqrt{2}=\frac{\sin \frac{{{30}^{o}}+{{\delta }_{m}}}{2}}{\sin 150}\]                 or            \[\sin \frac{{{30}^{o}}+{{\delta }_{m}}}{2}=\sqrt{2}\sin {{15}^{o}}\]                 or \[\sin \frac{{{30}^{o}}+{{\delta }_{m}}}{2}=1.414\times 0.2588\]                 or \[\sin \frac{{{30}^{o}}+{{\delta }_{m}}}{2}=0.3\,659=0.366\]                 or \[\frac{{{30}^{o}}+{{\delta }_{m}}}{2}={{\sin }^{-1}}0.366={{21.46}^{o}}={{21.5}^{o}}\]                 or            \[{{30}^{o}}+{{\delta }_{m}}={{43}^{o}}\]                 or            \[{{\delta }_{m}}={{43}^{o}}-{{30}^{o}}={{13}^{o}}\]


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