CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    A ray of light is travelling from glass to air. (refractive index of glass = 1.5). The angle of incidence is \[{{50}^{o}}\]. The deviation of the ray is

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

    B)  \[{{80}^{o}}\]

    C)  \[{{50}^{0}}-{{\sin }^{-1}}\left[ \frac{\sin {{50}^{0}}}{1.5} \right]\]                            

    D)  \[{{\sin }^{-1}}\left[ \frac{\sin {{50}^{0}}}{1.5} \right]-{{50}^{0}}\]

    Correct Answer: B

    Solution :

    Refractive index,              \[^{a}{{\mu }_{g}}=1.5\]                                 \[\frac{1}{\sin C}=1.5\] \[\Rightarrow \]                               \[C={{42}^{o}}\] Critical angle for glass \[={{42}^{o}}\] When the angle of incidence in the denser medium is greater than the critical angle, reflection takes place inside the denser medium. Hence, a ray of light incident at \[{{50}^{o}}\] in glass medium undergoes total internal reflection. Deviation \[(\delta )={{180}^{o}}-\text{(}{{50}^{o}}+{{50}^{o}}\text{)}\] (from the-figure) or            \[\delta ={{80}^{o}}\]


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