MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    A ray of light is travelling from glass to air. (refractive index of glass = 1.5). The angle of incidence is 50°. The deviation of the ray is

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

    B) \[{{80}^{0}}\]

    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, \[_{a}{{\mu }_{g}}=1.5\]            \[\frac{1}{\sin C}=1.5\] \[\Rightarrow \]               \[C=42{}^\circ \] Critical angle for glass\[=42{}^\circ .\] 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\[50{}^\circ \]in glass medium undergoes total internal reflection. Deviation\[(\delta )=180{}^\circ -(50{}^\circ +50{}^\circ )\] (from the figure) or          \[\delta =80{}^\circ \]


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