BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    The refractive index of water and glycerine are 1.33 and 1.47 respectively. What is the critical angle for a light ray going from the later to the former?

    A) \[60{}^\circ 48\]               

    B)         \[64{}^\circ 48\]                              

    C)        \[74{}^\circ 48\]                               

    D)  None of these

    Correct Answer: B

    Solution :

    Key Idea: When ray passes from denser to rarer medium, angle of refraction is greater than angle of incidence. When a ray of light passes from glycerine (denser,\[\mu =1.47\]) to water (rarer,\[\mu =1.33\]) the angle of refraction (r) is greater than angle of incidence (i), than from Snells law                                 \[\frac{\sin i}{\sin r}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}<1\] When \[r={{90}^{o}},\]corresponding angle of incidence is known as critical angle i.e., \[i={{\theta }_{c}}.\]                 \[\therefore \]  \[\frac{\sin {{\theta }_{c}}}{\sin {{90}^{o}}}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\]                 \[\Rightarrow \]               \[\sin {{\theta }_{c}}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\]                 \[\Rightarrow \]               \[{{\theta }_{c}}={{\sin }^{-1}}\left( \frac{{{\mu }_{2}}}{{{\mu }_{1}}} \right)\]                                 \[={{\sin }^{-1}}\left( \frac{1.33}{1.47} \right)\]                 \[{{\theta }_{c}}={{64}^{o}}\,48.\]


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