BCECE Medical BCECE Medical Solved Papers-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)  \[{{60}^{o}}48\]

    B)  \[{{64}^{o}}48\]

    C)  \[{{74}^{o}}48\]

    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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