JEE Main & Advanced Physics Ray Optics Sample Paper Topic Test - Refraction of Light Through Curved Surfaces

  • question_answer
    An electron revolves in air in a circle in a plane perpendicular to the principal axis. The air has a uniform magnetic field parallel to the principal axis and of magnitude 2 tesla with centre of circle at a distance 10 cm from the pole of the spherical curved glass \[(\mu =1.5)\] surface of radius of curvature 10 cm as shown in figure. An observer from P sees radius of circle to be 10 mm. The linear momentum of electron is

    A) \[0.4\times {{10}^{-21}}kg-m/s\]           

    B) \[0.8\times {{10}^{-21}}kg-m/s\]

    C) \[1.6\times {{10}^{-21}}kg-m/s\]

    D) \[2.4\times {{10}^{-21}}kg-m/s\]

    Correct Answer: C

    Solution :

    Applying refraction formula for curved surface \[\frac{{{\mu }_{2}}}{\upsilon }-\frac{{{\mu }_{1}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\]
    \[\Rightarrow \]   \[\frac{3}{2\upsilon }-\frac{1}{-10}=\frac{\frac{3}{2}-1}{+10}\] \[\Rightarrow \]   \[\upsilon =-30\,cm\]
    Magnification \[=\frac{{{h}_{i}}}{{{h}_{0}}}=\frac{{{\mu }_{1}}\upsilon }{{{\mu }_{2}}u}\]
    \[\Rightarrow \]   \[{{h}_{0}}=\frac{{{\mu }_{2}}u}{{{\mu }_{1}}\upsilon }{{h}_{i}}=\frac{\frac{3}{2}(-10)}{1\times (-30)}\times 10\,\,mm\]
    \[\therefore \]      \[{{h}_{0}}=5\,\,mm=\] radius of circle of \[{{e}^{-}}\]
    \[\Rightarrow \]   \[5\,mm=\frac{(m\upsilon )}{eB}\]  \[\Rightarrow \] \[m\upsilon =5\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}\times 2\,\,kg-m/s\]
    \[\therefore \]      Momentum \[=1.6\times {{10}^{-21}}\,kg-m/s\]

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