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EAMCET Medical EAMCET Medical Solved Paper-1996

  • question_answer
    In the determination of velocity of light by Focaults method, the minor is rotated at a speed of 300 rotations/second, and a concave mirror of radius of curvature 25 m is used. When the source is kept at 3 m from the lens and the rotating mirror at 10 m, a shift in the image is observed. If the velocity of light is \[3\times {{10}^{8}}m/s,\] the shift observed is:

    A)  1.346 mm                          

    B)  0.212 mm

    C)  4.13 mm                             

    D)  2.12 mm

    Correct Answer: A

    Solution :

     The velocity of light is determined by Focaults method is represented as follows \[C=\frac{8\pi n{{R}^{2}}b}{(R+a)x}\] Given: \[a=\]distance between lens and plane mirror \[R=\]radius of curvature of concave mirror \[b=\] distance between lens and source \[n=\]frequency of revolution of mirror \[x=\]displacement of image Hence,                  \[x=\frac{8\pi n{{R}^{2}}b}{(R+a)c}\] \[\left\{ \begin{align}   & \text{Given:}R=25m \\  & c=3\times {{10}^{8}}m/s \\  & a=10\,m \\  & b=3\,m \\  & n=300\,\text{rotation/sec} \\  & x=? \\ \end{align} \right\}\] Putting, the given values \[x=\frac{8\times 3.14\times 300\times {{(25)}^{2}}\times 3}{(10+25)\times 3\times {{10}^{8}}}\] \[=134571\times {{10}^{-8}}=1.346\,mm\]


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