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EAMCET Medical EAMCET Medical Solved Paper-2000

  • question_answer
    In Foucaults rotating mirror experiment for determining the velocity of light, the distance between the rotating mirror and the convex lens is negligible, when compared to the radius of curvature of the concave mirror. If the radius of curvature of the concave mirror is doubled, the image shift to:

    A)  halved                                

    B)  doubled

    C)  zero                     

    D)  independent of radius of curvature

    Correct Answer: B

    Solution :

                     Formula for speed of light in Foucaults experiment \[c=\frac{8\pi {{a}^{2}}nd}{x(a+b)}\] n = Frequency of rotating mirror a = Radius of curvature of concave mirror b = Distance between rotating mirror and convex lens \[x=\] Image shift According to question, \[a>>b\] So, b is neglected \[\therefore \]  \[c=\frac{8\pi {{a}^{2}}nd}{xa}\]                 \[c=\frac{8\pi a\,n\,d}{x}\]                 First case, \[x=\frac{8\pi a\,n\,d}{c}\]                 Second case, \[x=\frac{8\pi (2a)nd}{c}\] \[\therefore \]  \[\frac{x}{x}=\frac{2a}{a}\] \[x=2x\] So, the image shift becomes double.


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