CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    The time required for the light to pass through a glass slab (refractive index = 1.5) of thickness 4 mm is (\[c=3\times {{10}^{8}}m{{s}^{-1}}\] speed of light in free space)

    A)  \[{{10}^{-11}}s\]                             

    B)  \[2\times {{10}^{-11}}s\]

    C)  \[2\times {{10}^{11}}s\]                              

    D)  \[2\times {{10}^{-5}}s\]

    Correct Answer: B

    Solution :

    We know,           \[{{n}_{a}}{{c}_{a}}={{n}_{g}}\,{{C}_{g}}\]                                 \[\frac{{{n}_{g}}}{{{n}_{a}}}=\frac{{{c}_{a}}}{{{c}_{g}}}\]                                 \[\frac{3}{2}\times \frac{3\times {{10}^{8}}}{{{c}_{g}}}\]                                 \[{{c}_{g}}=2\times {{10}^{8}}\] We have.    Time \[=\frac{Distance}{Speed}\]                                 \[t=\frac{4\times {{10}^{-3}}}{2\times {{10}^{8}}}\] or                            \[t=2\times {{10}^{-11}}s\]


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