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Manipal Medical Manipal Medical Solved Paper-2005

  • question_answer
    In a Youngs experiment, one of the slit is covered with a transparent sheet of thickness \[3.6\times {{10}^{-3}}cm\]due to which position of central fringe shifts to a position originally occupied by 30th bright fringe. The refractive index of the sheet, if\[\lambda \text{= }6000\overset{o}{\mathop{\text{A}}}\,\]is:

    A)  1.5              

    B)  1.2

    C)  1.3              

    D)  1.7

    Correct Answer: A

    Solution :

     The position of 30th bright fringe \[{{y}_{30}}=\frac{30\lambda D}{d}\] Now, position shift of central fringe is \[{{y}_{0}}=\frac{30\lambda D}{d}\] But we know\[{{y}_{0}}=\frac{D}{d}(\mu -1)t\] \[\frac{30\lambda D}{d}=\frac{D}{d}(\mu -1)t\] \[(\mu -1)=\frac{30\lambda }{t}\] \[=\frac{30\times 6000\times {{10}^{-10}}}{3.6\times {{10}^{-5}}}=0.5\] \[\mu =1.5\]


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