CET Karnataka Medical CET - Karnataka Medical Solved Paper-2014

  • question_answer
     A -fringe width of a certain interference pattern is \[\beta =0.002cm\]. What is the distance of 5th dark fringe from centre?

    A)  \[1.1\times {{10}^{-2}}cm\]

    B)  \[1\times {{10}^{-2}}cm\]

    C)  \[3.28\times {{10}^{6}}cm\]

    D)  \[11\times {{10}^{-2}}cm\]

    E)  None of the Above

    Correct Answer: E

    Solution :

    : Here, Fringe width, \[\beta =0.002cm=2\times {{10}^{-3}}cm\] Distance of nth dark fringe from centre \[{{x}_{n}}=\left( n-\frac{1}{2} \right)\frac{\lambda D}{d}\]where \[n=1,2,3,....\] As  \[\beta =\frac{\lambda D}{d}\] \[\therefore \] \[{{x}_{n}}\left( n-\frac{1}{2} \right)\beta \] For 5th dark fringe, \[n=5\] \[\therefore \] \[{{x}_{5}}=\left( 5-\frac{1}{2} \right)\beta =\frac{9}{2}\beta =\frac{9}{2}\times 2\times {{10}^{-3}}cm\] \[=9\times {{10}^{-3}}cm\] * None of the given options is correct.


You need to login to perform this action.
You will be redirected in 3 sec spinner