2. Solved Papers
  3. CET Karnataka Medical

CET Karnataka Medical CET - Karnataka Medical Solved Paper-2012

  • question_answer
    In Youngs double slit experiment, fringes of width \[\beta \] are produced on a screen kept at a distance of 1 m from the slit. When the screen is moved away by\[5\times {{10}^{-2}}m\], fringe width changes by \[3\times {{10}^{-5}}m\]. The separation between the slits is\[1\times {{10}^{-3}}m\]. The wavelength of the light used is ... nm.

    A)  \[500\]   

    B)  \[600\]  

    C)  \[700\]  

    D)  \[400\]

    Correct Answer: B

    Solution :

    :  \[\beta =\frac{\lambda D}{d}\]                   ?..(i) \[\beta =\frac{\lambda D}{d}\] ??..(ii) Subtract (ii) from (i), we get \[\beta -\beta =\frac{\lambda (D-D)}{d}\] Substituting the given values, we get \[3\times {{10}^{-5}}m=\frac{\lambda \times 5\times {{10}^{-2}}m}{1\times {{10}^{-3}}m}\] \[\lambda =\frac{3\times {{10}^{-8}}}{5\times {{10}^{-2}}}m=0.6\times {{10}^{-6}}m\] \[=600\times {{10}^{-9}}m=600nm\]


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