JEE Main & Advanced Sample Paper JEE Main - Mock Test - 34

  • question_answer
    When wave of wavelength \[0.2\text{ }cm\]is made incident normally on a slit of width \[0.004\text{ }m,\] then the semi-angular width of central maximum of diffraction pattern will be

    A) \[60{}^\circ \]                   

    B)        \[30{}^\circ \]      

    C) \[90{}^\circ \]                   

    D)        \[0{}^\circ \]

    Correct Answer: B

    Solution :

    [b] \[\theta ={{\sin }^{-1}}\left( \frac{\lambda }{a} \right)\]                          ...(1) According to question \[\lambda =2\times {{10}^{-3}}m\] \[a=4\times {{10}^{-3}}m\]                              .....(2) From equation (1) and (2) \[\theta ={{\sin }^{-1}}\left( \frac{1}{2} \right)\] \[\theta =30{}^\circ ,\]Hence correct answer is (2).            


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