JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    Light wavelength \[550nm\] falls normally on a slit of width\[22.0\times {{10}^{-5}}cm\]. The angular position of the second minima from the central maximum will be (in radians)  [JEE Online 15-04-2018]

    A) \[\frac{\pi }{8}\]                                 

    B) \[\frac{\pi }{12}\]       

    C) \[\frac{\pi }{4}\]                                 

    D) \[\frac{\pi }{6}\]         

    Correct Answer: D

    Solution :

    For single slit, the angular position is given as \[\sin \theta =\frac{n\lambda }{d}\] Here, \[n=2\] \[\sin \theta =\frac{2\times 550\times {{10}^{-9}}}{22\times {{10}^{-7}}}\] \[\sin \theta =\frac{1}{2}\] \[\theta =\frac{\pi }{6}\]


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