JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    Visible light of wavelength \[6000\times {{10}^{8}}\text{ }cm\] falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at \[60{}^\circ \] from the central maximum. If the first minimum is produced at \[{{\theta }_{1}}\], then \[{{\theta }_{1}}\] is close to [JEE MAIN Held on 07-01-2020 Morning]

    A) \[25{}^\circ \]              

    B) \[30{}^\circ \]

    C) \[20{}^\circ \]  

    D) \[45{}^\circ \]

    Correct Answer: A

    Solution :

    [a] \[d\sin {{\theta }_{2}}=2\lambda \] (for 2nd minima) \[\Rightarrow d\times \left( \frac{\sqrt{3}}{2} \right)=2\lambda ...(i)\] For first minima, \[d\sin {{\theta }_{1}}=\lambda \] \[\Rightarrow \sin {{\theta }_{1}}=\frac{\lambda }{d}=\frac{\sqrt{3}}{4}=0.43\] \[\Rightarrow {{\theta }_{1}}<30{}^\circ ,\Rightarrow {{\theta }_{1}}\approx 25{}^\circ \]


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