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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    The first diffraction minimum due to the single slit diffraction is seen at \[\theta ={{30}^{o}}\]for a light of wavelength \[5000\overset{\text{o}}{\mathop{\text{A}}}\,\] falling perpendicularly on the slit. The width of the slit is   JEE Main Online Paper (Held On 12 May 2012)

    A) \[2.5\times {{10}^{-5}}cm\]                        

    B) \[1.25\times {{10}^{-5}}cm\]

    C) \[10\times {{10}^{-5}}cm\]         

    D) \[5\times {{10}^{-5}}cm\]

    Correct Answer: C

    Solution :

    For first minimum,\[d\sin \theta =\lambda \] \[\Rightarrow \]\[d=\frac{\lambda }{\sin \theta }=\frac{5000\times {{10}^{-8}}cm}{\sin {{30}^{o}}}\] \[=\frac{5000\times {{10}^{-8}}cm}{{}^{1}/{}_{2}}=10\times {{10}^{-5}}cm\]                


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