JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Self Evaluation Test - Wave Optics

  • question_answer
    The first diffraction minimum due to the single slit diffraction is seen at \[9=30{}^\circ \] for a light of wavelength 5000 A falling perpendicularly on the slit. The width of the slit is

    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 :

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


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