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JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Diffraction of Light

  • question_answer
    In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength \[{{\lambda }_{1}}\] is found to be coincident with the third maximum at \[{{\lambda }_{2}}\]. So

    A)            \[3{{\lambda }_{1}}=0.3{{\lambda }_{2}}\]                 

    B)            \[3{{\lambda }_{1}}={{\lambda }_{2}}\]

    C)            \[{{\lambda }_{1}}=3.5{{\lambda }_{2}}\]                    

    D)            \[0.3{{\lambda }_{1}}=3{{\lambda }_{2}}\]

    Correct Answer: C

    Solution :

               Position of first minima = position of third maxima i.e., \[\frac{1\times {{\lambda }_{1}}D}{d}=\frac{\left( 2\times 3+1 \right)}{2}\frac{{{\lambda }_{2}}D}{d}\,\,\Rightarrow \,{{\lambda }_{1}}=3.5{{\lambda }_{2}}\]


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