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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    A single slit of width b is illuminated by a coherent monochromatic light of wavelength \[\lambda \].If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e., distance between first minimum on either side of the central maximum)   [JEE Online 08-04-2017]

    A)  4.5 cm                 

    B)  1.5 cm

    C)  6.0 cm                 

    D)  3.0 cm

    Correct Answer: D

    Solution :

    Min. \[f\sin \theta =n\lambda \] \[\sin \theta =\frac{n\lambda }{6}\] n = 2 \[\sin \theta =\frac{n\lambda }{6}=\tan {{\theta }_{1}}=\frac{{{x}_{1}}}{D}\] x = 4 \[\sin {{\theta }_{2}}=\frac{4\lambda }{6}=\frac{{{x}_{2}}}{D}\] \[{{x}_{2}}-{{x}_{1}}=\frac{4\lambda }{6}-\frac{2\lambda }{6}=\frac{2\lambda }{6}=3cm\] width of central \[\max =\frac{2\lambda }{6}=3cm\]


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