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

  • question_answer
    The maximum number of possible interference maxima for slit separation equal to\[1.8\lambda \], where\[\lambda \] is the wavelength of light used, in a Young's double slit experiment is

    A) zero      

    B) 3    

    C) infinite  

    D) 5

    Correct Answer: B

    Solution :

    [b] As\[\sin \theta =\frac{n\lambda }{d}\]and\[\sin \theta \]cannot be\[1\] \[\therefore \,\,\,1=\frac{n\lambda }{1.8\lambda }\] or \[n=1.8\]    Hence maximum number of possible interference maximas, 0, \[~\pm 1\] i.e. 3


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