JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Critical Thinking

  • question_answer
    In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength l), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is                              [IIT-JEE (Screening) 2002]

    A)            2l  

    B)            \[\frac{2\lambda }{3}\]

    C)            \[\frac{\lambda }{3}\]      

    D)            l

    Correct Answer: A

    Solution :

               According to given condition \[(\mu -1)t=n\lambda \]for minimum t, n =1 So, \[(\mu -1){{t}_{\min }}=\lambda \] \[{{t}_{\min }}=\frac{\lambda }{\mu -1}=\frac{\lambda }{1.5-1}=2\lambda \]


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