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

  • question_answer
    A flake of glass (refractive index 1.5) is placed over one of the openings of a double slit apparatus. The interference pattern displaces itself through seven successive maxima towards the side where the flake is placed. if wavelength of the diffracted light is \[\lambda =600nm\], then the thickness of the flake is

    A)            2100 nm                                 

    B)            4200 nm

    C)            8400 nm                                 

    D)            None of these

    Correct Answer: C

    Solution :

               Shift \[=\frac{\beta }{\lambda }(\mu -1)\,t\] \[\Rightarrow 7\beta =\frac{\beta }{\lambda }(\mu -1)\,t\Rightarrow t=\frac{7\lambda }{(\mu -1)}\]\[=\frac{7\times 600}{(1.5-1)}=8400\,nm.\]


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