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

  • question_answer
    To produce a minimum reflection of wavelengths near the middle of visible spectrum (550 nm), how thick should a coating of\[Mg{{F}_{2}}\,(\mu =1.38)\]coated on a glass surface?

    A) \[{{10}^{-7}}\]

    B) \[{{10}^{-10}}\]

    C) \[{{10}^{-9}}m\]

    D) \[{{10}^{-8}}m\]

    Correct Answer: A

    Solution :

    [a] The optical path difference needed for destructive interference is \[2\mu d=(2n+1)\frac{\lambda }{2}\], n-0, 1, 2, ?.. Note that 2nd is the total optical path length that the rays traverse when n=0. \[\therefore \,\,\,d=\frac{\lambda /2}{2\mu }=\frac{\lambda }{4\mu }=\frac{350\times {{10}^{-9}}}{4\times 1.38}\] \[=100nm=1\times {{10}^{-7}}m\]


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