2. Solved Papers
  3. AMU Medical

AMU Medical AMU Solved Paper-2013

  • question_answer
    A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength 648 nm travelling in air is incident normally on the layer. It is partly reflected at upper and lower surfaces of the layer and the two reflected rays interfere. The least value of t for which the rays interfere constructively is

    A)  30 nm                  

    B)  60 nm

    C)  90 nm                  

    D)  120nm

    Correct Answer: C

    Solution :

                     Given \[\mu =\] refractive index of glass = 1.5 t = thickness of coating, \[\mu =\] refractive index of coating                 \[=1.8\,\lambda =648\times {{10}^{-9}}m\] For constructive interference                 \[2\mu t=(2n+1)\frac{\lambda }{2}\] \[2\times 1.8\times t=\frac{648\times {{10}^{-9}}}{2}\](t to be minimum n = 0)                 \[t=\frac{648}{2\times 2\times 1.8}\times {{10}^{-9}}m\]                 \[t=\frac{648}{7.2}\times {{10}^{-9}}m=90\,m\]


You need to login to perform this action.
You will be redirected in 3 sec spinner