A 1036 nm film with an index of refraction n=2.62 is placed on the surface of glass n=1.52. Light (λ=520.0 nm) falls hits the perpendicular

Question

A 1036 nm film with an index of refraction n=2.62 is placed on the surface of glass n=1.52. Light (λ=520.0 nm) falls hits the perpendicular to the surface from air. You want to increase the thickness so the reflected light cancels. What is the minimum thickness of the film that you must add?

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Hải Đăng 3 years 2021-09-03T03:08:20+00:00 2 Answers 6 views 0

Answers ( )

  1. Xavia
    0
    2021-09-03T03:09:57+00:00
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    Answer:

    55.64\ \text{nm}

    Explanation:

    \lambda = Wavelength falling on film = 520 nm

    n = Refractive index of film = 2.62

    T = Thickness of film

    m = Order

    We have the relation

    2T=\dfrac{m\lambda}{n}\\\Rightarrow T=\dfrac{m\lambda}{2n}\\\Rightarrow T=\dfrac{m\times 520}{2\times 2.62}\\\Rightarrow T=99.24m

    The thickness should be greater than 1036 nm. This means m=11

    T=99.24\times 11=1091.64\ \text{nm}

    Thickness of the film to be added would be

    \Delta T=1091.64-1036=55.64\ \text{nm}

    Thickness of the film to be added is 55.64\ \text{nm}.

  2. Giakhanh
    0
    2021-09-03T03:10:08+00:00
    Reply

    Answer:

    Explanation:

    The ray of light is passing from high refractive index medium to low refractive index medium so condition for cancellation of reflected light is as follows .

    2μt = (2n+1) λ/2

    where μ is refractive index of the medium , t is thickness , λ is wavelength of light and n is a integer .

    Putting n = 10

    2x 2.62 x t = 21 x 520 / 2 nm

    5.24 t = 5460 nm

    t = 1042 nm

    Thickness required to be added

    = 1042 – 1036 = 6 nm .

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