Radar consists of electromagnetic waves. A warplane is rendered invisible to enemy radar by applying an antireflective polymer coating. If r

Question

Radar consists of electromagnetic waves. A warplane is rendered invisible to enemy radar by applying an antireflective polymer coating. If radar waves have a frequency of 10 GHz (Giga = 1E9), and the index of refraction of the polymer for these waves is 1.50, what is the minimum thickness of the antireflective coating? (c = 3E8 m/s. The index of refraction of air is ~1, and that of the fuselage is larger than for the polymer coating)

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Nick 4 years 2021-09-04T05:29:30+00:00 1 Answers 5 views 0

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  1. Tryphena
    0
    2021-09-04T05:30:47+00:00
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    Answer:

    d = 0.015m

    Explanation:

    To find the thickness of the antireflective coating you take into account that waves must reflect and interfere destructively between them. A wave travels twice the thickness d of the coating, and for the destructive interference it is necessary that the reflected wave is (m+1/2) factor of the incident wave. Thus, you have:

    2d=(m+\frac{1}{2})\lambda_n

    d: thickness of the coating

    m: order of the interference (m=1 for the minimum thickness)

    λn: wavelength of light inside the coating

    You first calculate the wavelength of the wave:

    \lambda=\frac{c}{f}=\frac{3*10^8m/s}{10*10^9Hz}=0.03m

    \lambda_{coating}=\lambda_n=\frac{n_{air}}{n_{coating}}\lambda_{air}\\\\\lambda_n=\frac{1}{1.50}(0.03m)=0.06m

    Then, you replace the values of m and λn in order to calculate d:

    d=\frac{1}{2}(m+\frac{1}{2})\lambda_n

    d=\frac{1}{2}(0+\frac{1}{2})(0.06m)=0.015m=1.5cm

    hence, the thickness of the antireflective coating must be 0.015m

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