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29 May, 07:18

A thin film of MgF₂ (n = 1.38) coats a piece of glass. Constructive interference is observed for the reflection of light with wavelengths of 500 nm and 625 nm. What is the thinnest film for which this can occur?

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  1. 29 May, 09:44
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    t = 905.8 nm

    Explanation:

    Given:

    - Wavelength λ_1 = 500 nm

    - Wavelength λ_2 = 625 nm

    - MgF₂ refractive index n = 1.38

    Find:

    What is the thinnest film for which this can occur?

    Solution:

    - We have two different wavelength of light that constructively interfere at the surface of the film. We need the minimum thickness of film that would satisfy the condition of constructive interference for both!

    -Since the refractive index of glass is greater than that of MgF_2, the expression of constructive interference would be as follows:

    2*t = m*λ / n

    - Since, the orders m are unknown for each wavelength, also different. We will try to determine the first for each wavelength of light.

    - Construct two equation:

    t = m_1 * (500 nm) / (2*1.38)

    t = 181.1594203*m_1 nm

    t = m_2 * (625 nm) / (2*1.38)

    t = 226.4492754*m_2 nm

    - Now equate the two thicknesses which should be equal:

    226.4492754*m_2 = 181.1594203*m_1

    m_2 = 0.8*m_1

    - Now we know that m can only take integer values, and m is proportional to thickness t. So for thinnest thickness m's must take the least integer values. Hence, we have:

    m_2 = (4 / 5) * m_1

    So, m_1 = 5, m_2 = 4 ... Least integer values.

    - Now that we have m's we can compute the thickness t as follows:

    t = 181.1594203*m_1 nm

    - Substitute m_1 = 5, we have:

    t = 181.1594203*5 nm

    t = 905.8 nm

    - Substitute m_2 = 4 in:

    t = 226.4492754*m_2

    t = 226.4492754*4

    t = 905.8 nm

    - Our values of t = 905.8 nm matches for both wavelengths.
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