In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spot on a scree

Question

In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spot on a screen 1.6 m from the slits. Light of wavelength 670 nm is then projected through the same slits. Part A How far from the central bright spot will the second-order maximum of this light be located

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Nho 4 years 2021-08-06T01:09:15+00:00 1 Answers 12 views 0

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  1. Dau
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    2021-08-06T01:10:21+00:00
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    Answer:

    14.9 mm

    Explanation:

    We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

    Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

    So, sinθ = mλ/d

    Since θ is small, sinθ ≅ tanθ

    So,

    mλ/d = L/D

    d = mλD/L

    Substituting the values of the variables into the equation, we have

    d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

    d = 2448 × 10⁻⁹ m²/0.017 m

    d = 144000 × 10⁻⁹ m

    d = 1.44 × 10⁻⁴ m

    d = 0.144 × 10⁻³ m

    d = 0.144 mm

    Now, for the second-order maximum, m’ of the 670 nm wavelength of light,

    m’λ’/d = L’/D where m’ = order of maximum = 2, λ’ = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L’ = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

    So, L’ = m’λ’D/d

    So, substituting the values of the variables into the equation, we have

    L’ = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

    L’ = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

    L’ = 14888.89 × 10⁻⁶ m

    L’ = 0.01488 m

    L’ ≅ 0.0149 m

    L’ = 14.9 mm

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