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17 November, 02:27

Coherent light of frequency 6.34*1014 Hz passes through two thin slits and falls on a screen 82.0 cm away. You observe that the third bright fringe occurs at ± 3.16 cm on either side of the central bright fringeAt what distance from the central bright fringe will the third dark fringe occur?

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  1. 17 November, 06:13
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    y = 3.69 10⁻² m = 3.69 cm

    Explanation:

    The interference phenomenon is described by the equations.

    d sin θ = m λ for constrictive interference

    d sin θ = (m + ½) λ for destructive interference

    Where d is the distance between the slits, m is an integer representing the order of interference λ the wavelength

    Let's use the relationship

    c = λ f

    λ = c / f

    λ = 3 10⁸ / 6.34 10¹⁴

    λ = 4.732 10⁻⁷ m

    Let's use trigonometry to find the breast

    tan θ = y / x

    In interference experiments the angle is very small

    tan θ = sin θ / cos θ ≈ sin θ

    We replace

    d y / x = m λ

    d y / x = (m + ½) λ

    Let's reduce the magnitude to the SI system

    x = 82.0 cm (1m / 100 cm) = 0.820 m

    y = 3.16 cm = 0.0316 cm

    Let's use the first relationship to find the distance between the slits

    d = m λ x / y

    d = 3 4.732 10⁻⁷ 0.820 / 0.0316

    d = 368.38 10⁻⁷ m

    d = 3.6838 10⁻⁵ m

    Now we can use the relationship for destructive interference and clear the distance and

    d y / x = (m + ½) λ

    y = (m + ½) λ x / d

    y = (3 + ½) 4,732 10⁻⁷ 0.82 / 3.68 10⁻⁵

    y = 3.69 10⁻² m

    y = 3.69 cm
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