Monochromatic light falls on two very narrow slits 0.046 mm apart. Successive fringes on a screen 6.20 m away are 8.9 cm apart near the cent

Question

Monochromatic light falls on two very narrow slits 0.046 mm apart. Successive fringes on a screen 6.20 m away are 8.9 cm apart near the center of the pattern. Part A Determine the wavelength of the light.

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Doris 2 months 2021-07-28T08:52:58+00:00 1 Answers 8 views 0

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    2021-07-28T08:54:42+00:00

    Answer:

    λ = 6.602 x 10^(-7) m

    Explanation:

    In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is given as ;

    y = mλD/d

    Where;

    D is the distance of the screen from the slits = 6.2 m

    d is the distance between the two slits = 0.046 mm = 0.046 x 10^(-3) m

    The fringes on the screen are 8.9 cm = 0.089 m apart from each other, this means that the first maximum (m=1) is located at y = 0.089 m from the center of the pattern.

    Therefore, from the previous formula we can find the wavelength of the light:

    y = mλD/d

    So, λ = dy/mD

    Thus,

    λ = (0.046 x 10^(-3) x 0.089)/(1 x 6.2)

    λ = 6.602 x 10^(-7) m

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