In a double slit interference experiment, the distance between the slits is 0.0005m and the screen is 2 meters from the slits. Yellow light

Question

In a double slit interference experiment, the distance between the slits is 0.0005m and the screen is 2 meters from the slits. Yellow light from a sodium lamp is used and it has a wavelength of 5.89 x 10-7 m. What is the distance between the first and second bright fringes on the screen?

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Thiên Ân 6 months 2021-07-20T03:06:00+00:00 1 Answers 13 views 0

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    2021-07-20T03:07:41+00:00

    Answer:

    4.712 mm or 0.004712 m

    Explanation:

    Parameters given:

    Distance between the slits, d = 0.0005 m

    Distance between the screen and the slits, R = 2 m

    Wavelength of the light, λ = 5.89 * 10^{-7} m

    The condition for the maxima in a double slit experiment is given as:

    mλ = \frac{dY}{R}

    where Y = distance between the first order maxima and the mth order maxima

    In this case, m = 2.

    Therefore:

    Y = mλR/d

    Y = \frac{2 * 5.89 * 10{-7} * 2}{0.0005}

    Y = 0.004712 m = 4.712 mm

    The distance between the first and second bright fringes is 4.712 mm.

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