580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum (m=3)? (Remember, nan

Question

580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum (m=3)?
(Remember, nano means 10^-9.)
(Unit=deg)

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Amity 5 months 2021-09-04T07:29:23+00:00 1 Answers 4 views 0

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    2021-09-04T07:31:15+00:00

    Answer:

    0.66 degrees

    Explanation:

    The computation of the angle of the third dark interference is shown below:

    The condition of the minima is

    Path difference = (2n +1) × \lambda÷ 2

    For third minima, n = 2

    Now

    xd ÷ D = (2 × 2 + 1) × \lambda÷ 2

    d tan Q_3 = 5\lambda ÷ 2

    tan Q_3 = 5\lambda ÷ 2d

    Q_3 = tan^-1 × (5\lambda ÷2d)

    = tan^-1 × (5 × 580 × 10^-9) ÷ (2 × 0.000125)

    = 0.66 degrees

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