580 nm light shines on a double slit with d = 0.000125 m. What is the angle of the first dark interference minimum (m = 1)?

Question

580 nm light shines on a double slit
with d = 0.000125 m. What is the
angle of the first dark interference
minimum (m = 1)?
(Remember, nano means 10-9.).
(Unit = deg)​

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Delwyn 4 years 2021-08-24T03:00:14+00:00 2 Answers 124 views 0

Answers ( )

  1. nguyetanh
    0
    2021-08-24T03:01:57+00:00
    Reply

    Answer:

    .133

    Explanation:

    x=sin^(-1)(((1-(1)/(2))(580*10^(-9)))/(.000125)) took me a while to figure it out but I just took the quiz and I promise this answer is correct please mark brainliest

  2. taiduc
    0
    2021-08-24T03:01:58+00:00
    Reply

    Answer:

    Angular position of the first minimum is 0.27 degree

    Explanation:

    As we know by the formula of single slit diffraction pattern

    a sin\theta = m\lambda

    here we have

    a = d = 0.000125 m

    for m = 1

    \lambda = 580 nm

    now we have

    0.000125 sin\theta = 580 \times 10^{-9}

    sin\theta = 4.64 \times 10^{-3}

    \theta = 0.27 degree

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