Monochromatic (single-wavelength) light falling on two slits 0.016 mm apart produces the fifth-order bright spot at an 8.8◦angle. What is th

Question

Monochromatic (single-wavelength) light falling on two slits 0.016 mm apart produces the fifth-order bright spot at an 8.8◦angle. What is the wavelength of the light used?

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Xavia 5 months 2021-08-29T12:33:37+00:00 2 Answers 16 views 0

Answers ( )

  1. Jezebel
    0
    2021-08-29T12:35:02+00:00
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    Given Information:  

    distance = d = 0.016 mm = 0.0016 cm = 0.000016 m

    Angle = θ = 8.8°

    bright spot = m = 5

    Required Information:  

    Wavelength =  λ = ?

    Answer:

    Wavelength = 4.89×10⁻⁷ m

    Explanation:

    Monochromatic light passes through a double slit. The corresponding diffraction is given by the equation:

    dsinθ = mλ

    where d is the distance between two slits, m is the order of the diffraction, θ is the angle and λ is the wavelength of the light wave.

    λ = dsinθ/m

    λ = 0.016*sin(8.8°)/5

    λ = 4.89×10⁻⁷ m

    Therefore, the wavelength of the monochromatic light falling on two slits is 4.89×10⁻⁷ m

  2. minhkhang
    0
    2021-08-29T12:35:32+00:00
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    Answer:

    the wavelength of the light used is 4.8 × 10⁻⁷m

    Explanation:

    We have given distance between the two slits

    d=1.6\times 10^{-2}mm=1.6\times 10^{-5}m

    Angle \Theta =8.8^{\circ}

    We know that for nth order fringe

    dsin\Theta =n\lambda

    In question it is given that 5th order so n =5

    So 1.6\times 10^{-5}sin\ 8.8 =5\lambda\\\lambda = 4.8\times 10^{-7}m

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