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17 October, 01:15

A slit 0.380 mm wide is illuminated by parallel rays of light that have a wavelength of 600 nm. The diffraction pattern is observed on a screen that is 1.40 m from the slit. The intensity at the center of the central maximum (θ=0∘) is I0.

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  1. 17 October, 01:23
    0
    x = 2.19 mm

    I = 3.03*10^-9

    Explanation:

    step by step Explanation:

    diffraction is the overlapping of a wave particle when its made to pass through an aperture

    Sin θ = λ/a = 0.600nm / 380nm = 0.00158

    taking the sine inverse of bot sides

    θ = sin¯¹ (0.00158)

    angle of diffraction

    θ = 0.090°

    x = Rtanθ = 1.4 tan (0.090)

    x = 2.19 mm

    using small angle approximation,θ here will be half the previous value θ/2 = 0.090°/2

    θ = 0.0.45°

    so sinθ = 0.785*10¯³

    Using the formula

    β = 2πa sin θ/λ

    β = 2*3.142*380m*0.785*10¯³/0.600nm

    β = 3.124

    therefore we can write

    β/2 = 1.562

    substituting the parameters in to the equation

    I = Io (sin (β/2) / (β/2)) ²

    I = (10*10¯6) (0.0272/1.562) ²

    I = 3.03*10^-9
  2. 17 October, 01:38
    0
    x = 2.19 mm

    I = 3.03E-9

    Explanation:

    Sin θ = λ/a = 0.600µm / 380µm = 0.00158

    θ = sin¯¹ (0.00158)

    θ = 0.090°

    x = Rtanθ = 1.4 tan (0.090)

    x = 2.19 mm

    Approximating small amgle will be fine in this situation. By using small angle approximation, θ here will be 0.045°

    Therefore sinθ = 0.785*10¯³

    Using the formula

    β = 2πa sin θ/λ

    β = 2*3.142*380µm*0.785*10¯³/0.600µm

    β = 3.124

    So,

    β/2 = 1.562

    Utilizing above value in the formula

    I = Io (sin (β/2) / (β/2)) ²

    I = (10*10¯6) (0.0272/1.562) ²

    I = 3.03E-9
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