Two slits 0.200 mm apart. Light of wavelength 488 nm impinges on the slits and the interference pattern is observed on a surface 1.00 m from

Question

Two slits 0.200 mm apart. Light of wavelength 488 nm impinges on the slits and the interference pattern is observed on a surface 1.00 m from the slits. How far from the central axis is the fourth maxima? (9.8 mm)

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Thiên Thanh 4 years 2021-07-26T17:23:57+00:00 1 Answers 11 views 0

Answers ( )

  1. minhkhue
    0
    2021-07-26T17:25:43+00:00
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    Answer:

    Explanation:

    distance between two slits d = .2 x 10⁻³ m = 2 x 10⁻⁴

    Distance of screen D = 1.00 m

    wave length of light λ = 488 x 10⁻⁹ m

    distance of fourth maxima

    = 4 x  λ D / d

    = 4 x 488 x 10⁻⁹ x 1 / 2 x 10⁻⁴

    = 976 x 10⁻⁵ m .

    = 9.76 x 10⁻³ m

    = 9.8 mm .

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