Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15.

Question

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.
Which laser has its first maximum closer to the central maximum?
What is the distance Image for Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas las between the first maxima (on the same side of the central maximum) of the two patterns?
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas las= ______ m
What is the distance Deltay_max-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Deltay_max-min = ______ m

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Trúc Chi 5 years 2021-08-06T03:14:34+00:00 1 Answers 14 views 0

Answers ( )

  1. hongcuc2
    0
    2021-08-06T03:16:10+00:00
    Reply

    Answer:

    A)   Therefore laser1 has the maximum closest to the central maximum

    B) Δₓ = 0.8

    Explanation:

    A) The expression for the constructive interference of a double slit is

               d sin θ  = m λ

    let’s use trigonometry to find the angle

              tan θ = y / L

    in interference phenomena the angles are small

             tan θ = sin θ / cos θ = sin θ

             sin θ = y / L

    we subjugate

             d y / L = m λ

             y = m λ L / d

    let’s apply this equation for each case

    a) Lares 1 has a wavelength λ₁ = d / 20, the screen is at L = 6.00 m

    they ask us for the first axiom m = 1,

    let’s calculate

               y₁ = 1 (d / 20) 6.00 / d

               y₁ = 0.3

    Laser 2, λ₂ = d / 15

                λ₂ = 1 (d / 15) 6.00 / d

               λ₂ = 0.4

    Therefore laser1 has the maximum closest to the central maximum

    b) let’s find the distance of each requested value

    second maximum m = 2 of laser 1

                yi ‘= 2 (d / 20) 6 / d

                y1 ‘= 0.6

    3rd minimum of laser 2

    the expression for destructive interference is

                   d sinθ = (m + 1/2) lam

                   y = (m ) λ L / d

    in this case m = 3

    let’s calculate

                  y2 ‘= (3+0.5) (d / 15) 6 / d

                  y2 ‘=21/15

    They ask us for the dalt of these interference

                Δₓ = y3 -y2′          

               Δₓ = 21/15 – 0.6

               Δₓ = 0.8

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