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

Question

Two lasers are shining on a double slit, with slit separationd.Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patternson a screen a distance 5.00 {\rm m} away from the slits.
Part A
Which laser has its first maximum closer to the centralmaximum?
Part B
What is the distance Deltay_max-max between the first maxima (on the same side of thecentral maximum) of the two patterns?
Part B.1
Find the location of the first maximum for laser1
Part B.2
Find the location of the first maximum for laser2Part C
What is the distance Deltay_max-min between the second maximum of laser 1 and the thirdminimum of laser 2, on the same side of the central maximum?

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niczorrrr 5 years 2021-08-29T18:00:17+00:00 1 Answers 24 views 0

Answers ( )

  1. thachthao
    0
    2021-08-29T18:01:20+00:00
    Reply

    Answer:

    a) laser 1 has the maximum closest to the central maximum

    b) y₂ –y₁ = L 1.66 10⁻²

    Explanation:

    a), B1, B2)  The expression that describes the constructive interference for a double slit is

                d sin θ = m λ

    The pattern is observed on a screen

                tan θ = y / L

    Since the angles are very small

               tan θ = sin θ / cos θ = sin θ = y/L

     

               d y / L = m λ  

    In this case the laser has a wavelength

               λ
    ₁ = d/20

    We substitute

                d y / L = m d / 20

                m = 1

                y₁ = L / 20

    For the laser 2 λ
    ₂= d / 15

                 y₂ = L / 15

    When examining the two expressions, laser 1 has the maximum closest to the central maximum

    b) the difference between the two patterns is

                y₂- y₁ = L (1/15 – 1/20)

               y₂ –y₁ = L 1.66 10⁻²

    C) laser 1 second maximum

              y₁ ’= 2 L / 20

              y₁ ’= L 0.1

    Laser 2 third minimum

    To have a minimum, the equation must be satisfied

               d sin θ = (m + ½) λ  

               d y / L = (m + ½) λ  

               d y / L = (m + ½) d / 15

               y = L (m +1/2) / 15

               m = 3

               y₂’= L (3 + ½) / 15

               y₂’= L 0.2333

    The difference is

               y₁ ’- y₂’ = L (0.1 – 0.2333)

              y₁ ’–y₂’ = L (-0.133)

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