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 4.80m away from the slits.

Part A

Which laser has its first maximum closer to the central maximum?

Part B

What is the distance \Delta _y__max-max_ between the first maxima (on the same side of the central maximum) of the two patterns?

Express your answer in meters.

Part C

What is the distance \Delta _y__max-max_ between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Express your answer in meters.

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Thái Dương 4 years 2021-07-19T00:06:05+00:00 1 Answers 455 views 0

Answers ( )

  1. nguyetanh
    0
    2021-07-19T00:07:19+00:00
    Reply

    Answer:

    A) first laser

    B) 0.08m

    C) 0.64m

    Explanation:

    To find the position of the maximum you use the following formula:

    y=\frac{m\lambda D}{d}

    m: order of the maximum

    λ: wavelength

    D: distance to the screen = 4.80m

    d: distance between slits

    A) for the first laser you use:

    y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\

    for the second laser:

    y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m

    hence, the first maximum of the first laser is closer to the central maximum.

    B) The difference between the first maximum:

    \Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm

    hence, the distance between the first maximum is 0.08m

    C) you calculate the second maximum of laser 1:

    y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m

    and for the third minimum of laser 2:

    y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m

    Finally, you take the difference:

    1.12m-0.48m=0.64m

    hence, the distance is 0.64m

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