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 4.90 m away from the slits.

a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

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Khoii Minh 3 years 2021-08-13T12:19:50+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-13T12:21:15+00:00

    Answer:

    a)Δy = 81.7mm

    b)Δy = 32.7cm

    Explanation:

    To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

    Tan \theta = \frac{y}{D}

    where  D  is the separation between the slits and the screen where the interference pattern is observed.

    a) In this case:

    Δy  =  |y1max  (λ1) −  y1max  (λ2)|

    Δy = |\frac{D\lambda _1}{d}  - \frac{D\lambda _2}{d} |

    Δy = D |\frac{d/20}{d} - \frac{d/15}{d}  |

    Δy = D |\frac{1}{20} - \frac{1}{15}  |

    Δy = 4.90 |\frac{1}{20}- \frac{1}{15}  |

    Δy = 81.7mm

    The separation between these maxima is 81.7 mm

    b)

    Δy  =  |y₂max  (λ1) −  y₂max  (λ2)|

    Δy = D|\frac{2(d/20)}{d} - \frac{5(d/15)}{2d} |

    Δy = 4.90|\frac{1}{10} - \frac{1}{6} |

    Δy = 32.7cm

    The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

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