7. A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)

Question

7. A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a) What is the distance between first-order and second-order bright fringes?

in progress 0
Orla Orla 4 years 2021-07-19T05:25:20+00:00 1 Answers 31 views 0

Answers ( )

  1. bonexptip
    0
    2021-07-19T05:26:45+00:00
    Reply

    Answer:

    The distance between first-order and second-order bright fringes is 12.66mm.

    Explanation:

    The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

    \Lambda x = L\frac{\lambda}{d}  (1)

    Where \Lambda x is the distance between two adjacent maxima, L is the distance of the screen from the slits, \lambda is the wavelength and d is the separation between the slits.  

    The values for this particular case are:

    L = 2.0m

    \lambda = 633nm

    d = 0.100mm

    Notice that is necessary to express L and \lambda in units of milimeters.

    L = 2.0m \cdot \frac{1000mm}{1m} ⇒ 2000mm

    \lambda = 633nm \cdot \frac{1mm}{1x10^{6}nm} ⇒ 6.33x10^{-4}mm

    Finally, equation 1 can be used:

    \Lambda x = (2000mm)\frac{(6.33x10^{-4}mm)}{(0.100mm)}

    \Lambda x = 12.66mm

    Hence, the distance between first-order and second-order bright fringes is 12.66mm.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )