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?

Answer 1

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.