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

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

Step-by-step 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(\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 (1000mm)/(1m)` ⇒
`2000mm`

`\lambda = 633nm \cdot (1mm)/(1x10^(6)nm)` ⇒
`6.33x10^(-4)mm`

Finally, equation 1 can be used:

`\Lambda x = (2000mm)((6.33x10^(-4)mm))/((0.100mm))`

`\Lambda x = 12.66mm`

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