Question

Monochromatic light is incident on a pair of slits that are separated by 0.230 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.) (a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.57 cm, find the wavelength of the incident light

Answer 1

The wavelength of incident light is
`1.38x10^(-6)m`

Step-by-step explanation:

The physicist Thomas Young established, through his double slit experiment, a relation 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 = 2.60m`

`d = 0.230mm`

`\Lambda x = 1.57cm`

Then,
`\lambda` can be isolated from equation 1

`\lambda = (d \Lambda x)/(L)` (2)

However, before equation 2 can be used, it is necessary to express
`\Lambda x` and d in units of meters.

`\Lambda x= 1.57cm \cdot (1m)/(100cm)` ⇒
`0.0157m`

`d = 0.230mm \cdot (1m)/(1000mm)` ⇒
`2.3x10^(-4)m`

Finally, equation 2 can be used.

`\lambda = ((2.3x10^(-4)m)(0.0157m)/((2.60m))`

`\lambda = 1.38x10^(-6)m`

Hence, the wavelength of incident light is
`1.38x10^(-6)m`