Question

Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen that is 4 m from the slits. Near the center of the secreen the separation between adjacent maxima is 2 mm. What is the distance between the two slits?

Answer 1

The distance between the two slits is 1.2mm.

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.

If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.

Therefore, d can be isolated from equation 1.

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

Notice that it is necessary to express L and
`\lambda` in units of millimeters.

`L = 4m \cdot (1000mm)/(1m)` ⇒
`4000mm`

`\lambda = 600nm \cdot (1mm)/(1x10^(6)nm)` ⇒
`0.0006mm`

`d = (4000mm)(0.0006mm)/(2mm)`

`d = 1.2mm`

Hence, the distance between the two slits is 1.2mm.