Question

Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light.

Answer 1

The wavelength is
`\lambda = 6.28 *10^(-7)=628 nm`

The frequency is
`f = 4.78 Hz`

Step-by-step explanation:

From the question we are told that

The slit distance is
`d = 0.048 \ mm = 4.8 *0^(-5)\ m`

The distance from the screen is
`D = 6.50 \ m`

The distance between fringes is
`Y = 8.5 \ cm = 0.085 \ m`

Generally the distance between the fringes for a two slit interference is mathematically represented as

`Y = (\lambda * D)/(d)`

=>
`\lambda = (Y * d )/(D)`

substituting values

`\lambda = (0.085 * 4.8*10^(-5) )/(6.50 )`

`\lambda = 6.27 *10^(-7)=628 nm`

Generally the frequency of the light is mathematically represented as

`f = (c)/(\lambda )`

where c is the speed of light with values

`c = 3.0 *10^(8) \ m/s`

substituting values

`f = (3.0*10^8)/(6.28 *10^(-7))`

`f = 4.78 Hz`