Question

A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright red fringes spaced 6.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 6.19 mm apart. What is the wavelength of light produced by the pointer?

Answer 1

The wavelength is
`\lambda_R = 649 *10^(-9)\ m`

Step-by-step explanation:

From the question we are told that

The wavelength of the red laser is
`\lambda_r = 632.8 \ nm = 632.8 *10^(-9)\ m`

The spacing between the fringe is
`y_r = 6.00\ mm = 6.00*10^(-3) \ m`

The spacing between the fringe for smaller laser point is
`y_R = 6.19 \ mm = 6.19 *10^(-3) \ m`

Generally the spacing between the fringe is mathematically represented as

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

Here
`D` is the distance to the screen

and d is the distance of the slit separation

Now for both laser red light light and small laser point D and d are same for this experiment

So

`(y_r)/(\lambda_r) = (D)/(d)`

=>
`(y_r)/(\lambda_r) = (y_R)/(\lambda_R)`

Where
`\lambda_R` is the wavelength produced by the small laser pointer

So

`(6.0 *10^(-3))/( 632.8*10^(-9)) = ( 6.15 *10^(-9))/(\lambda_R)`

=>
`\lambda_R = 649 *10^(-9)\ m`