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 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 6.30 mm apart.

Required:
What is the wavelength of light produced by the pointer?

Answer 1

Wavelength =
`\lambda_p = 3.986 * 10^(-6)` m

Step-by-step explanation:

As we know

Fringe width (w) =
`(D*\lambda)/(d)`

where

`\lambda` is the wavelength

D is distance between source and screen

d is the distance between two slits

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

`(D)/(d) = (y_r)/(\lambda_r) = (y_p)/(\lambda_p)\\(y_r)/(\lambda_r) = (y_p)/(\lambda_p)\\\lambda_p = (y_p* \lambda_r)/(y_r) \\\lambda_p =(6.30 * 10^(-3) * 632.8 * 10^(-9))/(6 *10^(-3)) \\\lambda_p = 3.986 * 10^(-6)`m