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

Answer:

Wavelength = $\lambda_p = 3.986 * 10^{-6}$ m

Explanation:

As we know

Fringe width (w) = $\frac{D*\lambda}{d}$

where

$\lambda$ is the wavelength

D is distance between source and screen

d is the distance between two slits

$\frac{D}{d} = \frac{y}{\lambda}$

$\frac{D}{d} = \frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\lambda_p = \frac{y_p* \lambda_r}{y_r} \\\lambda_p =\frac{6.30 * 10^{-3} * 632.8 * 10^{-9}}{6 *10^{-3}} \\\lambda_p = 3.986 * 10^{-6}$m