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

Answer:

The wavelength is  $\lambda_R = 649 *10^{-9}\ m$

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

       $y = \frac{D * \lambda }{d}$

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

         $\frac{y_r}{\lambda_r} = \frac{D}{d}$

=>      $\frac{y_r}{\lambda_r} = \frac{y_R}{\lambda_R}$

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

  So

           $\frac{6.0 *10^{-3}}{ 632.8*10^{-9}} = \frac{ 6.15 *10^{-9}}{\lambda_R}$

=>       $\lambda_R = 649 *10^{-9}\ m$