Question

In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?

Answer 1

Answer:

The wavelength is  $\lambda = 419 \ nm$

Explanation:

From the question we are told that

   The  distance of separation is   $d = 3.34 *10^{-5} \ m$

   The  distance of the screen is  $D = 3.30 \ m$

      The  order of the fringe is  n =  7

     The distance of separation of  fringes is y =  29.0 cm = 0.29 m

   

Generally the wavelength of the light is mathematically represented as

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

substituting values

         $\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}$

        $\lambda = 4.19*10^{-7}\ m$

        $\lambda = 419 \ nm$