Question

Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light.

Answer 1

Answer:

The wavelength is  $\lambda = 6.28 *10^{-7}=628 nm$

The frequency is  $f = 4.78 Hz$

Explanation:

From the question we are told that  

      The slit distance is $d = 0.048 \ mm = 4.8 *0^{-5}\ m$

       The distance from the screen is  $D = 6.50 \ m$

       The distance between fringes is  $Y = 8.5 \ cm = 0.085 \ m$

Generally the distance between the fringes for a two slit interference is  mathematically represented as

           $Y = \frac{\lambda * D}{d}$

=>       $\lambda = \frac{Y * d }{D}$

substituting values      

           $\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }$

           $\lambda = 6.27 *10^{-7}=628 nm$

Generally the frequency of the light is mathematically represented as

          $f = \frac{c}{\lambda }$

where  c is  the speed of light with  values  

         $c = 3.0 *10^{8} \ m/s$

substituting values  

      $f = \frac{3.0*10^8}{6.28 *10^{-7}}$

      $f = 4.78 Hz$