Question

A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by 0.180 mm. The two overlapping interference patterns, one from each wavelength, are observed on a screen 1.53 m from the slits. What is the minimum distance (in cm) from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light?

Answer 1

Answer:

The maximun distance is  $z_1 = z_2 = 0.0138m$

Explanation:

    From the question we are told that

       The wavelength are  $\lambda _ 1 = 540nm (green) = 540 *10^{-9}m$

                                           $\lambda_2 = 450nm(blue) = 450 *10^{-9}m$

        The distance of seperation of the two slit is $d = 0.180mm = 0.180 *10^{-3}m$

        The distance from the screen is $D = 1.53m$

Generally the distance of the bright fringe to the center of the screen is mathematically represented as

           $z = \frac{m \lambda D}{d}$

   Where m is  the order of the fringe

For the first wavelength  we have

        $z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}$

             $z_1=0.00459m_1 m$

                 $z_1= 4.6*10^{-3}m_1 m ----(1)$

For the second  wavelength  we have              

        $z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}$

        $z_2 = 0.003825m_2$

        $z_2 = 3.825 *10^{-3} m_2 m$  ----(2)

From the question we are told that the two sides coincides with one another so

            $zy_1 =z_2$

         $4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m$

          $\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}$

Hence for this equation to be solved

       $m_1 = 3$

and  $m_2 = 4$

Substituting this into the  equation

                      $z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}$

      Hence $z_1 = z_2 = 0.0138m$