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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 :

Answer:

The maximun distance is  [tex]z_1 = z_2 = 0.0138m[/tex]

Explanation:

    From the question we are told that

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

                                           [tex]\lambda_2 = 450nm(blue) = 450 *10^{-9}m[/tex]

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

        The distance from the screen is [tex]D = 1.53m[/tex]

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

           [tex]z = \frac{m \lambda D}{d}[/tex]

   Where m is  the order of the fringe

For the first wavelength  we have

        [tex]z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}[/tex]

             [tex]z_1=0.00459m_1 m[/tex]

                 [tex]z_1= 4.6*10^{-3}m_1 m ----(1)[/tex]

For the second  wavelength  we have              

        [tex]z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}[/tex]

        [tex]z_2 = 0.003825m_2[/tex]

        [tex]z_2 = 3.825 *10^{-3} m_2 m[/tex]  ----(2)

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

            [tex]zy_1 =z_2[/tex]

         [tex]4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m[/tex]

          [tex]\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}[/tex]

Hence for this equation to be solved

       [tex]m_1 = 3[/tex]

and  [tex]m_2 = 4[/tex]

Substituting this into the  equation

                      [tex]z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}[/tex]

      Hence [tex]z_1 = z_2 = 0.0138m[/tex]

                       

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