Working on your car you spill oil (index of refraction = 1.55) on the ground into a puddle of water (n = 1.33). You notice a rainbow pattern appear across the oil slick. You realize you can calculate where constructive and destructive interference occurs based on the thickness of the oil slick. (Assume that the average wavelength is 524 nm.)

Required:
a. What are the first three thicknesses necessary for constructive interference?
b. What are the first three thicknesses necessary for destructive interference?

Respuesta :

Answer:

For constructive interference

    The first three thicknesses are

                          [tex]t_1 =85nm[/tex]  

                         [tex]t_2 =254nm[/tex]

                        [tex]t_3 =423nm[/tex]  

For destructive interference

    The first three thicknesses are

                         [tex]t_D__{1}} = 169nm[/tex]

                       [tex]t_D__{2}} = 338nm[/tex]

                       [tex]t_D__{3}} = 507nm[/tex]

Explanation:

The diagram for this  question is shown on the first uploaded image

    From the question we are told that

       The refractive index of oil is [tex]n_o = 1.55[/tex]

         The refractive index of  water is  [tex]n_w = 1.33[/tex]

           The average wavelength is [tex]\lambda_a = 524 nm = 524*10^{-9}m[/tex]

For constructive interference the thickness is mathematically represented as

          [tex]t = (m + \frac{1}{2} ) \frac{\lambda }{2 n_w}[/tex]

Where m is the order of the interference with a value from [tex]m = 0,1 , -1 , 2 , -2 , ...[/tex]

For  m = 0

The thickness of the oil slick would be  

          [tex]t_1 = 0 + \frac{1}{2} * \frac{\lambda }{2 * n_w}[/tex]

Substituting value  

           [tex]t_1 = 0 + \frac{1}{2} * \frac{524 *10^{-9}}{2 * 1.55}[/tex]

            [tex]t_1 =85nm[/tex]

 For  m = 1

   The thickness of the oil slick would be  

          [tex]t_2 = 1 + \frac{1}{2} * \frac{\lambda }{2 * n_w}[/tex]

   Substituting value  

           [tex]t_2 = 1 + \frac{1}{2} * \frac{524 *10^{-9}}{2 * 1.55}[/tex]

            [tex]t_2 =254nm[/tex]

 For  m = 2

   The thickness of the oil slick would be  

          [tex]t_3 = 2 + \frac{1}{2} * \frac{\lambda }{2 * n_w}[/tex]

   Substituting value  

           [tex]t_3= 2 + \frac{1}{2} * \frac{524 *10^{-9}}{2 * 1.55}[/tex]

            [tex]t_3 =423nm[/tex]      

For destructive interference the thickness is mathematically represented as

          [tex]t = \frac{m \lambda }{2 n_w}[/tex]

  For  m = 1

        [tex]t_D__{1}} = \frac{\lambda }{2 * n_w}[/tex]

    Substituting value  

          [tex]t_D__{1}} = \frac{524 *10^{-9} }{2 * 1.55}[/tex]

         [tex]t_D__{1}} = 169nm[/tex]

For  m = 2

        [tex]t_D__{2}} = \frac{\lambda }{2 * n_w}[/tex]

    Substituting value  

          [tex]t_D__{2}} = \frac{ 2 * 524 *10^{-9} }{2 * 1.55}[/tex]

         [tex]t_D__{2}} = 338nm[/tex]

For  m = 3

        [tex]t_D__{3}} = \frac{\lambda }{2 * n_w}[/tex]

    Substituting value  

          [tex]t_D__{3}} = \frac{ 3 * 524 *10^{-9} }{2 * 1.55}[/tex]

         [tex]t_D__{3}} = 507nm[/tex]

Ver imagen okpalawalter8