{"id":292,"date":"2021-06-02T22:20:22","date_gmt":"2021-06-02T20:20:22","guid":{"rendered":"https:\/\/www.astroartu.studio\/en\/?p=292"},"modified":"2021-06-02T23:09:04","modified_gmt":"2021-06-02T21:09:04","slug":"the-lights-mysteries-refraction","status":"publish","type":"post","link":"https:\/\/www.astroartu.studio\/en\/2021\/06\/02\/the-lights-mysteries-refraction\/","title":{"rendered":"The light’s mysteries: refraction"},"content":{"rendered":"\n

Bonjour guys, <\/p>\n\n\n\n

today I want to tell you about a discovery I came across while I was in Versailles. I went to see a particular painting, entitled \u201cChat Angora Blanc\u201d by Jean-Jacques Bachelier, which portrays Madame Brillant. According to my family tree, she could be an illustrious ancestor of mine. Here you can see the painting I am referring to and, in fact, it seems to exist a great similarity between us.<\/p>\n\n\n\n

\"\"
Chat Angora Blanc Jean-Jacques Bachelier, commons.wikimedia.org<\/em><\/figcaption><\/figure>\n\n\n\n

As I was telling you, I had gone to visit the places where my ancestor lived and I was struck by the light of the chandeliers in the hall of mirrors: the light was breaking up into many rays of different colors and while this thought was spinning in my head, I went out to admire the garden.<\/p>\n\n\n\n

Immersed in my thoughts, I almost didn\u2019t end up immersing myself in the Latona fountain, where I caught a glimpse of a rainbow admist the water droplets from the splashes of the fountain. Back home, I decided to resume my studies on optics, a subject I\u2019m very passionate about, and to write this article, the first of a long and interesting series dedicated to the light and its mysteries.<\/p>\n\n\n\n

Here we see the passage of the light beam from material 1 to material 2.<\/p>\n\n\n\n

Since the materials have different optical properties, a variation in the angle of the light beam is obtained.<\/p>\n\n\n\n

\"\"
created by astroartu.studio<\/em><\/figcaption><\/figure><\/div>\n\n\n\n

Where: <\/p>\n\n\n\n

  • n1<\/sub> is the refractive coefficient of the medium 1<\/li>
  • n2<\/sub> is the refractive coefficient of the medium 2<\/li>
  • v1<\/sub>=c<\/sup>\/n1<\/sub>, speed of light in the medium 1\u00a0<\/li>
  • v2<\/sub>=c<\/sup>\/n2<\/sub>, speed of light in the medium 2\u00a0<\/li>
  • c speed of light in vacuum<\/li><\/ul>\n\n\n\n

     <\/p>\n\n\n\n\n\n\n\n

    From Fermat’s principle, we know that light tends to travels the shortest path. <\/p>\n\n\n\n

    Let\u2019s write the total time it takes for the ray of light to cross the space:<\/p>\n\n\n\n

    <\/p>\t\n

    \n T =
    \n\n\t
    \n \u221a<\/span>\n \n
    \n
    x2<\/sup>+y1<\/sub>2<\/sup><\/div>\n\n
    <\/div>\n\n <\/div>\n <\/div>\n\t
    v1<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div> +\n\n
    \n\n\t
    \n \u221a<\/span>\n \n
    \n
    y2<\/sub>2<\/sup>+(h-x)2<\/sup><\/div>\n\n
    <\/div>\n\n <\/div>\n <\/div>\n\t
    v2<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div> \n<\/div>\n

    <\/p>\n\n\n\n

    We set the derivative of T to zero, in order to minimize the time it takes for the ray to travel through space:<\/p>\n\n\n\n

    <\/p>\t\n

    \n
    \n\n\t
    dT<\/div>\n\t
    dx<\/div>\n\t
    <\/div>\n\n\n <\/div> =\n\n\n\n
    \n\n\t
    x<\/div>\n\t
    v1<\/sub>\n\n \u221a<\/span>\n
    \n
    x2<\/sup>+y1<\/sub>2<\/sup><\/div>\n\n
    <\/div>\n\n <\/div>\n\n\n <\/div>\n\t
    <\/div>\n\n\n <\/div> +\n\n
    \n\n\t
    -(h-x)<\/div>\n\t
    v2<\/sub>\n\n \u221a<\/span>\n
    \n
    y2<\/sub>2<\/sup>+(h-x)2<\/sup><\/div>\n\n
    <\/div>\n\n <\/div>\n\n\n <\/div>\n\t
    <\/div>\n\n\n <\/div>=0\n\n<\/div>\n

    <\/p>\n\n\n\n

    <\/p>\n\n\n\n

    being by definition: <\/p>\n\n\n\n

    <\/p>\t\n

    \n
    \n\n
    x<\/div>\n\t
    \n\n \u221a<\/span>\n
    \n
    x2<\/sup>+y1<\/sub>2<\/sup><\/div>\n\n
    <\/div>\n\n <\/div>\n\n\n <\/div>\n\t
    <\/div>\n\n\n <\/div> = sin(\u03b81<\/sub>)\n<\/div>\n

    <\/p>\n\n\n\n

    and:<\/p>\n\n\n\n

    <\/p>\t\n

    \n
    \n\n
    h-x<\/div>\n\t
    \n\n \u221a<\/span>\n
    \n
    y2<\/sub>2<\/sup>+(h-x)2<\/sup><\/div>\n\n
    <\/div>\n\n <\/div>\n\n\n <\/div>\n\t
    <\/div>\n\n\n <\/div> = sin(\u03b82<\/sub>)\n<\/div>\n

    <\/p>\n\n\n\n

    Replacing in the previous equation we have: <\/p>\n\n\n\n

    <\/p>\t\n

    \n
    \n\n\t
    dT<\/div>\n\t
    dx<\/div>\n\t
    <\/div>\n\n\n <\/div> =\n\n
    \n\n\t
    sin(\u03b81<\/sub>)<\/div>\n\t
    v1<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div> –\n\n
    \n\n\t
    sin(\u03b82<\/sub>)<\/div>\n\t
    v2<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div> = 0\n\n\n<\/div>\n

    <\/p>\n\n\n\n

    We have then:<\/p>\n\n\n\n

    <\/p>\t\n

    \n\n\n
    \n\n\t
    sin(\u03b81<\/sub>)<\/div>\n\t
    v1<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div> =\n\n
    \n\n\t
    sin(\u03b82<\/sub>)<\/div>\n\t
    v2<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div>\n\n\n<\/div>\n

    <\/p>\n\n\n\n

    replacing v1<\/sub> and v2<\/sub> with the definitions: <\/p>\n\n\n\n

    <\/p>\t\n

    \n\n\n
    \n\n\t
    n1<\/sub>sin(\u03b81<\/sub>)<\/div>\n\t
    c<\/div>\n\t
    <\/div>\n\n\n <\/div> =\n\n
    \n\n\t
    n2<\/sub>sin(\u03b82<\/sub>)<\/div>\n\t
    c<\/div>\n\t
    <\/div>\n\n\n <\/div>\n\n\n<\/div>\n

    <\/p>\n\n\n\n

    and, finally, simplifying with respect to the speed of light c we have Snell’s law, which defines the angle of refraction due to the passage of light from one material to another:<\/p>\n\n\n\n

    n1<\/sub>sin(\u03b81<\/sub>)=n2<\/sub>sin(\u03b82<\/sub>)<\/p>\n\n\n\n

    We\u2019ll therefore have that the new angle of the light will be: <\/p>\n\n\n\n

    <\/p>\t\n

    \n\n\n \u03b82<\/sub>=arcsin(
    \n\n\t
    n1<\/sub><\/div>\n\t
    n2<\/sub><\/div>\n\t
    <\/div>\n\n\n <\/div> sin(\u03b81<\/sub>) )\n\n\n\n\n<\/div>\n

    <\/p>\n\n\n\n

    But this doesn\u2019t explain why light breaks down into different colors: in fact a bit of physics is missing.<\/p>\n\n\n\n

    The refractive index depends on the frequency<\/span>, this is the phenomenon that leads the light to break down into different colors, that run through the material at different angles.<\/p>\n\n\n\n

    Here the formula that explain this:<\/p>\n\n\n\n

    n=n(\u03bb)<\/p>\n\n\n\n

    where \u03bb is the wavelength of the electromagnetic radiation of the color considered.<\/p>\n\n\n\n

    Here I report the intervals of the wavelengths of the different colors: <\/p>\n\n\n\n

    • red: ~700\u2013630 nm<\/li>
    • orange: ~630\u2013590 nm <\/li>
    • yellow: ~ 590\u2013560 nm <\/li>
    • green: ~ 560\u2013490 nm <\/li>
    • blue: ~ 490\u2013450 nm <\/li>
    • purple: ~ 450\u2013400 nm<\/li><\/ul>\n\n\n\n

       
      The refractive coefficient can be approximated using the Cauchy equation:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n n( \u03bb )= A +
      \n\n\t
      B<\/div>\n\t
      \u03bb2<\/sup><\/div>\n\t
      <\/div>\n\n\n <\/div>\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      note: in the Cauchy equation, the wavelength is expressed in \u03bcm (micro-meters, where 1 \u03bcm = 1000 nm). <\/p>\n\n\n\n

      For our calculations, I considered a glass with a large refractive index used to build prisms, that is flint glass SF10, with values of the parameters of the Cauchy equation of:<\/p>\n\n\n\n

      A=1.7280 
      B=0.01342<\/p>\n\n\n\n

      I show below the application of this law for the case of red and blue light.<\/p>\n\n\n\n

      1. Let\u2019s calculate the refractive index for red light, taking an intermediate wavelength of 660 nm. We\u2019ll have:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n n( 0.66 \u03bcm)= 1.7280 +
      \n\n\t
      0.01342<\/div>\n\t
      0.662<\/sup><\/div>\n\t
      <\/div>\n\n\n <\/div> = 1.75881\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      2. Let\u2019s calculate the refractive index for blue light, taking an intermediate wavelength of 470 nm. We\u2019ll have:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n n( 0.47 \u03bcm)= 1.7280 +
      \n\n\t
      0.01342<\/div>\n\t
      0.472<\/sup><\/div>\n\t
      <\/div>\n\n\n <\/div> = 1.78875\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      Now let’s consider the geometry of the prism: <\/p>\n\n\n\n

      \"\"
      created by astroartu.studio<\/em><\/figcaption><\/figure><\/div>\n\n\n\n

      The ray of light hits the prism at point A with an angle \u03b1 = 30 \u00b0; for the geometry of the prism (which is an equilateral triangle), this corresponds to the angle \u03b81<\/sub> = 60 \u00b0 (angle with respect to the tangent of the prism surface).<\/p>\n\n\n\n

      We then calculate \u03b82<\/sub> for the red light:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n \u03b82,r<\/sub>= arcsin(
      \n\n\t
      1<\/div>\n\t
      1.75881<\/div>\n\t
      <\/div>\n\n\n <\/div> sin(60\u00b0) ) = 29.4980\u00b0\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      And for the blue color:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n \u03b82,b<\/sub>= arcsin(
      \n\n\t
      1<\/div>\n\t
      1.78875<\/div>\n\t
      <\/div>\n\n\n <\/div> sin(60\u00b0) ) = 28.9569\u00b0\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      Respect to the horizontal plane in a clockwise direction we have:<\/p>\n\n\n\n

      \u03b2r<\/sub> = 30-\u03b82,r<\/sub> = 0.502\u00b0<\/p>\n\n\n\n

      \u03b2b<\/sub> = 30-\u03b82,b<\/sub> = 1.0431\u00b0<\/p>\n\n\n\n

      Now let’s consider the point B where the light beam exits the prism:<\/p>\n\n\n\n

      \"\"
      created by astroartu.studio<\/em><\/figcaption><\/figure><\/div>\n\n\n\n

      For the geometry of the prism, the angle of incidence \u03b8’2<\/sub> is equal to \u03b2 + 30, so we have:<\/p>\n\n\n\n

      \u03b8’2,r<\/sub> = \u03b2r<\/sub>+30 = 30.502\u00b0<\/p>\n\n\n\n

      \u03b8’2,b<\/sub> = \u03b2b<\/sub>+30 = 31.0431\u00b0<\/p>\n\n\n\n

      Now, let’s calculate the angle \u03b8’1<\/sub> of the ray exiting the prism at point B for the red light:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n \u03b8’1,r<\/sub>= arcsin(
      \n\n\t
      1.75881<\/div>\n\t
      1<\/div>\n\t
      <\/div>\n\n\n <\/div> sin(30.502\u00b0) ) = 63.2166\u00b0\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      And for the blue color:<\/p>\n\n\n\n

      <\/p>\t\n

      \n\n\n \u03b8’1,b<\/sub>= arcsin(
      \n\n\t
      1.78875<\/div>\n\t
      1<\/div>\n\t
      <\/div>\n\n\n <\/div> sin(31.0431\u00b0) ) = 67.2836\u00b0\n\n\n\n\n<\/div>\n

      <\/p>\n\n\n\n

      Taking all in function of the angle \u03b4 we have:<\/p>\n\n\n\n

      \u03b4r<\/sub>=90-\u03b8’2,r<\/sub>= 26.7834\u00b0<\/p>\n\n\n\n

      \u03b4b<\/sub>=90-\u03b8’2,b<\/sub>= 22.7164\u00b0<\/p>\n\n\n\n

      We therefore have that the angle between the red light and the blue light is 4.067 \u00b0 -> the light is broken down into the various colors that make it up by the prism!<\/p>\n\n\n\n

      Below we have a fun interactive representation of refraction in a prism, with which you can experience refraction for yourself:<\/p>\n\n\n\n

      <\/p>\n\n\n\n

      \n<\/canvas>\n<\/div>\n