double-slit-experimentinterferenceopticsvisible-light
In a double slit experiment, if monochromatic light is replaced by white light what effect does it have on the fringes.
Answer given is below. Can someone please explain the answer.?
White light consists of many wavelengths. Hence, the interference pattern using white light appears different than that for monochromatic light. At the center point, all the waves travel the same length and hence no path difference is produced at the center point. Thus at the center point we get the maxima of all wavelengths and we obtain the maximum for white light at the center point. Slightly above or below the center point the fringes will be coloured. Watch this video for visualising this: http://www.youtube.com/watch?v=GEqwRBjHdwc
I will give you the mathematical details at the last, first just visualize what is going on. I am assuming that slits are like pinholes because if you took into account that they are thin rectangles then fringes won't be exactly hyperbolic and it will be mixed problem of interference and diffraction.
In all the diagrams and animations, blue axis is Z, red is X and green is Y.
This is how a fringe looks on a screen (Pink) perpendicular to the y-axis(green) when two slits A and B are placed along the z-axis(Blue).
Now a lot of people explain this by saying that: "A fringe is obtained when path difference of interfering waves is constant. So, the locus of that point from which the difference between the distances from A and B to that point is constant is a Hyperbola. Hence, the fringes are Hyperbolic." This is true if we are looking, for the locus of those points, on a plane which passes through A & B. As the gif below shows different planes passing through A and B contain those hyperbolas:
But, that's not true, in general, if you are looking for the points in a plane which doesn't pass through points A and B. For example, if you look for fringes in a plane that is perpendicular to the line joining A and B, then the fringes will be CIRCULAR.
Because, it is true that for each plane passing through A and B locus of those points is a hyperbola, and whole 3D space can be thought as made up of infinite planes passing through points A & B, the locus in 3D space will be all those hyperbolas combined to form a hyperboloid; as shown below:
Now if we want to find locus on any plane we can just take the intersection of this hyperboloid with the plane.
Mathematical Details: I will first prove that the locus is hyperboloid and then will take the intersections to find fringe shapes on desired planes. Let A be $(0,0,a)$ and B be $(0,0,-a)$. Coordinates are in the order (x,y,z). Locus of points which have a constant path difference is: $\sqrt{(z+a)^2+(x^2)+(y^2)}-\sqrt{(z-a)^2+(x^2)+(y^2)} = d $
$\sqrt{(z+a)^2+(x^2)+(y^2)} =\sqrt{(z-a)^2+(x^2)+(y^2)} + d $
Squaring both sides:
${(z+a)^2+(x^2)+(y^2)} ={(z-a)^2+(x^2)+(y^2)} + 2\sqrt{(z-a)^2+(x^2)+(y^2)}d + d^2 $
$4za = 2\sqrt{(z-a)^2+(x^2)+(y^2)}d + d^2$
$(2za - \frac{d^2}{2})^2 = ((z-a)^2+(x^2)+(y^2)) d^2 $
Re-arrange to get:
$z^2 - \frac{x^2 + y^2}{\frac{4a^2}{d^2} -1} = \frac{d^2}{4}$
Let $\frac{4a^2}{d^2} -1=k^2$
So you get: $z^2 - \frac{x^2 + y^2}{k^2} = \frac{d^2}{4}$
Now, $x^2 + y^2 = {\rho}^2$
,where $\rho$ is the distance of any point from z-axis or if you know cylindrical coordinates, it is the cylindrical radius. Our equation now becomes: $z^2 - \frac{{\rho}^2}{k^2} = \frac{d^2}{4}$
So, it is a hyperboloid.
If the screen is perpendicular to y-axis(green axis), the equation of the plane would be $y = h$. Substitute $y=h$ in the equation of hyperboloid, you will get an equation of hyperbola (in x and z):
$z^2 - \frac{x^2}{k^2} = \frac{d^2}{4} - \frac{h^2}{k^2} $
It is shown below as as the black curve of intersection:
If the screen is perpendicular to line joining A and B (or perpendicular to z-axis) then substitute $z = h$ in the equation of the hyperboloid and you will get a circle in x and y, giving you the circular fringes $x^2 + y^2 = (h^2 - \frac{d^2}{4})k^2$ as shown below as the black curve of intersection:
Best Answer
I would describe the colors closest to the white central maximum as yellow (absence of blue) and magenta (absence of green). There is no "farthest" fringe, but next one sees a band of cyan (absence of red). These are complementary colors of the primary ones.
It has to do with the pigments in our eyes. Cameras and displays try to match these with their RGB channels. Here is an image I made, using a camera without a lens, using a $20\ \mu$m slit in the lens cap. I also decomposed the slit image in the separate color channels: