- #1
Jam132
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Hey all!,
I'm working on a project in which I have to design the shape of a mirror to focus X-rays. This is basically what I've done so far so far.
I've found that if you use a parabola mirror it will focus parallal rays, but not off-axis ones. If you then add a second mirror in the form of a hyperbola mirror it will also focus paralellel rays but not off-axis rays. This is the problem. From my research I know that in practice scientists use a design develped by Hans Wolter. What he did was stick a parabola onto a hyperbola mirror, but I can't work out how he got rid of the blurring. I've came up with an idea but I'm not sure if its correct, I was wondering if anyone here who knows anything about this subject can help me out.
Heres my method for getting rid of the blur:
I have devived an equation giving the point at which rays cross the focal plane:
y(c) = (Pheta)x(a)[x(b) - L] / x(b)
Where L equals the focus point of the two mirros combined, Pheta is the angle by which the ray if off axis, x(a) is the point the ray hits the parabola mirror, and x(b) is the point the ray hits the hyperbola, y(c) is where the ray crosses the focal plane.
A ray hitting the front of the parabola mirror will focus at the point:
x(a) = x(1) = x(2) + (delta)x(1) = 2L + (delta)x(1)
So, y(c) = (pheta)L[1 + (delta)x(1)/2L - (delta)x(2)/2L] eqn (1)
Where, (delta)x(1) = length of the parabola mirror, (delta)x(2) = length of the hyperbola mirror. We have also set the point at which the two mirrors join equal to 2L.
If the rays are to focus equation(1) must equal zero:
y(c) = (Pheta)L[ 1 + ... - ...] = 0
For this to happen either Pheta must equal zero, in other words, the rays must be parallel, which I've already proven. Or, the sum of everything inside the square brackets must equal zero:
So,
1+ (delta)x(1)/2L - (delta)x(2)/2L = 0
The only way this can happen is if:
(delta)x(1)/2L - (delta)x(2)/2L = -1
When you rearrange this equation you find that:
(delta)x(1) = (delta)x(2) - x(2) and (delta)x(2) = (delta)x(1) + x(2)
So, (delta)x(2) - (delta)x(1) = x(2)
Where (delta)x(1), is the length of the parabola mirror, and (delta)x(2) is the length of the hyperbola mirror and x(2) is the point at which the two mirrors join.
Therefore, if we set the mirrors lengths so that the diffence in the length between the two mirrors is equal to the point at which the two mirrors join the rays should focus. Example - If we set x(2) equal to 2L, then that would mean that the length of the hyperbola mirror must be longer than the parabola mirror by 2L.
This will focus rays hitting the front of the parabola mirror and I asume will work for rays hitting all parts of the mirror, apart form the point at which the two mirrors meet.
For the back of the parabola mirror, the point at which the two mirrors join, x(2):
x(a) = x(b) = x(2) = 2L
Since we have set x(2) = 2L.
So, y(c) = (pheta)L eqn (2) for the back of the parabola mirror.
The only way eqn(2) will equal zero (focus), is if pheta = 0 , in other words only for parallel rays. Off axis rays will cause a blur. To get round this problem you can layer the mirrors like in the XMM Newton, this will collect more X-rays, but can also be used to block rays hitting the point x(2) off axis.
Thanks for your time and reading all of this, hope I made it clear enough.
Thanks James.
I'm working on a project in which I have to design the shape of a mirror to focus X-rays. This is basically what I've done so far so far.
I've found that if you use a parabola mirror it will focus parallal rays, but not off-axis ones. If you then add a second mirror in the form of a hyperbola mirror it will also focus paralellel rays but not off-axis rays. This is the problem. From my research I know that in practice scientists use a design develped by Hans Wolter. What he did was stick a parabola onto a hyperbola mirror, but I can't work out how he got rid of the blurring. I've came up with an idea but I'm not sure if its correct, I was wondering if anyone here who knows anything about this subject can help me out.
Heres my method for getting rid of the blur:
I have devived an equation giving the point at which rays cross the focal plane:
y(c) = (Pheta)x(a)[x(b) - L] / x(b)
Where L equals the focus point of the two mirros combined, Pheta is the angle by which the ray if off axis, x(a) is the point the ray hits the parabola mirror, and x(b) is the point the ray hits the hyperbola, y(c) is where the ray crosses the focal plane.
A ray hitting the front of the parabola mirror will focus at the point:
x(a) = x(1) = x(2) + (delta)x(1) = 2L + (delta)x(1)
So, y(c) = (pheta)L[1 + (delta)x(1)/2L - (delta)x(2)/2L] eqn (1)
Where, (delta)x(1) = length of the parabola mirror, (delta)x(2) = length of the hyperbola mirror. We have also set the point at which the two mirrors join equal to 2L.
If the rays are to focus equation(1) must equal zero:
y(c) = (Pheta)L[ 1 + ... - ...] = 0
For this to happen either Pheta must equal zero, in other words, the rays must be parallel, which I've already proven. Or, the sum of everything inside the square brackets must equal zero:
So,
1+ (delta)x(1)/2L - (delta)x(2)/2L = 0
The only way this can happen is if:
(delta)x(1)/2L - (delta)x(2)/2L = -1
When you rearrange this equation you find that:
(delta)x(1) = (delta)x(2) - x(2) and (delta)x(2) = (delta)x(1) + x(2)
So, (delta)x(2) - (delta)x(1) = x(2)
Where (delta)x(1), is the length of the parabola mirror, and (delta)x(2) is the length of the hyperbola mirror and x(2) is the point at which the two mirrors join.
Therefore, if we set the mirrors lengths so that the diffence in the length between the two mirrors is equal to the point at which the two mirrors join the rays should focus. Example - If we set x(2) equal to 2L, then that would mean that the length of the hyperbola mirror must be longer than the parabola mirror by 2L.
This will focus rays hitting the front of the parabola mirror and I asume will work for rays hitting all parts of the mirror, apart form the point at which the two mirrors meet.
For the back of the parabola mirror, the point at which the two mirrors join, x(2):
x(a) = x(b) = x(2) = 2L
Since we have set x(2) = 2L.
So, y(c) = (pheta)L eqn (2) for the back of the parabola mirror.
The only way eqn(2) will equal zero (focus), is if pheta = 0 , in other words only for parallel rays. Off axis rays will cause a blur. To get round this problem you can layer the mirrors like in the XMM Newton, this will collect more X-rays, but can also be used to block rays hitting the point x(2) off axis.
Thanks for your time and reading all of this, hope I made it clear enough.
Thanks James.