If the laser beam is focussed directly at a point then the simplified expression is: $$I_1(i,j) = 8\frac{P}{\pi d^2} e^{-\frac{8}{d^2}\left(x^2(i)+y^2(j)\right)}$$

If the laser is focused at same point but directed at an angle theta then what will be the intensity expression?

I am asked to work on a project related to quadrant detector. In this I have to plot the current variations of the four photodiodes. For finding the current I need the intensity of laser beam reaching the four photodiodes. So in case I focus the laser beam directly at a point then intensity is same in all four quadrants and that is given by expression above. If I focus the laser beam at an angle theta then what will be the expression of intensity. I want to know the intensity expression, so that I can plot it in Matlab.

I tried to replace d in above expression with d/cos(theta) but that is not correct. Next, I replaced x(i)^2+y(j)^2 with (x(i)^2+y(j)^2)*cos(theta)^2 but that is also not correct.

Please help me out with the correct result.

  • $\begingroup$ Thank you. Do you have any idea about the answer to the question. $\endgroup$ – user39869 Jan 5 '19 at 10:41
  • $\begingroup$ yes. just that it is not written properly $\endgroup$ – user39869 Jan 5 '19 at 10:44
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    $\begingroup$ Have you considered applying an orthographic projection onto arbitrary points of this circularly symmetric function, whilst realizing that your plane circles of equal intensity become ellipses of equal intensity in the projection? $\endgroup$ – Marcus Müller Jan 5 '19 at 10:48
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    $\begingroup$ Ah, so there's an answer coming in! $\endgroup$ – Marcus Müller Jan 5 '19 at 11:07
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    $\begingroup$ I'm quick, but I too need time to write. Your pressure just earned me a short coffee break. See you in an hour! $\endgroup$ – Marcus Müller Jan 5 '19 at 11:19


We can apply simple ray optics to this problem, and it gets easy!

What you're describing is an Orthographic Projection (OP).

So, what you'll do is

  • realize circles subject to OP become ellipses
  • take the center of the circles of equal intensity (in your case, $x^2+y^2=\text{const.}$), and project that. It'll be the center of all the ellipses of equal intensity.
  • figure out what the major and minor axes of the ellipses are going to be
  • simply write down the formula for points on ellipses with given axes.

Orthogonal Projection

For the math behind OP itself, I'd like to refer you to the wikipedia link given above. It's really just a single matrix multiplication, if your original (circular) image plane and the projected plane both contain the origin $(x,y,z)=(0,0,0)$; it becomes especially easy if the intersection of these two planes is e.g. the $y$-coordinate axis, which we can assume here; you can calculate the normal vector of the slanted plane simply by multiplying the cosine (or sine, depending on your construction) of the incident angle to unit vector on the $z$ axis, then.

Major axis of the ellipses of equal intensity

For the question of what the major axis is going to be: it's the projection of the diameter of the original circle that's parallel to the plane intersection (source). Luckily for you, you've defined that intersection to be a diameter of the circle! Since all points on the intersection of original and image plane are invariant to the OP, you have to do nothing to find the major axis of any circle of given intensity: it's simply the distance on the $y$-axis of the original!

Minor axis

The minor axis is simply perpendicular. You'll find that the projection is also especially easy, and is simply the "complementary" trigonometric function to what you've used to find the normal of the projection plane.

Overall formula

Now that we know how to convert perpendicular incidence to slanted incidence circles, we just have to figure out how to do the inverse. Luckily, since the projection is such a well-formed matrix, that's just taking the inverse– and that's trivial, given the structure of the matrix!

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  • $\begingroup$ Can you please tell me what will be the final expression. Actually I am having problem in deriving it. I will be highly thankful to you. $\endgroup$ – user39869 Jan 7 '19 at 10:30
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    $\begingroup$ So, what specifically would you need help with? Please edit your question to contain your current efforts and point out where you're stuck! $\endgroup$ – Marcus Müller Jan 7 '19 at 10:35
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    $\begingroup$ well, major axis == intersection of original circle with coordinate axis that is the intersection of original and projection plane. Minor axis == projection of the other coordinate axis $\endgroup$ – Marcus Müller Jan 7 '19 at 10:38
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    $\begingroup$ The projection plane is the plane you project onto. It has angle $\theta$ relative to the original plane, and intersects the same at the center of the dot; the intersection of these planes is either the x or the y axis, as I tried to explain in my answer. $\endgroup$ – Marcus Müller Jan 7 '19 at 10:48
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    $\begingroup$ Then you've got 1 week of progress and attempts that you're not showing us in your question. Please edit your question to explain what exactly you need help with. $\endgroup$ – Marcus Müller Jan 7 '19 at 10:56

I have a suggestion.

Use the geometry in the following image


Since you know theta, and AD is the "waist" of the laser beam, the major axis of the ellipse will be AC.

Using the the radius of the "waist" of the laser beam as the minor axis, and AC as the major axis, you can calculate the area of the ellipse.

Then scale the power using a ratio of the area of "waist" of the laser beam to the area of the ellipse.

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