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I have an detector of $20\mathrm{cm}\times 10\mathrm{cm}$, how can I calculate the solid angle subtended by the detector if the detector is placed at $30\ \mathrm{cm}$ apart? Because directly I can not use the formula $\frac{area}{distance^2}$. Results obtained from this method are approximation and not exact.

Harish Chandra Rajpoot
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john
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  • Integrate $$\frac{\text{(effective) area}}{\text{distance}^2}$$ over the rectangle. – Travis Willse Apr 21 '15 at 05:04
  • What will be the effective area in the case of rectangle? please suggest – john Apr 21 '15 at 06:43
  • The effective area of an infinitesimal piece of the rectangle varies across the rectangle. In general, the infinitesimal effective area should be something like $\cos \theta ,dA$, where $dA$ is the infinitesimal area and $\theta$ is the (acute) angle between the normal to the rectangle and the displacement vector from the infinitesimal rectangle to the source. – Travis Willse Apr 21 '15 at 07:29
  • If you don't want to set up the integral yourself, you can split the rectangle into two triangles and use the formula given at http://en.wikipedia.org/wiki/Solid_angle#Tetrahedron , in the display equation in the sentence that begins, "An efficient algorithm..." – Travis Willse Apr 21 '15 at 07:31

3 Answers3

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As you have not mentioned the location of the point with respect to the given point hence let's assume that the given point lies on the vertical axis passing through the center of rectangle. The solid angle $(\omega)$ subtended by any rectangle of size $l\times b$ at any point lying on the perpendicular axis at a distance $d$ from the center is given by Standard formula [1] $$\omega=4\sin^{-1}\left(\frac{lb}{\sqrt{(l^2+4d^2)(b^2+4d^2)}}\right)$$ Hence, by substituting $l=20cm$, $b=10cm$ & $d=30cm$ we can easily get the solid angle subtended by rectangle $$\omega=4\sin^{-1}\left(\frac{20\cdot10}{\sqrt{(20^2+4\cdot30^2)(10^2+4\cdot30^2)}}\right)$$$$\omega=4\sin^{-1}\left(\frac{200}{\sqrt{14800000}}\right)$$ $$\omega=4\sin^{-1}\left(\frac{1}{\sqrt{370}}\right)\approx 0.208043883 \ \mathrm{sr}$$ [1]: https://www.academia.edu/8747694

Harish Chandra Rajpoot
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Equations for the solid angle subtended by rectangles, including off-center rectangles, are derived and reported here: https://vixra.org/abs/2001.0603

If the detector is squarely facing the origin:

$$\alpha = \frac{20cm}{2*30cm} = 1/3$$

$$\beta = \frac{10cm}{2*30cm} = 1/6$$

$$\Omega = 4 \arccos \left(\sqrt{\frac{1+ \alpha^2 + \beta^2}{(1+\alpha^2)(1+\beta^2) }}\right) \approx 0.208\, steradian$$ (Identical to Mr. Rajpoot's answer)

Leading order approximation: $$\Omega \approx 4\alpha\beta \approx 0.222\, steradian$$

Next to leading order approximation: $$\Omega \approx 4\alpha\beta - 2\alpha\beta(\alpha^2+\beta^2) \approx 0.207\, steradian$$

Schroeder
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Schroeder's suggestion works, but there is a more elegant solution. Eriksson's formula for a tetrahedron works for any oblique angle, because it projects the triangular base onto a spherical triangle on the unit sphere. Just take your rectangle base and divide along the diagonal, thus dividing the solid angle into two tetrahedra. You need to calculate the solid angle for both of them, they are not equal.

Thad Harroun
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