Solid angle

The solid angle under which an object is seen from an observation point\(O\) is the ratio between the area of ​​the conical projection of the apparent contour of this object on a sphere centered at \(O\), by the square of the radius of the sphere ( figure 2.1).

This quantity, which is the ratio between a surface and the square of a distance, is expressed in steradians (sr). It represents the extension in space of the notion of angle which is generally defined in a plane. We have

Ω = S d 2 size 12{ %OMEGA = { {S} over {d rSup { size 8{2} } } } } {}
Figure 2.1 : definition of solid angle

If the object is planar and if its transverse dimensions are small compared to its distance from point \(O\), the elementary solid angle will be expressed

d Ω = dS cos θ d 2 size 12{d %OMEGA = { { ital "dS""cos"θ} over {d rSup { size 8{2} } } } } {}

\(dS\) being the real surface of the object and \(\theta\) being the angle between the normal of the object and the direction of observation (figure 2.2). The term \(dscos\text{ }\theta\) is the apparent surface of the object in the direction of obliquity whose factor is \(dscos\text{ }\theta\) .

Figure 2.2 : elementary solid angle

If the object is perceived in the form of a disk of radius \(R(dS=\pi R^{2}) \)whose angular radius \(\alpha\) (half angle at the vertex) is small then the solid angle for this object is given by (figure 2.3)

Ω = πR 2 d 2 ≃ πα 2 size 12{ %OMEGA = { {πR rSup { size 8{2} } } over {d rSup { size 8{2} } } } simeq ital "πα" rSup { size 8{2} } } {}
Figure 2.3 : solid angle for a disk

Let us now consider an elementary solid angle defined by a ring of radius \(R\) whose average angular radius is \(\alpha\) and the angular width \(d\alpha\). The disk of radius \(R\) has a surface \(S=\pi R^{2} \)and for a variation \(dR\) of the radius, the surface of the crown \(dS=2\pi RdR\) (figure 2.3). The solid angle is therefore

d Ω = dS cos α d 2 = 2 π RdR cos ( α ) d 2 = 2 π R d dR d cos ( α ) size 12{d %OMEGA = { { ital "dS""cos"α} over {d rSup { size 8{2} } } } = { {2π ital "RdR""cos" left (α right )} over {d rSup { size 8{2} } } } =2π { {R} over {d} } { { ital "dR"} over {d} } "cos" left (α right )} {}

as \(\text{tan} (\alpha )=\frac{R}{d}\) and \(d\alpha =\frac{dR}{d}\), he comes :

d Ω = 2 π sin ( α ) dα size 12{d %OMEGA =2π"sin" left (α right )dα} {}
Figure 2.4 : solid angle for a crown

For a cone of revolution with half angle at the vertex \(\alpha_{M}\), the solid angle is given by

Ω = ∫ 0 α M d Ω = ∫ 0 α M 2 π sin ( α ) dα = 2 π ( 1 − cos α M ) size 12{ %OMEGA = Int rSub { size 8{``0} } rSup { size 8{``α rSub { size 6{M} } } } {d %OMEGA } = Int rSub {``0} rSup {``α rSub { size 6{M} } } {2π"sin" left (α right )dα} size 12{ {}=2π left (1 - "cos"α rSub {M} right )}} {}
Figure 2.5 : solid angle for a cone of revolution

The solid angle corresponding to a half space is given by

Ω half − space = ∫ 0 π / 2 d Ω = 2 π sr size 12{ %OMEGA rSub { size 8{ ital "half" - ital "space"} } = Int rSub { size 8{`0} } rSup { size 8{`π/2} } {d %OMEGA } =2π`"sr"} {}

and for the entire space we have

Ω space = ∫ 0 π d Ω = 4 π sr size 12{ %OMEGA rSub { size 8{ ital "space"} } = Int rSub { size 8{`0} } rSup { size 8{`π} } {d %OMEGA } =4π`"sr"} {}