# Solid Angle Subtended by a Disk at a Point on its Axis

## Solid Angle Subtended by a Disk at a Point on its Axis

Solid angle is the cone subtended by an area at the point of interest. The magnitude of solid angle subtended by an area S at a point is defined as its center. Therefore, total solid angle around a point in space is the solid angle subtended by entire spherical surface on its center.

$${{\Omega }_{0}}=\frac{4\pi {{R}^{2}}}{{{R}^{2}}}=4\pi \,Steradian$$.

Consider a coaxial area element of radius x and thickness dx.

dS = 2πx dx

Solid angle subtended by this element at point P is:

$$d\Omega =\frac{dS\,\cos \theta }{\left( {{x}^{2}}+{{a}^{2}} \right)}=\frac{2\pi x\,dx\,a}{\left( {{x}^{2}}+{{a}^{2}} \right)\sqrt{\left( {{a}^{2}}+{{x}^{2}} \right)}}$$,

Hence, the total solid angle subtended by the disk is: $$\Omega =\pi a\int\limits_{0}^{R}{\frac{2x\,dx}{{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{3/2}}}}$$,

$$\Omega =2\pi a\left[ -\frac{1}{\sqrt{\left( {{a}^{2}}+{{x}^{2}} \right)}} \right]_{0}^{R}=2\pi \left( 1-\frac{a}{\sqrt{\left( {{a}^{2}}+{{R}^{2}} \right)}} \right)$$,

$$\Omega =2\pi \left( 1-\frac{a}{\sqrt{\left( {{a}^{2}}+{{R}^{2}} \right)}} \right)$$,

∴ Ω = 2π (1 – cosα); Where α is the semi vertical angle of the cone subtended by the disk at P.