Question

Consider an area element $dS$ at a distance $r$ from a point P. Let $\hat{r}$ be the unit vector along the outward normal to $dS$. If $\alpha$ is the angle between $\hat{r}$ and $dS$, the element of the solid angle subtended by the area element at P is defined as

• A. $d\Omega =\frac{dS\sin \alpha }{r^2}$
• B. $d\Omega =\frac{dS\tan \alpha }{r^2}$
• C. $d\Omega =\frac{dS\cos \alpha }{r^2}$
• D. $d\Omega =\frac{dS\alpha }{r^2}$

Answer 1

The correct option is C $d\Omega =\frac{dS\cos \alpha }{r^2}$

Solid angle is defined as the component of area normal to radius per unit $r^2$.

$d\Omega =\frac{dS\cos \alpha }{r^2}$

Answer-(C)