Analysis of Light Propagation

We have an isotropic point source located at a distance $r$ from the center of a homogeneous sphere of radius $R$ and refractive index $\mu$. For all light emitted by the source to exit the sphere, no Total Internal Reflection (TIR) must occur at the spherical boundary.

The condition for light to escape is that the angle of incidence $i$ at the surface must be less than the critical angle $C$ for all possible rays emitted by the source.

$$ i < C \implies \sin i < \sin C $$

Since the sphere is in air (implied), $\sin C = \frac{1}{\mu}$. Thus, the condition becomes:

$$ \sin i < \frac{1}{\mu} \quad \text{for all rays.} $$

Geometry of Maximum Incidence

Consider a triangle formed by the center of the sphere $O$, the source $S$, and a point of incidence $P$ on the surface.

• Side $OS = r$
• Side $OP = R$
• Angle $\angle OPS = i$ (angle of incidence)
• Angle $\angle OSP = \alpha$ (angle of emission relative to radius)

Using the Sine Rule in $\triangle OSP$:

$$ \frac{R}{\sin \alpha} = \frac{r}{\sin i} $$ $$ \sin i = \frac{r}{R} \sin \alpha $$

The angle of incidence $i$ is maximum when $\sin \alpha$ is maximum. Since the source is isotropic, it emits rays in all directions, so the maximum value of $\sin \alpha$ is 1 (when the ray is emitted perpendicular to the radius vector).

$$ (\sin i)_{\text{max}} = \frac{r}{R} $$

Conclusion

To ensure no light is trapped inside the sphere, the maximum angle of incidence must be less than the critical angle:

$$ (\sin i)_{\text{max}} < \sin C $$ $$ \frac{r}{R} < \frac{1}{\mu} $$ $$ \mu < \frac{R}{r} $$