Solution to Question 7
1. Ray Geometry and TIR Check
Light from the source $S$ enters the top face of the cylinder. Consider the extreme ray that enters the cylinder at the edge of the top face (radius $R$).
Angle of Incidence ($i$): From the geometry, $\tan i = \frac{R}{h}$.
Refraction: Using Snell’s law at the top surface ($n=1.5$):
$$ \sin i = 1.5 \sin r $$
The ray travels into the cylinder. The maximum angle of refraction $r_{max}$ occurs for the grazing ray. Even for grazing incidence ($i=90^\circ$), $\sin r_{max} = 1/1.5$, so $r_{max} = \sin^{-1}(2/3) \approx 41.8^\circ$.
Side Wall Interaction: Rays hitting the vertical side walls have an angle of incidence $\theta = 90^\circ – r$. Since $r < 41.8^\circ$, $\theta > 48.2^\circ$. The critical angle for glass-air is $C = \sin^{-1}(1/1.5) \approx 41.8^\circ$. Since $\theta > C$, Total Internal Reflection (TIR) occurs at the side walls.
This means no light emerges from the curved surface of the cylinder. The cylinder acts as an opaque obstruction for rays trying to exit the sides.
Figure 2: The cylinder prevents light from exiting the sides due to TIR, creating a shadow region between the base and the direct light boundary.
2. Calculation of Unilluminated Area
Since no light exits the sides, the area on the table immediately outside the cylinder base is unilluminated (shadowed). This shadow extends until the “direct” rays from the source, which miss the cylinder entirely, strike the table.
Consider the ray that just grazes the top edge of the cylinder (radius $R$). This ray travels in a straight line to the table. Let $R_{shadow}$ be the radius where this ray hits the tabletop.
Using similar triangles formed by the source height $h$ and the total height $H+h$: $$ \frac{R}{h} = \frac{R_{shadow}}{h + H} $$ $$ R_{shadow} = R \left( \frac{h + H}{h} \right) = R \left( 1 + \frac{H}{h} \right) $$
The unilluminated area is the annular region between the cylinder’s base (radius $R$) and this outer boundary ($R_{shadow}$). $$ A = \pi R_{shadow}^2 – \pi R^2 $$ $$ A = \pi \left[ R^2 \left( 1 + \frac{H}{h} \right)^2 – R^2 \right] $$ $$ A = \pi R^2 \left[ \left( 1 + \frac{H}{h} \right)^2 – 1 \right] $$ $$ A = \pi R^2 \left[ 1 + \frac{2H}{h} + \frac{H^2}{h^2} – 1 \right] $$ $$ A = \pi R^2 \left[ \frac{2H}{h} + \frac{H^2}{h^2} \right] $$ $$ A = \frac{\pi R^2 H (2h + H)}{h^2} $$