Mystic Darkness

Consider a point $P$ on the curved surface at an angle $\theta$ to the horizontal. Since at the first interface the light undergoes normal incidence, it will suffer no deviation. As a result the light will strike the curved surface horizontally and hence at $P$ also. Using simple geometry, we can observe and obtain that such a ray will strike at an angle $\theta$ to the normal at $P$ . Using Snell's law, we obtain the equation $\mu_{glass} sin(\theta) = \mu_{air} sin(r)$ , where $r$ is the angle of refraction in air. Clearly, as $\theta$ increases, $r$ will also increase. Also, the distance from the center of the cylinder (and indirectly from where distance $x$ is measured) where the ray strikes after refraction strikes the surface (on which the cylinder rests) decreases as $\theta$ increases.

As $\theta$ goes on increasing, we would obtain a value such that $r = 90^{\circ}$ , i.e., critical angle for the given pair of optical media. For points having $\theta$ greater than this value, the light rays would suffer Total Internal Reflection and hence will not come outside the cylinder. Hence, we would obtain a dark patch.

At $r = 90^{\circ}, \theta = i_{c} \Rightarrow \mu_{glass} \sin(i_{c}) = 1.\sin(90^{\circ})$ .

$\displaystyle \frac{2}{\sqrt{3}} \sin(i_{c}) = 1 \Rightarrow \sin(i_{c}) = \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow i_{c} = 60^{\circ}$

Clearly, at the point where this specific refracted ray will strike the ground will be the point to the left of which the dark region would exist. By elementary geometry, we get,

$R + x = R \sec(i_{c}) \Rightarrow x = R\sec(60^{\circ}) - R = 2R - R = R$

$\boxed{ x= 5 cm }$