Answer
See the detailed answer below.
Work Step by Step
From the geometry of the figure below, we can see that angle 1 is equal to angle $\phi$ since the same diagonal line cuts the two parallel lines. And angle 1 is equal to angle 2 since the incident angle is equal to the reflected angle.
Thus, the triangle ACF is an isosceles one. And hence, segments FC and fA are equal.
$$\overline{FC}=\overline{FA}$$
now if we proved that $\overline{CF}= \overline{CA}/2$, then $f=R/2$.
Now we can use the law of cosines, where
$$(\overline{FA})^2 =(\overline{CA})^2 +(\overline{CF})^2 -2(\overline{CA})(\overline{CF})\cos\phi $$
Noting that $R=\overline{CA}=\overline{CB}$,
$f=\overline{BF}=\overline{CB}-\overline{CF}=R-\overline{FC}$, so $\overline{CF}=R-f$
and that, $\overline{CF}=\overline{FA}$
$$0 =R^2 -2R(R-f)\cos\phi $$
So,
$$R^{ \color{red}{\bf\not} 2}=2 \color{red}{\bf\not} R (R-f)\cos\phi $$
$$R=2R\cos\phi -2f\cos\phi$$
Hence,
$$f=\dfrac{2R\cos\phi -R}{2\cos\phi}$$
And when $\phi\lt \lt 1$, then $\cos\phi$ approach 1.
Thus,
$$\boxed{f=\dfrac{R}{2}}$$
Now we know that $\phi$ must be small for some limits to use the law of $f=R/2$
