Cylindrical mirror

The light rays hit the cylinder at an angle $\alpha = \arcsin x$ . The reflected beam then has an angle $2 \alpha$ to the incident beam. The corresponding point on the wall then gives $\begin{aligned} f(x) = y &= -\cos \alpha - (3 - \sin \alpha) \cot 2 \alpha \\ &= -\cos \alpha - (3 - \sin \alpha) \frac{2 \sin \alpha \cos \alpha}{\cos^2 \alpha - \sin^2 \alpha} \\ &= - \sqrt{1 - x^2} - (3 - x) \frac{2 x \sqrt{1 - x^2}}{(1 - x^2) - x^2} \\ &= \frac{6 x^2 - x - 3}{2 x \sqrt{1 - x^2}} \end{aligned}$ This function has poles for the points $x = 0$ and $x = 1$ , so that the function value diverges here ( $\lim_{x \to 0} f(x) = - \infty$ , $\lim_{x \to 1} f(x) = + \infty$ ) . Physically these cases correspond to reflections with the angles $\alpha = 0^\circ$ and $180^\circ$ , so that the reflected rays run parallel to the wall. In the domain $x \in (0,1)$ the function is well-behaved, so that we can calculate the derivative of the function $f (x)$ results according to the quotient rule: $\begin{aligned} f'(x) &= \frac{(12 x - 1) \cdot (2 x \sqrt{1 - x^2}) - (6 x^2 - x - 3) \cdot (2 \sqrt{1 - x^2} - 2 x^2/\sqrt{1 - x^2})}{(2 x \sqrt{1 - x^2})^2} \\ &= \frac{(12 x^2 - x)(1 - x^2) - (6 x^2 - x - 3) ( 1 - 2 x^2) }{2 x^2 (1 - x^2)^{3/2}}\\ &= \frac{(- 12 x^4 + x^3 + 12 x^2 - x) - (- 12 x^4 + 2 x^3 + 12 x^2 - x - 3) }{2 x^2 (1 - x^2)^{3/2}} \\ &= \frac{3 - x^3}{2 x^2 (1 - x^2)^{3/2}} \end{aligned}$ For $x \in (0,1)$ , the derivative $f'(x)$ is positive, so that the function $f(x)$ is monotone increasing. Therefore, the function can be inverted within the domain, so that each point $y$ can be assigned a unique point $x = f^{-1}(y)$ . However, the inverse function can only be determined graphically or numerically so that we first plot the function and the derivative: For the points $A$ , $B$ , $C$ , $D$ and $E$ , which correspond to the function values $f(x) = -3, -2, -1, 0$ and $1$ , the corresponding x-values and derivatives can be determined numerically: $\begin{aligned} x_A &= f^{-1}(-3) \approx 0.429 & & \Rightarrow & f'(x_A) &\approx 10.77\\ x_B &= f^{-1}(-2) \approx 0.536 & & \Rightarrow & f'(x_B) &\approx 8.23\\ x_C &= f^{-1}(-1) \approx 0.668 & & \Rightarrow & f'(x_C) &\approx 7.35\\ x_D &= f^{-1}(0) \approx 0.795 & & \Rightarrow & f'(x_D) &\approx 8.85\\ x_E &= f^{-1}(1) \approx 0.886 & & \Rightarrow & f'(x_E) &\approx 14.72 \end{aligned}$ It is noticeable, that the point $x_C$ corresponds to the point of lowest slope. The slope $f '(x)$ is a measure of how much the reflected rays diverge. Two rays, each incident at points $x$ and $x + dx$ , strike the wall at points $f (x)$ and $f (x) + f '(x) dx$ , respectively. If the slope $f '(x)$ is high, the rays diverge and the light intensity is small. If the slope $f '(x)$ is small, then the rays are focused and the light intensity is large. Therefore, at point $C$ , the highest light intensity will be observed.