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| ### Statement |
| ### Statement |
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| $13.1.20.$ [Insert the problem statement] |
| $13.1.20.$ What shape should the reflecting surface be so that it collects all parallel rays at one point, regardless of the beam width? |
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| ### Solution |
| ### Solution |
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| We show that a parabolic surface will collect all light rays parallel to its axis of symmetry to its focal point as follows. Suppose that the equation of the parabola in the figure is $y^2=4cx$, so its focal point is at $F(c,0)$. Let $P(a^2/(4c),a)$ be any point on the parabola. (In the figure, $a$ is negative.) Then the slope of the tangent line (shown in blue) at $P$ is |
| We show that a parabolic surface will collect all light rays parallel to its axis of symmetry to its focal point as follows. Suppose that the equation of the parabola in the figure is $y^2=4cx$, so its focal point is at $F(c,0)$. Let $P(a^2/(4c),a)$ be any point on the parabola. (In the figure, $a$ is negative.) Then the slope of the tangent line (shown in blue) at $P$ is |
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| \[\left.\frac{dy}{dx}\right|_{y=a}=\frac{2c}{a},\] |
| \[\left.\frac{dy}{dx}\right|_{y=a}=\frac{2c}{a},\] |
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| and the slope of the normal line (shown in green) at $P$ is $-a/(2c)$. The slope of $PF$ is |
| and the slope of the normal line (shown in green) at $P$ is $-a/(2c)$. The slope of $PF$ is |
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| \[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2)}=\tan(2\angle APB.)\] |
| \[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2}=\tan(2\cdot\angle APB.)\] |
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| In other words, the red lines obey the law of reflection, so an arbitrary light ray parallel to the axis of symmetry of the parabola is collected at its focal point. |
| Consequently, $\angle APF=2\cdot\angle APB$; that is, the red lines obey the law of reflection. Therefore, an arbitrary light ray parallel to the axis of symmetry of the parabola is collected at its focal point. |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| Parabolic surface |
