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en/13.1.20.md
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| ### 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? | ||
| + | $13.1.20^*.$ What shape should the reflecting surface be so that it collects all parallel rays at one point, regardless of the beam width? | ||
| ### Solution | |||
|  | |||
| 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 | |||
| \[\left.\frac{dy}{dx}\right|_{y=a}=\frac{2c}{a},\] | |||
| and the slope of the normal line (shown in green) at $P$ is $-a/(2c)$. The slope of $PF$ is | |||
| \[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2}=\tan(2\cdot\angle APB.)\] | |||
| 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. | |||
| #### Answer | |||
| Parabolic surface | |||
| unchanged lines 15 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### 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? | $13.1.20^*.$ What shape should the reflecting surface be so that it collects all parallel rays at one point, regardless of the beam width? | ||
| ### Solution | ### Solution | ||
|  |  | ||
| 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 | ||
| \[\left.\frac{dy}{dx}\right|_{y=a}=\frac{2c}{a},\] | \[\left.\frac{dy}{dx}\right|_{y=a}=\frac{2c}{a},\] | ||
| 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 | ||
| \[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2}=\tan(2\cdot\angle APB.)\] | \[\tan\angle APF=\frac{a}{a^2/(4c)-c}=\frac{2(-a/(2c))}{1-(-a/(2c))^2}=\tan(2\cdot\angle APB.)\] | ||
| 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. | 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. | ||
| #### Answer | #### Answer | ||
| Parabolic surface | Parabolic surface | ||
| unchanged lines 15 | |||