Edits to “Statement”, “Answer”
en/13.1.12.md
+6 −2
| @@ -1,6 +1,10 @@ | |||
| ### Statement | |||
| − | $13.1.12.$ | ||
| + | $13.1.12.$ | ||
| + | From the base of a hollow cone of height h with a small angle at the top, a | ||
| + | small ring was cut off and placed in a parallel beam of light, with a wide part | ||
| + | in the direction of the beam. At what distance from the ring will the reflected | ||
| + | light rays focus? | ||
| ### Solution | |||
| The ring is cut from the base of a hollow cone of height h and semi-angle$ \alpha \ll 1$. Its radius is approximately $R = h \tan\alpha \approx h\alpha. $ | |||
| When placed with the wide part facing the beam, light hits the inner conical surface parallel to the axis. | |||
| Each ray arrives with an angle of incidence$ \alpha$ with respect to the normal to the surface. Reflection deflects the ray by an angle$ 2\alpha $towards the axis. The distance f from the point of incidence to the focus on the axis satisfies$ R = f \tan 2\alpha$ | |||
| Therefore, | |||
| $f = \frac{R}{\tan 2\alpha} = \frac{h\tan\alpha}{\tan 2\alpha} = \frac{h}{2}\bigl(1 - \tan^2\alpha\bigr)$ | |||
| Since the angle at the vertex is small, $\tan^2\alpha \ll 1$, and the focus is located at a distance | |||
| $\boxed{\dfrac{h}{2}}$ | |||
| from the plane of the ring. | |||
| @@ -21,4 +25,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\boxed{\dfrac{h}{2}}$ | ||
| @@ -1,6 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $13.1.12.$ |
$13.1.12.$ | ||
| From the base of a hollow cone of height h with a small angle at the top, a | |||
| small ring was cut off and placed in a parallel beam of light, with a wide part | |||
| in the direction of the beam. At what distance from the ring will the reflected | |||
| light rays focus? | |||
| ### Solution | ### Solution | ||
| The ring is cut from the base of a hollow cone of height h and semi-angle$ \alpha \ll 1$. Its radius is approximately $R = h \tan\alpha \approx h\alpha. $ | The ring is cut from the base of a hollow cone of height h and semi-angle$ \alpha \ll 1$. Its radius is approximately $R = h \tan\alpha \approx h\alpha. $ | ||
| When placed with the wide part facing the beam, light hits the inner conical surface parallel to the axis. | When placed with the wide part facing the beam, light hits the inner conical surface parallel to the axis. | ||
| Each ray arrives with an angle of incidence$ \alpha$ with respect to the normal to the surface. Reflection deflects the ray by an angle$ 2\alpha $towards the axis. The distance f from the point of incidence to the focus on the axis satisfies$ R = f \tan 2\alpha$ | Each ray arrives with an angle of incidence$ \alpha$ with respect to the normal to the surface. Reflection deflects the ray by an angle$ 2\alpha $towards the axis. The distance f from the point of incidence to the focus on the axis satisfies$ R = f \tan 2\alpha$ | ||
| Therefore, | Therefore, | ||
| $f = \frac{R}{\tan 2\alpha} = \frac{h\tan\alpha}{\tan 2\alpha} = \frac{h}{2}\bigl(1 - \tan^2\alpha\bigr)$ | $f = \frac{R}{\tan 2\alpha} = \frac{h\tan\alpha}{\tan 2\alpha} = \frac{h}{2}\bigl(1 - \tan^2\alpha\bigr)$ | ||
| Since the angle at the vertex is small, $\tan^2\alpha \ll 1$, and the focus is located at a distance | Since the angle at the vertex is small, $\tan^2\alpha \ll 1$, and the focus is located at a distance | ||
| $\boxed{\dfrac{h}{2}}$ | $\boxed{\dfrac{h}{2}}$ | ||
| from the plane of the ring. | from the plane of the ring. | ||
| @@ -21,4 +25,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\boxed{\dfrac{h}{2}}$ | ||