Edit to “Solution”
en/13.1.16.md
+2 −2
| ### Statement | |||
| $13.1.16.$ A converging conical beam of light rays falls on a concave mirror. At what distance from the focus will the reflected rays intersect, if the mirror radius is $80$ cm, and the continuation of the rays intersects the main optical axis at a distance of $40$ cm from the mirror. | |||
| ### Solution | |||
| @@ -6,12 +6,12 @@Solution | |||
|  | |||
| − | Conversely, we may ask where a light source should be placed in order that its image is formed $40$ cm behind the mirror. Using half the radius of curvature of the mirror as its focal length, we have | ||
| + | Conversely, we may ask where a light source should be placed in order that its image is formed $40$ cm behind the mirror. Using half the radius of curvature of the mirror as its focal length, we have the position of the source | ||
| \[u=\frac{vf}{v-f}=\frac{-40\cdot40}{-40-40}=20.\] | |||
| − | Therefore, the object should be placed $20$ cm from the focal point toward the mirror. | ||
| + | Therefore, the object should be placed $20$ cm from the mirror and hence $20$ cm from the focal point toward the mirror. | ||
| #### Answer | |||
| $20$ cm from the focal point toward the mirror. | |||
| ### Statement | ### Statement | ||
| $13.1.16.$ A converging conical beam of light rays falls on a concave mirror. At what distance from the focus will the reflected rays intersect, if the mirror radius is $80$ cm, and the continuation of the rays intersects the main optical axis at a distance of $40$ cm from the mirror. | $13.1.16.$ A converging conical beam of light rays falls on a concave mirror. At what distance from the focus will the reflected rays intersect, if the mirror radius is $80$ cm, and the continuation of the rays intersects the main optical axis at a distance of $40$ cm from the mirror. | ||
| ### Solution | ### Solution | ||
| @@ -6,12 +6,12 @@Solution | |||
|  |  | ||
| Conversely, we may ask where a light source should be placed in order that its image is formed $40$ cm behind the mirror. Using half the radius of curvature of the mirror as its focal length, we have | Conversely, we may ask where a light source should be placed in order that its image is formed $40$ cm behind the mirror. Using half the radius of curvature of the mirror as its focal length, we have the position of the source | ||
| \[u=\frac{vf}{v-f}=\frac{-40\cdot40}{-40-40}=20.\] | \[u=\frac{vf}{v-f}=\frac{-40\cdot40}{-40-40}=20.\] | ||
| Therefore, the object should be placed $20$ cm from the focal point toward the mirror. | Therefore, the object should be placed $20$ cm from the mirror and hence $20$ cm from the focal point toward the mirror. | ||
| #### Answer | #### Answer | ||
| $20$ cm from the focal point toward the mirror. | $20$ cm from the focal point toward the mirror. | ||