Edits to “Statement”, “Solution”, “Answer”
en/13.3.4.md
+3 −3
| @@ -1,15 +1,15 @@ | |||
| ### Statement | |||
| − | $13.3.4.$ | ||
| + | $13.3.4.$ At what distance from the lens is the object located, if the distance between the object and its actual image is minimal? The focal length of the lens is $f$. | ||
| ### Solution | |||
| For a thin len,\ | |||
| − | $\frac{1}{s}+\frac{1}{s'} = \frac{1 | ||
| + | $\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$\ | ||
| as the distance between the object and its actual image is minimal, $s = s'$, | |||
| $\frac{2}{s} = \frac{1}{f}$\ | |||
| $s = 2f$ | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $s = 2f$ | ||
| @@ -1,15 +1,15 @@ | |||
| ### Statement | ### Statement | ||
| $13.3.4.$ |
$13.3.4.$ At what distance from the lens is the object located, if the distance between the object and its actual image is minimal? The focal length of the lens is $f$. | ||
| ### Solution | ### Solution | ||
| For a thin len,\ | For a thin len,\ | ||
| $\frac{1}{s}+\frac{1}{s'} = \frac{1 |
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$\ | ||
| as the distance between the object and its actual image is minimal, $s = s'$, | as the distance between the object and its actual image is minimal, $s = s'$, | ||
| $\frac{2}{s} = \frac{1}{f}$\ | $\frac{2}{s} = \frac{1}{f}$\ | ||
| $s = 2f$ | $s = 2f$ | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $s = 2f$ | ||