Новое решение
en/13.3.4.md
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| + | ### Statement | ||
| + | |||
| + | $13.3.4.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | For a thin len,\ | ||
| + | $\frac{1}{s}+\frac{1}{s'} = \frac{1/f}$\ | ||
| + | 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] | ||
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| ### Statement | |||
| $13.3.4.$ [Insert the problem statement] | |||
| ### Solution | |||
| For a thin len,\ | |||
| $\frac{1}{s}+\frac{1}{s'} = \frac{1/f}$\ | |||
| 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] | |||