13.3.4 | before revision #18639 | Savchenko Solutions

Statement

13.3.4. [Insert the problem statement]

Solution

For a thin len,
$Missing argument for \frac\frac{1}{s}+\frac{1}{s'} = \frac{1/f}$\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]