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en/13.3.20.md
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| + | ### Statement | ||
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| + | $13.3.20.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Suppose a light source is placed at the focal point of the optical system at $A$ which is at the distance $d$ from the left lens. The (virtual) image produced by the left lens must be at the focal point of the right lens at $B$ in order for parallel light rays to leave the right lens. Thus, | ||
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| + | \[\frac{1}{f}=\frac{1}{d}-\frac{1}{f-l}\qquad\Rightarrow\qquad d=\frac{f(f-l)}{2f-l}\] | ||
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| + | Consequently, | ||
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| + | \[d=\frac{f(1-l/f)}{2(1-l/2f)}\approx\frac{f}{2}(1-\frac{l}{f})(1+\frac{l}{2f})\approx\frac{f}{2}-\frac{l}{4}.\] | ||
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| + | Therefore, the focal length of the system is $d+l=\frac{f}{2}+\frac{3l}{4}$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $13.3.20.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| Suppose a light source is placed at the focal point of the optical system at $A$ which is at the distance $d$ from the left lens. The (virtual) image produced by the left lens must be at the focal point of the right lens at $B$ in order for parallel light rays to leave the right lens. Thus, | |||
| \[\frac{1}{f}=\frac{1}{d}-\frac{1}{f-l}\qquad\Rightarrow\qquad d=\frac{f(f-l)}{2f-l}\] | |||
| Consequently, | |||
| \[d=\frac{f(1-l/f)}{2(1-l/2f)}\approx\frac{f}{2}(1-\frac{l}{f})(1+\frac{l}{2f})\approx\frac{f}{2}-\frac{l}{4}.\] | |||
| Therefore, the focal length of the system is $d+l=\frac{f}{2}+\frac{3l}{4}$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||