Новое решение
en/13.3.17.md
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| + | ### Statement | ||
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| + | $13.3.17.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | The magnification is a result of the planoconvex water lens in front of an observer. Using the refractive index of $4/3$ for water, this lens has the focal length of $f=R/(4/3-1)=3R$, which is behind the aquarium. As a result, the magnification is largest when the fish is closest to the focal point, i.e. at Point $A$ farthest from the observer. The water between the fish and the lens serves only to bring the first image of the fish to Point $B$ with the apparent distance of $u=2R/(4/3)=3R/2$ from the lens. Thus, the magnification by the lens is | ||
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| + | \[m=\frac{f}{u-f}=\frac{3R}{3R/2-3R}=-2.\] | ||
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| + | Note that $v=mu=-3R$, so the (virtual) image of the fish at Point $C$ is in fact where the focal point of the lens is. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $13.3.17.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| The magnification is a result of the planoconvex water lens in front of an observer. Using the refractive index of $4/3$ for water, this lens has the focal length of $f=R/(4/3-1)=3R$, which is behind the aquarium. As a result, the magnification is largest when the fish is closest to the focal point, i.e. at Point $A$ farthest from the observer. The water between the fish and the lens serves only to bring the first image of the fish to Point $B$ with the apparent distance of $u=2R/(4/3)=3R/2$ from the lens. Thus, the magnification by the lens is | |||
| \[m=\frac{f}{u-f}=\frac{3R}{3R/2-3R}=-2.\] | |||
| Note that $v=mu=-3R$, so the (virtual) image of the fish at Point $C$ is in fact where the focal point of the lens is. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||