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en/13.3.27.md
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| + | ### Statement | ||
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| + | $13.3.27.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Typically, an object of a telescope is far away, so the first image from the objective is near its focal point. This image serves as the object for the eyepiece and is near its focal point as well so that the final image is at infinity. As a result, the ratio of the angle subtended by the final image and the angle subtended by the original object (i.e. the angular magnification) equals $f/f_e$, where $f_e$ is the focal length of the eyepiece. | ||
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| + | When the object is moved to the distance $a$ from the objective instead, the first image is at the distance $af/(a-f)>f$, so the eyepiece has to be moved back so that this image is near its focal point as well. Thus, the new angular magnification is $(af/(a-f))/f_e$, which is $a/(a-f)$ times the original magnification. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $13.3.27.$ [Insert the problem statement] | |||
| ### Solution | |||
| Typically, an object of a telescope is far away, so the first image from the objective is near its focal point. This image serves as the object for the eyepiece and is near its focal point as well so that the final image is at infinity. As a result, the ratio of the angle subtended by the final image and the angle subtended by the original object (i.e. the angular magnification) equals $f/f_e$, where $f_e$ is the focal length of the eyepiece. | |||
| When the object is moved to the distance $a$ from the objective instead, the first image is at the distance $af/(a-f)>f$, so the eyepiece has to be moved back so that this image is near its focal point as well. Thus, the new angular magnification is $(af/(a-f))/f_e$, which is $a/(a-f)$ times the original magnification. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||