Новое решение
en/13.3.28.md
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| + | ### Statement | ||
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| + | $13.3.28.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | The total magnification of a microscope is the product of the magnification at the objective $L/f_1$, where $L$ is the distance between the objective and the eyepiece, and the magnification at the eyepiece ($25$ cm)$/f_2$, where $25$ cm is the near-point distance of human eyes. Thus, if $L$ is increased by $l$, the total magnification will increase by $25l/(f_1f_2)$ cm. | ||
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| + | If the microscope is enlarged to $k$ times its original size, each of $L$, $f_1$, $f_2$ will be $k$ times its original value (as the radius of curvature of each lens also increases). Consequently, the magnification of the microscope will be $1/k$ times its original value. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $13.3.28.$ [Insert the problem statement] | |||
| ### Solution | |||
| The total magnification of a microscope is the product of the magnification at the objective $L/f_1$, where $L$ is the distance between the objective and the eyepiece, and the magnification at the eyepiece ($25$ cm)$/f_2$, where $25$ cm is the near-point distance of human eyes. Thus, if $L$ is increased by $l$, the total magnification will increase by $25l/(f_1f_2)$ cm. | |||
| If the microscope is enlarged to $k$ times its original size, each of $L$, $f_1$, $f_2$ will be $k$ times its original value (as the radius of curvature of each lens also increases). Consequently, the magnification of the microscope will be $1/k$ times its original value. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||