Edits to “Statement”, “Answer”
en/13.3.15.md
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| ### Statement | |||
| − | $13.3.15.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $13.3.15.$ The rim of a magnifying glass bears the inscription "$\times 10$", meaning the magnifying glass increases the angular size of the object being viewed by ten times. Determine its focal length. | |||
| ### Solution | |||
| The commercial magnification of a magnifying glass, indicated on its rim, is standardly calculated for the case when the eye is accommodated at infinity (a relaxed eye). In this case, the angular magnification $\Gamma$ is determined by the formula: | |||
| $$\Gamma = \frac{d_0}{F}$$ | |||
| where $d_0 = 25 \text{ cm} = 0.25 \text{ m}$ is the distance of most distinct vision for a normal eye, and $F$ is the focal length of the magnifying glass. | |||
| Expressing the required focal length $F$ from this formula: | |||
| $$F = \frac{d_0}{\Gamma}$$ | |||
| Substituting the known values ($\Gamma = 10$): | |||
| $$F = \frac{25 \text{ cm}}{10} = 2.5 \text{ cm}$$ | |||
| @@ -22,7 +16,3 @@Solution | |||
| #### Answer | |||
| $F = 2.5 \text{ cm}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $13.3.15.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $13.3.15.$ The rim of a magnifying glass bears the inscription "$\times 10$", meaning the magnifying glass increases the angular size of the object being viewed by ten times. Determine its focal length. | $13.3.15.$ The rim of a magnifying glass bears the inscription "$\times 10$", meaning the magnifying glass increases the angular size of the object being viewed by ten times. Determine its focal length. | ||
| ### Solution | ### Solution | ||
| The commercial magnification of a magnifying glass, indicated on its rim, is standardly calculated for the case when the eye is accommodated at infinity (a relaxed eye). In this case, the angular magnification $\Gamma$ is determined by the formula: | The commercial magnification of a magnifying glass, indicated on its rim, is standardly calculated for the case when the eye is accommodated at infinity (a relaxed eye). In this case, the angular magnification $\Gamma$ is determined by the formula: | ||
| $$\Gamma = \frac{d_0}{F}$$ | $$\Gamma = \frac{d_0}{F}$$ | ||
| where $d_0 = 25 \text{ cm} = 0.25 \text{ m}$ is the distance of most distinct vision for a normal eye, and $F$ is the focal length of the magnifying glass. | where $d_0 = 25 \text{ cm} = 0.25 \text{ m}$ is the distance of most distinct vision for a normal eye, and $F$ is the focal length of the magnifying glass. | ||
| Expressing the required focal length $F$ from this formula: | Expressing the required focal length $F$ from this formula: | ||
| $$F = \frac{d_0}{\Gamma}$$ | $$F = \frac{d_0}{\Gamma}$$ | ||
| Substituting the known values ($\Gamma = 10$): | Substituting the known values ($\Gamma = 10$): | ||
| $$F = \frac{25 \text{ cm}}{10} = 2.5 \text{ cm}$$ | $$F = \frac{25 \text{ cm}}{10} = 2.5 \text{ cm}$$ | ||
| @@ -22,7 +16,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $F = 2.5 \text{ cm}$ | $F = 2.5 \text{ cm}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||