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| ### Statement |
| ### Statement |
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| $13.3.1.$ [Insert the problem statement] |
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| ### Solution |
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| ### Statement |
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| $13.3.1.$ The figures show objects and lens foci to a certain scale. Construct the images of these objects. What is the magnification in each case? |
| $13.3.1.$ The figures show objects and lens foci to a certain scale. Construct the images of these objects. What is the magnification in each case? |
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| ### Solution |
| ### Solution |
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| To analytically verify the magnification $\Gamma$ obtained from the geometrical construction, we can use the thin lens formula and the expression for magnification $\Gamma = \frac{|f|}{|d \pm f|}$, where $d$ is the distance from the object to the lens, and $f$ is the focal length. By determining the ratio of $d$ to $f$ directly from the scale of the given figures, we can calculate the exact values: |
| To analytically verify the magnification $\Gamma$ obtained from the geometrical construction, we can use the thin lens formula and the expression for magnification $\Gamma = \frac{f}{|d \pm f|}$, where $d$ is the distance from the object to the lens, and $f$ is the positive focal length. By determining the ratio of $d$ to $f$ directly from the scale of the given figures, we can calculate the exact values: |
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| 1\. **Figure (a)**: Converging lens. The distance to the object is $d = 3f$. |
| 1\. Figure (a): Converging lens. The distance to the object is $d = 3f$. |
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| $$ |
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| \Gamma = \frac{f}{d - f} = \frac{f}{3f - f} = \frac{1}{2} |
| \Gamma = \frac{f}{d - f} = \frac{f}{3f - f} = \frac{1}{2} |
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| 2\. **Figure (b)**: Converging lens. The object is located between $f$ and $2f$, specifically at $d = \frac{5}{3}f$. |
| 2\. Figure (b): Converging lens. The object is located between $0$ and $f$, specifically at $d = \frac{1}{3}f$. |
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| $$ |
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| \Gamma = \frac{f}{d - f} = \frac{f}{\frac{5}{3}f - f} = \frac{f}{\frac{2}{3}f} = \frac{3}{2} |
| \Gamma = \frac{f}{f-d} = \frac{f}{f-\frac{1}{3}f} = \frac{f}{\frac{2}{3}f} = \frac{3}{2} |
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| 3\. **Figure (c)**: Diverging lens. For a diverging lens, the magnification formula is $\Gamma = \frac{|f|}{d + |f|}$. The distance to the object is $d = 3|f|$. |
| 3\. Figure (c): Diverging lens. For a diverging lens, the magnification formula is $\Gamma = \frac{f}{d + f}$. The distance to the object is $d = 3f$. |
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| $$ |
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| \Gamma = \frac{|f|}{3|f| + |f|} = \frac{1}{4} |
| \Gamma = \frac{f}{3f + f} = \frac{1}{4} |
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| 4\. **Figure (d)**: Diverging lens. The distance to the object is $d = \frac{1}{3}|f|$. |
| 4\. Figure (d): Diverging lens. The distance to the object is $d = \frac{1}{3}f$. |
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| $$ |
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| \Gamma = \frac{|f|}{\frac{1}{3}|f| + |f|} = \frac{|f|}{\frac{4}{3}|f|} = \frac{3}{4} |
| \Gamma = \frac{f}{\frac{1}{3}f + f} = \frac{f}{\frac{4}{3}f} = \frac{3}{4} |
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| #### Answer |
| #### Answer |
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| 1. $\Gamma = 1/2$ |
| 1. $\Gamma = 1/2$ |
| 2. $\Gamma = 3/2$ |
| 2. $\Gamma = 3/2$ |
| 3. $\Gamma = 1/4$ |
| 3. $\Gamma = 1/4$ |
| 4. $\Gamma = 3/4$ |
| 4. $\Gamma = 3/4$ |
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| #### Answer |
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| [Insert a concise answer or boxed result] |
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