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en/13.4.22.md
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| − | ### Statement | ||
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| − | $13.4.22.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| ### Problem | |||
| $13.4.22.$ What exposure time is needed when photographing a drawing with a linear magnification of \(k_1\), if an exposure time of \(t_2\) is used when photographing with magnification \(k_2\)? | |||
| ### Solution | |||
| Linear magnification \(k\) is the ratio of the image size to the object size. For a thin lens, linear magnification can be expressed as the ratio of the image distance \(b\) (from the lens to the image) to the object distance \(a\) (from the object to the lens): | |||
| $$k = \frac{b}{a} \tag{1}$$ | |||
| The lens equation is: | |||
| $$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \tag{2}$$ | |||
| where \(f\) is the focal length of the lens. From equation (1), we find \(a = \frac{b}{k}\). Substituting this into equation (2): | |||
| $$\frac{k}{b} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{k + 1}{b} = \frac{1}{f} \Rightarrow b = f(k + 1)$$ | |||
| The illumination on the film is inversely proportional to the square of the distance from the lens to the film (according to the inverse square law): | |||
| $$E \sim \frac{1}{b^2} = \frac{1}{f^2(k + 1)^2}$$ | |||
| If the focal length \(f\) is fixed, then illumination is inversely proportional to \((k + 1)^2\): | |||
| $$E \sim \frac{1}{(k + 1)^2}$$ | |||
| To maintain the same exposure when illumination decreases, the exposure time must be increased proportionally: | |||
| $$t \sim (k + 1)^2$$ | |||
| So, for two magnifications: | |||
| $$\frac{t_1}{t_2} = \frac{(k_1 + 1)^2}{(k_2 + 1)^2}$$ | |||
| Hence: | |||
| $$t_1 = t_2 \left( \frac{k_1 + 1}{k_2 + 1} \right)^2$$ | |||
| #### Answer | |||
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| $$ | |||
| t_1 = t_2 \left( \dfrac{k_1 + 1}{k_2 + 1} \right)^2 | |||
| $$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,9 +1,3 @@ | |||
| ### Statement | |||
| $13.4.22.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | ### Problem | ||
| $13.4.22.$ What exposure time is needed when photographing a drawing with a linear magnification of \(k_1\), if an exposure time of \(t_2\) is used when photographing with magnification \(k_2\)? | $13.4.22.$ What exposure time is needed when photographing a drawing with a linear magnification of \(k_1\), if an exposure time of \(t_2\) is used when photographing with magnification \(k_2\)? | ||
| ### Solution | ### Solution | ||
| Linear magnification \(k\) is the ratio of the image size to the object size. For a thin lens, linear magnification can be expressed as the ratio of the image distance \(b\) (from the lens to the image) to the object distance \(a\) (from the object to the lens): | Linear magnification \(k\) is the ratio of the image size to the object size. For a thin lens, linear magnification can be expressed as the ratio of the image distance \(b\) (from the lens to the image) to the object distance \(a\) (from the object to the lens): | ||
| $$k = \frac{b}{a} \tag{1}$$ | $$k = \frac{b}{a} \tag{1}$$ | ||
| The lens equation is: | The lens equation is: | ||
| $$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \tag{2}$$ | $$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \tag{2}$$ | ||
| where \(f\) is the focal length of the lens. From equation (1), we find \(a = \frac{b}{k}\). Substituting this into equation (2): | where \(f\) is the focal length of the lens. From equation (1), we find \(a = \frac{b}{k}\). Substituting this into equation (2): | ||
| $$\frac{k}{b} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{k + 1}{b} = \frac{1}{f} \Rightarrow b = f(k + 1)$$ | $$\frac{k}{b} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{k + 1}{b} = \frac{1}{f} \Rightarrow b = f(k + 1)$$ | ||
| The illumination on the film is inversely proportional to the square of the distance from the lens to the film (according to the inverse square law): | The illumination on the film is inversely proportional to the square of the distance from the lens to the film (according to the inverse square law): | ||
| $$E \sim \frac{1}{b^2} = \frac{1}{f^2(k + 1)^2}$$ | $$E \sim \frac{1}{b^2} = \frac{1}{f^2(k + 1)^2}$$ | ||
| If the focal length \(f\) is fixed, then illumination is inversely proportional to \((k + 1)^2\): | If the focal length \(f\) is fixed, then illumination is inversely proportional to \((k + 1)^2\): | ||
| $$E \sim \frac{1}{(k + 1)^2}$$ | $$E \sim \frac{1}{(k + 1)^2}$$ | ||
| To maintain the same exposure when illumination decreases, the exposure time must be increased proportionally: | To maintain the same exposure when illumination decreases, the exposure time must be increased proportionally: | ||
| $$t \sim (k + 1)^2$$ | $$t \sim (k + 1)^2$$ | ||
| So, for two magnifications: | So, for two magnifications: | ||
| $$\frac{t_1}{t_2} = \frac{(k_1 + 1)^2}{(k_2 + 1)^2}$$ | $$\frac{t_1}{t_2} = \frac{(k_1 + 1)^2}{(k_2 + 1)^2}$$ | ||
| Hence: | Hence: | ||
| $$t_1 = t_2 \left( \frac{k_1 + 1}{k_2 + 1} \right)^2$$ | $$t_1 = t_2 \left( \frac{k_1 + 1}{k_2 + 1} \right)^2$$ | ||
| #### Answer | #### Answer | ||
| @@ -47,7 +41,3 @@Answer | |||
| $$ | $$ | ||
| t_1 = t_2 \left( \dfrac{k_1 + 1}{k_2 + 1} \right)^2 | t_1 = t_2 \left( \dfrac{k_1 + 1}{k_2 + 1} \right)^2 | ||
| $$ | $$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||