Edits to “Solution”, “Answer”
en/13.4.21.md
+1 −10
| @@ -1,9 +1,3 @@ | |||
| − | ### Statement | ||
| − | |||
| − | $13.4.21.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| ### Problem | |||
| $13.4.21.$ A beetle is photographed at two different scales by placing the camera at distances equal to three and five times the focal length of the lens, respectively. How should the diameter of the lens aperture be changed so that the illumination of the image on the film remains the same in both cases? The diameter of the aperture in both cases is much smaller than the distance to the beetle. | |||
| ### Solution | |||
| Since the aperture diameter is much smaller than the distance to the beetle, the beetle can be considered a point source, with light reaching the lens in nearly parallel rays. | |||
| The larger the area of the aperture, the more light will reach the film from the beetle, and thus the greater the illumination of the film. That is, \(E \sim d^2\), where \(d\) is the aperture diameter. On the other hand, the illumination decreases with the square of the distance from the lens to the image. For a point source emitting flux \(\Phi\), in an isotropic, homogeneous medium, the illumination on a sphere of radius \(b\) is: | |||
| $$E = \frac{\Phi}{S} = \frac{\Phi}{4\pi b^2} \sim \frac{1}{b^2}$$ | |||
| In this problem, \(b\) is the distance from the lens to the image. Thus, the illumination of the film is directly proportional to the square of the aperture diameter and inversely proportional to the square of the image distance: | |||
| $$E \sim \left( \frac{d}{b} \right)^2$$ | |||
| @@ -23,6 +17,7 @@Solution | |||
| According to the problem, the illumination is the same in both shooting conditions: | |||
| $$E_1 = E_2$$ | |||
| + | |||
| $$\left( \frac{d_1}{b_1} \right)^2 = \left( \frac{d_2}{b_2} \right)^2$$ | |||
| Solving for \(d_2\): | |||
| $$d_2 = \frac{b_2}{b_1} d_1 \tag{1}$$ | |||
| To find \(b_1\) and \(b_2\), we use the lens equation: | |||
| $$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{1}{b} = \frac{1}{f} - \frac{1}{a}$$ | |||
| For \(a = a_1 = 3f\): | |||
| $$\frac{1}{b_1} = \frac{1}{f} - \frac{1}{3f} = \frac{2}{3f} \Rightarrow b_1 = \frac{3f}{2}$$ | |||
| For \(a = a_2 = 5f\): | |||
| $$\frac{1}{b_2} = \frac{1}{f} - \frac{1}{5f} = \frac{4}{5f} \Rightarrow b_2 = \frac{5f}{4}$$ | |||
| Substitute these into equation (1): | |||
| $$d_2 = \frac{\frac{5f}{4}}{\frac{3f}{2}} d_1 = \frac{5}{6} d_1$$ | |||
| @@ -48,7 +43,3 @@Solution | |||
| #### Answer | |||
| \(d_2 = \frac{5}{6} d_1\) | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,9 +1,3 @@ | |||
| ### Statement | |||
| $13.4.21.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | ### Problem | ||
| $13.4.21.$ A beetle is photographed at two different scales by placing the camera at distances equal to three and five times the focal length of the lens, respectively. How should the diameter of the lens aperture be changed so that the illumination of the image on the film remains the same in both cases? The diameter of the aperture in both cases is much smaller than the distance to the beetle. | $13.4.21.$ A beetle is photographed at two different scales by placing the camera at distances equal to three and five times the focal length of the lens, respectively. How should the diameter of the lens aperture be changed so that the illumination of the image on the film remains the same in both cases? The diameter of the aperture in both cases is much smaller than the distance to the beetle. | ||
| ### Solution | ### Solution | ||
| Since the aperture diameter is much smaller than the distance to the beetle, the beetle can be considered a point source, with light reaching the lens in nearly parallel rays. | Since the aperture diameter is much smaller than the distance to the beetle, the beetle can be considered a point source, with light reaching the lens in nearly parallel rays. | ||
| The larger the area of the aperture, the more light will reach the film from the beetle, and thus the greater the illumination of the film. That is, \(E \sim d^2\), where \(d\) is the aperture diameter. On the other hand, the illumination decreases with the square of the distance from the lens to the image. For a point source emitting flux \(\Phi\), in an isotropic, homogeneous medium, the illumination on a sphere of radius \(b\) is: | The larger the area of the aperture, the more light will reach the film from the beetle, and thus the greater the illumination of the film. That is, \(E \sim d^2\), where \(d\) is the aperture diameter. On the other hand, the illumination decreases with the square of the distance from the lens to the image. For a point source emitting flux \(\Phi\), in an isotropic, homogeneous medium, the illumination on a sphere of radius \(b\) is: | ||
| $$E = \frac{\Phi}{S} = \frac{\Phi}{4\pi b^2} \sim \frac{1}{b^2}$$ | $$E = \frac{\Phi}{S} = \frac{\Phi}{4\pi b^2} \sim \frac{1}{b^2}$$ | ||
| In this problem, \(b\) is the distance from the lens to the image. Thus, the illumination of the film is directly proportional to the square of the aperture diameter and inversely proportional to the square of the image distance: | In this problem, \(b\) is the distance from the lens to the image. Thus, the illumination of the film is directly proportional to the square of the aperture diameter and inversely proportional to the square of the image distance: | ||
| $$E \sim \left( \frac{d}{b} \right)^2$$ | $$E \sim \left( \frac{d}{b} \right)^2$$ | ||
| @@ -23,6 +17,7 @@Solution | |||
| According to the problem, the illumination is the same in both shooting conditions: | According to the problem, the illumination is the same in both shooting conditions: | ||
| $$E_1 = E_2$$ | $$E_1 = E_2$$ | ||
| $$\left( \frac{d_1}{b_1} \right)^2 = \left( \frac{d_2}{b_2} \right)^2$$ | $$\left( \frac{d_1}{b_1} \right)^2 = \left( \frac{d_2}{b_2} \right)^2$$ | ||
| Solving for \(d_2\): | Solving for \(d_2\): | ||
| $$d_2 = \frac{b_2}{b_1} d_1 \tag{1}$$ | $$d_2 = \frac{b_2}{b_1} d_1 \tag{1}$$ | ||
| To find \(b_1\) and \(b_2\), we use the lens equation: | To find \(b_1\) and \(b_2\), we use the lens equation: | ||
| $$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{1}{b} = \frac{1}{f} - \frac{1}{a}$$ | $$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{1}{b} = \frac{1}{f} - \frac{1}{a}$$ | ||
| For \(a = a_1 = 3f\): | For \(a = a_1 = 3f\): | ||
| $$\frac{1}{b_1} = \frac{1}{f} - \frac{1}{3f} = \frac{2}{3f} \Rightarrow b_1 = \frac{3f}{2}$$ | $$\frac{1}{b_1} = \frac{1}{f} - \frac{1}{3f} = \frac{2}{3f} \Rightarrow b_1 = \frac{3f}{2}$$ | ||
| For \(a = a_2 = 5f\): | For \(a = a_2 = 5f\): | ||
| $$\frac{1}{b_2} = \frac{1}{f} - \frac{1}{5f} = \frac{4}{5f} \Rightarrow b_2 = \frac{5f}{4}$$ | $$\frac{1}{b_2} = \frac{1}{f} - \frac{1}{5f} = \frac{4}{5f} \Rightarrow b_2 = \frac{5f}{4}$$ | ||
| Substitute these into equation (1): | Substitute these into equation (1): | ||
| $$d_2 = \frac{\frac{5f}{4}}{\frac{3f}{2}} d_1 = \frac{5}{6} d_1$$ | $$d_2 = \frac{\frac{5f}{4}}{\frac{3f}{2}} d_1 = \frac{5}{6} d_1$$ | ||
| @@ -48,7 +43,3 @@Solution | |||
| #### Answer | #### Answer | ||
| \(d_2 = \frac{5}{6} d_1\) | \(d_2 = \frac{5}{6} d_1\) | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||