| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine the magnification of the image formed by a convex mirror, we need to use the mirror formula and the magnification formula. The mirror formula for a convex mirror is given by: 1/f = 1/v - 1/u where f is the focal length of the mirror, v is the image distance, and u is the object distance. The magnification formula is given by: m = -v/u where m is the magnification of the image. First, we need to find the focal length of the convex mirror. The focal length of a convex mirror is given by: f = R/2 where R is the radius of curvature of the mirror. Substituting the given value of R, we get: f = 45.0 cm / 2 = 22.5 cm Next, we need to find the image distance, v. We can use the mirror formula to do this. Substituting the values of f and u, we get: 1/22.5 = 1/v - 1/(-10) Simplifying this equation, we get: v = -15 cm Note that the negative sign indicates that the image is virtual and behind the mirror. Finally, we can find the magnification of the image using the magnification formula. Substituting the values of v and u, we get: m = -(-15 cm) / (-10 cm) = 1.5 Therefore, the magnification of the image formed by the convex mirror is 1.5. This means that the image is 1.5 times larger than the object and is upright (since the magnification is positive). |
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| deepseekr1 | To determine the magnification of the image formed by a convex mirror, follow these steps: 1. **Calculate the focal length (f):** The radius of curvature (R) is 45.0 cm. For a convex mirror, \[ f = -\frac{R}{2} = -\frac{45.0}{2} = -22.5 \, \text{cm}. \] 2. **Apply the mirror equation:** The mirror equation is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, \] where \(d_o = 10 \, \text{cm}\) (object distance). Solving for \(d_i\) (image distance): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-22.5} - \frac{1}{10}. \] Convert to fractions: \[ \frac{1}{d_i} = -\frac{2}{45} - \frac{1}{10} = -\frac{4}{90} - \frac{9}{90} = -\frac{13}{90}. \] Thus, \[ d_i = -\frac{90}{13} \approx -6.923 \, \text{cm}. \] 3. **Calculate magnification (m):** Magnification is given by: \[ m = -\frac{d_i}{d_o}. \] Substituting \(d_i = -\frac{90}{13} \, \text{cm}\) and \(d_o = 10 \, \text{cm}\): \[ m = -\left(\frac{-90/13}{10}\right) = \frac{9}{13} \approx 0.692. \] **Final Answer:** The magnification of the image is \(\boxed{0.692}\). |
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| gpto1 | To determine the magnification of the image formed by the convex mirror, we'll use the mirror equation and the magnification formula. Let's proceed step by step. **Given:** - Object height, \( h_o = +5.0 \) cm (positive since it's upright). - Object distance, \( s = +10.0 \) cm (positive because the object is in front of the mirror). - Radius of curvature of the mirror, \( R = -45.0 \) cm (negative for a convex mirror, since the center of curvature is behind the mirror). **1. Calculate the focal length (\( f \)) of the convex mirror:** For a spherical mirror, the focal length is half the radius of curvature: \[ f = \frac{R}{2} \] Since it's a convex mirror: \[ f = \frac{-45.0\ \text{cm}}{2} = -22.5\ \text{cm} \] **2. Use the mirror equation to find the image distance (\( s' \)):** The mirror equation relates the object distance (\( s \)), the image distance (\( s' \)), and the focal length (\( f \)): \[ \frac{1}{f} = \frac{1}{s} + \frac{1}{s'} \] Rearranged to solve for \( \frac{1}{s'} \): \[ \frac{1}{s'} = \frac{1}{f} - \frac{1}{s} \] Plugging in the known values: \[ \frac{1}{s'} = \frac{1}{-22.5\ \text{cm}} - \frac{1}{+10.0\ \text{cm}} = \left( -\frac{1}{22.5} - \frac{1}{10} \right) \ \text{cm}^{-1} \] Compute the values: \[ \frac{1}{s'} = \left( -0.04444 - 0.1 \right) \ \text{cm}^{-1} = -0.14444\ \text{cm}^{-1} \] Thus: \[ s' = \frac{1}{-0.14444\ \text{cm}^{-1}} \approx -6.923\ \text{cm} \] The negative sign indicates that the image is virtual and located behind the mirror. **3. Calculate the magnification (\( m \)):** The magnification is given by: \[ m = -\frac{s'}{s} \] Plug in the values: \[ m = -\left( \frac{-6.923\ \text{cm}}{+10.0\ \text{cm}} \right) = +0.6923 \] **Interpretation:** - The positive magnification indicates that the image is upright relative to the object. - Since \( |m| < 1 \), the image is smaller than the object. **Answer:** The magnification of the image is approximately **+0.692**. |
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| GPT-4o | To determine the magnification of the image formed by a convex mirror, we need to use the mirror formula and the magnification formula. Here's the step-by-step solution: ### Step 1: Understand the given values - Height of the object (h_o): 5 cm - Distance of the object from the mirror (u): -10 cm (Since the object is in front of the mirror, the distance is taken as negative in mirror calculations) - Radius of curvature (R): 45 cm ### Step 2: Determine the focal length (f) The focal length (f) of a convex mirror is given by: \[ f = \frac{R}{2} \] Since the mirror is convex, the focal length is taken as positive: \[ f = \frac{45}{2} = 22.5 \text{ cm} \] ### Step 3: Use the mirror formula to find the image distance (v) The mirror formula is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Rearrange to solve for \( \frac{1}{v} \): \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] Substitute the values of \( f \) and \( u \): \[ \frac{1}{v} = \frac{1}{22.5} - \frac{1}{-10} \] \[ \frac{1}{v} = \frac{1}{22.5} + \frac{1}{10} \] Convert to a common denominator to simplify: \[ \frac{1}{v} = \frac{10 + 22.5}{225} \] \[ \frac{1}{v} = \frac{32.5}{225} \] Calculate \( v \): \[ v = \frac{225}{32.5} \] \[ v \approx 6.92 \text{ cm} \] Since \( v \) is positive, the image is formed behind the mirror. ### Step 4: Calculate the magnification (m) The magnification \( m \) is given by: \[ m = \frac{h_i}{h_o} = -\frac{v}{u} \] Substitute the values of \( v \) and \( u \): \[ m = -\frac{6.92}{-10} \] \[ m = 0.692 \] ### Conclusion The magnification of the image is approximately 0.692. This means the image is smaller than the object and is upright (as indicated by the positive magnification value for a convex mirror). |
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| llama3.1-70B-Instruct | A classic optics problem! To determine the magnification of the image, we need to use the mirror equation and the magnification equation. The mirror equation is: 1/f = 1/do + 1/di where: f = focal length of the mirror do = object distance (distance between the object and the mirror) di = image distance (distance between the mirror and the image) The magnification equation is: m = -di/do where: m = magnification of the image First, we need to find the focal length of the mirror. Since it's a convex mirror, the focal length is negative and equal to half the radius of curvature: f = -R/2 = -45.0 cm / 2 = -22.5 cm Now, we can plug in the values we know into the mirror equation: 1/(-22.5 cm) = 1/10 cm + 1/di Simplifying and solving for di, we get: di = -15 cm Now, we can find the magnification: m = -di/do = -(-15 cm) / 10 cm = 1.5 So, the magnification of the image is 1.5, which means the image is 1.5 times larger than the object. |
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| llama3.1-405B-Instruct | To determine the magnification of the image formed by a convex mirror, we can use the mirror equation and the magnification formula. The mirror equation is given by: 1/f = 1/do + 1/di where f is the focal length of the mirror, do is the object distance, and di is the image distance. The magnification formula is given by: m = -di/do First, we need to find the focal length of the mirror. The focal length of a convex mirror is half the radius of curvature, so: f = R/2 = 45.0 cm / 2 = 22.5 cm Now, we can use the mirror equation to find the image distance: 1/f = 1/do + 1/di 1/22.5 cm = 1/10 cm + 1/di Solving for di, we get: di = -15 cm The negative sign indicates that the image is virtual and behind the mirror. Now, we can find the magnification: m = -di/do = -(-15 cm) / 10 cm = 1.5 The magnification is 1.5, which means that the image is 1.5 times larger than the object. Since the magnification is positive, the image is upright. |
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| DeepSeek-Coder-V2 | To determine the magnification of the image formed by a convex mirror, we need to follow these steps: 1. **Identify the given values:** - Height of the object, \( h_o = 5 \) cm - Distance of the object from the mirror, \( d_o = 10 \) cm - Radius of curvature of the mirror, \( R = 45.0 \) cm 2. **Calculate the focal length of the mirror:** The focal length \( f \) of a convex mirror is half of its radius of curvature. \[ f = \frac{R}{2} = \frac{45.0 \text{ cm}}{2} = 22.5 \text{ cm} \] 3. **Use the mirror formula to find the image distance \( d_i \):** The mirror formula is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Substituting the known values: \[ \frac{1}{22.5} = \frac{1}{10} + \frac{1}{d_i} \] Solving for \( \frac{1}{d_i} \): \[ \frac{1}{d_i} = \frac{1}{22.5} - \frac{1}{10} \] To subtract these fractions, find a common denominator: \[ \frac{1}{22.5} = \frac{10}{225} = \frac{2}{45} \] \[ \frac{1}{10} = \frac{4.5}{45} \] \[ \frac{1}{d_i} = \frac{2}{45} - \frac{4.5}{45} = \frac{2 - 4.5}{45} = \frac{-2.5}{45} = -\frac{1}{18} \] Therefore: \[ d_i = -18 \text{ cm} \] 4. **Calculate the magnification \( m \):** The magnification formula is: \[ m = -\frac{d_i}{d_o} \] Substituting the values: \[ m = -\frac{-18 \text{ cm}}{10 \text{ cm}} = \frac{18}{10} = 1.8 \] Thus, the magnification of the image is: \[ \boxed{1.8} \] |
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| Qwen2-72B-Instruct | To determine the magnification of the image formed by a convex mirror, we need to use the mirror formula and the magnification formula. The mirror formula for a convex mirror is given by: 1/f = 1/v - 1/u where f is the focal length of the mirror, v is the image distance, and u is the object distance. The magnification formula is given by: m = -v/u where m is the magnification of the image. First, we need to find the focal length of the convex mirror. The focal length of a convex mirror is given by: f = R/2 where R is the radius of curvature of the mirror. Substituting the given value of R, we get: f = 45.0 cm / 2 = 22.5 cm Next, we need to find the image distance, v. We can use the mirror formula to do this. Substituting the values of f and u, we get: 1/22.5 = 1/v - 1/(-10) Simplifying this equation, we get: v = -15 cm Note that the negative sign indicates that the image is virtual and behind the mirror. Finally, we can find the magnification of the image using the magnification formula. Substituting the values of v and u, we get: m = -(-15 cm) / (-10 cm) = 1.5 Therefore, the magnification of the image formed by the convex mirror is 1.5. This means that the image is 1.5 times larger than the object and is upright (since the magnification is positive). |
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| baichuan4 | To determine the magnification of the image, we need to use the formula for magnification (m) in a convex mirror, which is: m = -h' / h where: - h is the height of the object (5 cm) - h' is the height of the image, which we need to calculate. First, we need to find the position of the image (p') using the mirror equation: 1/f = 1/p + 1/p' where: - f is the focal length of the mirror, which is half the radius of curvature (45.0 cm / 2 = 22.5 cm) - p is the object distance (10 cm) Substituting the values: 1/22.5 = 1/10 + 1/p' Solving for p': 1/p' = 1/22.5 - 1/10 1/p' = 2/45 - 4.5/45 1/p' = -2.5/45 p' = -45/2.5 p' = -18 cm The negative sign indicates that the image is virtual (as expected for a convex mirror). Now, we can use the formula for calculating the height of the image (h') in a convex mirror: h' = -m * h But we need to find m first. Since the image is upright in a convex mirror, the magnification (m) is also given by: m = p' / p Substituting the values: m = -18 / 10 m = -1.8 Now, we can calculate h': h' = -(-1.8) * 5 h' = 9 cm So, the magnification of the image is -1.8, and the height of the image is 9 cm. |
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| doubao | For a convex mirror, the focal length (f) is half the radius of curvature, so f = 45/2 = 22.5 cm. Using the mirror equation 1/f = 1/u + 1/v (where u is the object distance and v is the image distance), we can find the image distance. 1/22.5 = 1/10 + 1/v 1/v = 1/22.5 - 1/10 1/v = (2 - 4.5)/45 1/v = -2.5/45 v = -18 cm. The negative sign indicates that the image is virtual and behind the mirror. The magnification (m) is given by -v/u, so m = -(-18)/10 = 1.8. The magnification of the image is 1.8. |
1.8 |
| minimax | To solve this problem, we will use the mirror equation and the magnification formula for mirrors. The mirror equation relates the object distance (do), the image distance (di), and the focal length (f) of a mirror: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] The focal length (f) of a convex mirror is half the radius of curvature (R), so: \[f = \frac{R}{2}\] Given that the radius of curvature (R) is 45.0 cm, the focal length (f) is: \[f = \frac{45.0 \text{ cm}}{2} = 22.5 \text{ cm}\] Now, we can plug in the values for the object distance (do = 10 cm) and the focal length (f = 22.5 cm) into the mirror equation to find the image distance (di): \[\frac{1}{22.5 \text{ cm}} = \frac{1}{10 \text{ cm}} + \frac{1}{d_i}\] Solving for di: \[\frac{1}{d_i} = \frac{1}{22.5 \text{ cm}} - \frac{1}{10 \text{ cm}}\] \[\frac{1}{d_i} = \frac{10 - 22.5}{225 \text{ cm}^2}\] \[\frac{1}{d_i} = \frac{-12.5}{225 \text{ cm}^2}\] \[\frac{1}{d_i} = -\frac{1}{18 \text{ cm}}\] \[d_i = -18 \text{ cm}\] The negative sign indicates that the image is virtual and on the same side as the object. Next, we calculate the magnification (m) using the magnification formula: \[m = -\frac{d_i}{d_o}\] Plugging in the values: \[m = -\frac{-18 \text{ cm}}{10 \text{ cm}}\] \[m = \frac{18}{10}\] \[m = 1.8\] The magnification of the image is 1.8. This means the image is 1.8 times larger than the object, and since the magnification is positive, the image is upright and virtual. |
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| kimi | To determine the magnification of the image in a convex mirror, we can use the mirror formula and the magnification formula. The mirror formula is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Where: - \( f \) is the focal length of the mirror - \( d_o \) is the object distance from the mirror - \( d_i \) is the image distance from the mirror For a convex mirror, the radius of curvature \( R \) is positive, and the focal length \( f \) is half of the radius of curvature: \[ f = \frac{R}{2} = \frac{45.0 \text{ cm}}{2} = 22.5 \text{ cm} \] Given: - \( d_o = 10 \text{ cm} \) (object distance) - \( R = 45.0 \text{ cm} \) (radius of curvature) Now, we can find the image distance \( d_i \) using the mirror formula: \[ \frac{1}{22.5} = \frac{1}{10} + \frac{1}{d_i} \] \[ \frac{1}{d_i} = \frac{1}{22.5} - \frac{1}{10} \] \[ \frac{1}{d_i} = \frac{10 - 22.5}{10 \times 22.5} \] \[ \frac{1}{d_i} = \frac{-12.5}{225} \] \[ d_i = -\frac{225}{12.5} \] \[ d_i = -18 \text{ cm} \] The negative sign indicates that the image is virtual and erect. The magnification \( m \) is given by: \[ m = -\frac{d_i}{d_o} \] \[ m = -\frac{-18 \text{ cm}}{10 \text{ cm}} \] \[ m = 1.8 \] So, the magnification of the image is \( 1.8 \), which means the image is 1.8 times the size of the object and erect. |
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| wenxinyiyan | To solve this problem, we will use the mirror equation and the magnification equation for convex mirrors. The mirror equation relates the object distance (do), the image distance (di), and the focal length (f) of the mirror: 1/do + 1/di = 1/f. The magnification (m) is given by the ratio of the image height (hi) to the object height (ho): m = hi/ho = -di/do. For a convex mirror, the focal length (f) is positive and is related to the radius of curvature (R) by f = R/2. Given that the radius of curvature is 45.0 cm, the focal length is f = 45.0 cm / 2 = 22.5 cm. The object is placed at a distance do = 10 cm in front of the mirror. Using the mirror equation, we can solve for the image distance (di): 1/do + 1/di = 1/f 1/10 + 1/di = 1/22.5 Solving for di, we get: di = -45.0 cm The negative sign indicates that the image is virtual, as expected for a convex mirror. Now, we can calculate the magnification (m) using the magnification equation: m = hi/ho = -di/do m = -(-45.0 cm) / 10 cm m = 4.5 Therefore, the magnification of the image is 4.5. This means that the image appears to be 4.5 times taller than the object. Since the image is virtual, it is located behind the mirror and is upright. |
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| zhipu | To determine the magnification of the image formed by a convex mirror, we can use the mirror equation and the magnification equation. Let's go through the steps: ### Step 1: Gather Known Values - Object height (\( h_o \)) = 5 cm - Object distance (\( d_o \)) = -10 cm (negative because the object is in front of the mirror) - Radius of curvature (\( R \)) = 45.0 cm ### Step 2: Find the Focal Length The focal length (\( f \)) of a convex mirror is half the radius of curvature, and it is negative for a convex mirror: \[ f = \frac{R}{2} = \frac{45.0 \, \text{cm}}{2} = -22.5 \, \text{cm} \] ### Step 3: Use the Mirror Equation The mirror equation is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where \( d_i \) is the image distance. Rearrange to solve for \( d_i \): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \] Substitute the known values: \[ \frac{1}{d_i} = \frac{1}{-22.5} - \frac{1}{-10} \] \[ \frac{1}{d_i} = -\frac{1}{22.5} + \frac{1}{10} \] \[ \frac{1}{d_i} = -\frac{1}{22.5} + \frac{2.25}{22.5} \] \[ \frac{1}{d_i} = \frac{2.25 - 1}{22.5} \] \[ \frac{1}{d_i} = \frac{1.25}{22.5} \] \[ d_i = \frac{22.5}{1.25} \] \[ d_i = 18 \, \text{cm} \] Since it's a convex mirror, the image distance \( d_i \) is positive, indicating that the image is virtual and located behind the mirror. ### Step 4: Calculate Magnification The magnification (\( m \)) is given by: \[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \] Substitute the known values: \[ m = -\frac{18}{-10} \] \[ m = \frac{18}{10} \] \[ m = 1.8 \] ### Conclusion The magnification of the image is 1.8. This means the image is 1.8 times smaller than the object, and since the magnification is positive, the image is upright. So, the magnification of the image formed by the convex mirror is **1.8**. |
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| gpto1 | To determine the magnification of the image formed by the convex mirror, we'll use the mirror equation and the magnification formula. Let's proceed step by step. **Given:** - Object height, \( h_o = +5.0 \) cm (positive since it's upright). - Object distance, \( s = +10.0 \) cm (positive because the object is in front of the mirror). - Radius of curvature of the mirror, \( R = -45.0 \) cm (negative for a convex mirror, since the center of curvature is behind the mirror). **1. Calculate the focal length (\( f \)) of the convex mirror:** For a spherical mirror, the focal length is half the radius of curvature: \[ f = \frac{R}{2} \] Since it's a convex mirror: \[ f = \frac{-45.0\ \text{cm}}{2} = -22.5\ \text{cm} \] **2. Use the mirror equation to find the image distance (\( s' \)):** The mirror equation relates the object distance (\( s \)), the image distance (\( s' \)), and the focal length (\( f \)): \[ \frac{1}{f} = \frac{1}{s} + \frac{1}{s'} \] Rearranged to solve for \( \frac{1}{s'} \): \[ \frac{1}{s'} = \frac{1}{f} - \frac{1}{s} \] Plugging in the known values: \[ \frac{1}{s'} = \frac{1}{-22.5\ \text{cm}} - \frac{1}{+10.0\ \text{cm}} = \left( -\frac{1}{22.5} - \frac{1}{10} \right) \ \text{cm}^{-1} \] Compute the values: \[ \frac{1}{s'} = \left( -0.04444 - 0.1 \right) \ \text{cm}^{-1} = -0.14444\ \text{cm}^{-1} \] Thus: \[ s' = \frac{1}{-0.14444\ \text{cm}^{-1}} \approx -6.923\ \text{cm} \] The negative sign indicates that the image is virtual and located behind the mirror. **3. Calculate the magnification (\( m \)):** The magnification is given by: \[ m = -\frac{s'}{s} \] Plug in the values: \[ m = -\left( \frac{-6.923\ \text{cm}}{+10.0\ \text{cm}} \right) = +0.6923 \] **Interpretation:** - The positive magnification indicates that the image is upright relative to the object. - Since \( |m| < 1 \), the image is smaller than the object. **Answer:** The magnification of the image is approximately **+0.692**. |
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| deepseekv3 | To determine the magnification of the image formed by the convex mirror, we need to follow these steps: --- ### Step 1: Understand the given information - Object height (\(h_o\)) = 5 cm - Object distance (\(u\)) = -10 cm (negative because it is in front of the mirror) - Radius of curvature (\(R\)) = 45.0 cm --- ### Step 2: Find the focal length (\(f\)) of the mirror The focal length of a mirror is related to its radius of curvature by the formula: \[ f = \frac{R}{2} \] For a convex mirror, the focal length is **negative** because the focus is behind the mirror. Thus: \[ f = \frac{45.0}{2} = 22.5 \, \text{cm} \] Since it's a convex mirror: \[ f = -22.5 \, \text{cm} \] --- ### Step 3: Use the mirror formula to find the image distance (\(v\)) The mirror formula is: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] Substitute the known values: \[ \frac{1}{-22.5} = \frac{1}{-10} + \frac{1}{v} \] Solve for \(v\): \[ \frac{1}{v} = \frac{1}{-22.5} - \frac{1}{-10} \] \[ \frac{1}{v} = -\frac{1}{22.5} + \frac{1}{10} \] \[ \frac{1}{v} = \frac{-10 + 22.5}{225} \] \[ \frac{1}{v} = \frac{12.5}{225} \] \[ v = \frac{225}{12.5} = 18 \, \text{cm} \] The positive sign of \(v\) indicates that the image is formed **behind the mirror**, which is consistent with the behavior of a convex mirror. --- ### Step 4: Calculate magnification (\(m\)) The magnification formula for a mirror is: \[ m = -\frac{v}{u} \] Substitute the known values: \[ m = -\frac{18}{-10} = 1.8 \] The positive sign of \(m\) indicates that the image is **upright**. --- ### Step 5: Interpret the magnification The magnification \(m = 1.8\) means the image is 1.8 times larger than the object. --- ### Final Answer: The magnification of the image is **1.8**. |
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