Pregunta
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9. An object is 15 cm from a concave mirror that has a focal length of 10 cm . Determine the image location, magnification, real or virtual, and upright or inverted for the object. a. Using the equation:

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Feb 24,2025

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The image is located 30 cm behind the mirror, is inverted and twice the size of the object, and is virtual.

Solución

To solve this problem, we will use the mirror formula and the magnification formula for concave mirrors. ### Known Conditions: - Object distance (\(d_o\)) = -15 cm (the object distance is taken as negative in mirror conventions) - Focal length (\(f\)) = -10 cm (the focal length is negative for concave mirrors) ### Step 1: Use the Mirror Formula The mirror formula is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Where: - \(d_i\) = image distance (what we want to find) ### Step 2: Rearranging the Mirror Formula We can rearrange the formula to solve for \(d_i\): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \] ### Step 3: Substitute the Known Values Now, substituting the known values into the equation: \[ \frac{1}{d_i} = \frac{1}{-10} - \frac{1}{-15} \] ### Step 4: Calculate \(d_i\) Let's calculate \(d_i\) using the above expression. \[ \frac{1}{d_i} = -\frac{1}{10} + \frac{1}{15} \] To perform this calculation, we will find a common denominator and simplify. Calculating \(d_i\): \[ \frac{1}{d_i} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30} \] Now, taking the reciprocal to find \(d_i\): \[ d_i = -30 \text{ cm} \] ### Step 5: Determine Magnification The magnification (\(m\)) is given by the formula: \[ m = -\frac{d_i}{d_o} \] Substituting the values we have: \[ m = -\frac{-30}{-15} = -2 \] ### Step 6: Determine Image Characteristics - **Image Location**: \(d_i = -30 \text{ cm}\) (the negative sign indicates that the image is on the same side as the object, which is typical for virtual images). - **Magnification**: \(m = -2\) (the negative sign indicates that the image is inverted, and the magnitude indicates that the image is twice the size of the object). - **Real or Virtual**: Since \(d_i\) is negative, the image is virtual. - **Upright or Inverted**: The negative magnification indicates that the image is inverted. ### Final Summary: - **Image Location**: -30 cm - **Magnification**: -2 - **Real or Virtual**: Virtual - **Upright or Inverted**: Inverted

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To find the image location, you can use the mirror formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Plugging in the values, where \( f = -10 \) cm (negative since it's a concave mirror) and \( d_o = -15 \) cm, you get \( \frac{1}{-10} = \frac{1}{-15} + \frac{1}{d_i} \). Solving for \( d_i \) gives you an image distance of 6 cm from the mirror, meaning the image is real and located on the same side as the object. Next, the magnification \( m \) can be calculated using the formula \( m = -\frac{d_i}{d_o} \). By substituting in the values, \( m = -\frac{6}{-15} = 0.4 \). This positive magnification indicates that the image is upright, while the fact that it's a real image means it will appear inverted with respect to the object in a physical sense. So, we have an image located at 6 cm, magnified by 0.4, and inverted!

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