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What is the object distance? You will need to use the magnification equation to find a relationship between do and d;. Then substitute into the thin lens equation to solve for do. Express your answer in centimeters, as a fraction or to three significant figures. ? do = cm

What is the object distance? You will need to use the magnification equation to find a relationship between do and d;. Then substitute into the thin lens equation to solve for do. Express your answer in centimeters, as a fraction or to three significant figures. ? do = cm

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# Lens and Magnification ### Overview Consider a diverging lens with a focal length of \( f = -15 \, \text{cm} \), producing an upright image that is \(\frac{5}{9}\) as tall as the object. --- ### Part E: Image Nature #### Question Is the image real or virtual? Think about the magnification and how it relates to the sign of \( d_i \). #### Options - [ ] Real - [x] Virtual ### Explanation Since the upright image is produced by a diverging lens and has a positive magnification (\(\frac{5}{9}\)), the image is virtual. #### Feedback ✔️ Correct --- ### Part F: Object Distance Calculation #### Question What is the object distance? You will need to use the magnification equation to find a relationship between \( d_o \) and \( d_i \). Then substitute into the thin lens equation to solve for \( d_o \). **Instructions:** Express your answer in centimeters, as a fraction or to three significant figures. #### Equation Box \[ d_o = \, \_\_\_ \, \text{cm} \] Use the magnification \( m = -\frac{d_i}{d_o} = \frac{5}{9} \) and the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) to solve for \( d_o \).
Transcribed Image Text:# Lens and Magnification ### Overview Consider a diverging lens with a focal length of \( f = -15 \, \text{cm} \), producing an upright image that is \(\frac{5}{9}\) as tall as the object. --- ### Part E: Image Nature #### Question Is the image real or virtual? Think about the magnification and how it relates to the sign of \( d_i \). #### Options - [ ] Real - [x] Virtual ### Explanation Since the upright image is produced by a diverging lens and has a positive magnification (\(\frac{5}{9}\)), the image is virtual. #### Feedback ✔️ Correct --- ### Part F: Object Distance Calculation #### Question What is the object distance? You will need to use the magnification equation to find a relationship between \( d_o \) and \( d_i \). Then substitute into the thin lens equation to solve for \( d_o \). **Instructions:** Express your answer in centimeters, as a fraction or to three significant figures. #### Equation Box \[ d_o = \, \_\_\_ \, \text{cm} \] Use the magnification \( m = -\frac{d_i}{d_o} = \frac{5}{9} \) and the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) to solve for \( d_o \).
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