Answer:
To determine the magnification produced by the lens, we can use the magnification formula:
\[ M = \frac{-d_i}{d_o} \]
Where:
- \( M \) = Magnification
- \( d_i \) = Image distance (distance of the image from the lens)
- \( d_o \) = Object distance (distance of the object from the lens)
Given:
- \( f = 12.5 \) cm (focal length)
- \( d_i = 20.0 \) cm (image distance)
- Since the image is virtual, it is formed on the same side as the object, so \( d_o \) is negative.
First, let's find the object distance \( d_o \) using the lens formula:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
\[ \frac{1}{12.5} = \frac{1}{d_o} + \frac{1}{20.0} \]
\[ \frac{1}{d_o} = \frac{1}{12.5} - \frac{1}{20.0} \]
\[ \frac{1}{d_o} = \frac{8}{100} - \frac{5}{100} \]
\[ \frac{1}{d_o} = \frac{3}{100} \]
\[ d_o = \frac{100}{3} \]
\[ d_o = -33.3 \, \text{cm} \]
Now, let's calculate the magnification \( M \):
\[ M = \frac{-d_i}{d_o} \]
\[ M = \frac{-20.0}{-33.3} \]
\[ M ≈ \frac{2}{3} \]
Now, to determine whether the image is upright or inverted, we need to check the sign of the magnification. Since the magnification \( M \) is positive, the image is upright.