The distance between an object and its upright image is `20 cm.` if the magnification is `0.5,` what is the focal length of the lens that is being used to form the image?

Explanation

To find the focal length of the lens used to form an upright image, we can follow these steps:

### Step 1: Understand the relationship between object distance (u), image distance (v), and magnification (m).
Given:
- Magnification (m) = 0.5
- The distance between the object and the image = 20 cm

### Step 2: Set up the equations based on the information provided.
Since the image is upright and diminished, we can conclude that the lens is a concave lens. For a concave lens, the magnification is given by:
\[ m = \frac{h'}{h} = \frac{-v}{u} \]
Where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object,
- \( v \) is the image distance,
- \( u \) is the object distance.

From the magnification, we have:
\[ 0.5 = \frac{-v}{u} \]
This implies:
\[ v = -0.5u \]

### Step 3: Use the distance between the object and image.
The distance between the object and its image is given by:
\[ |u| - |v| = 20 \text{ cm} \]

Substituting \( v \) from the magnification equation:
\[ |u| - |-0.5u| = 20 \]
This simplifies to:
\[ u - 0.5u = 20 \]
\[ 0.5u = 20 \]
Thus:
\[ u = 40 \text{ cm} \]

### Step 4: Calculate the image distance (v).
Using the relationship \( v = -0.5u \):
\[ v = -0.5 \times 40 = -20 \text{ cm} \]

### Step 5: Use the lens formula to find the focal length (f).
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{-20} + \frac{1}{-40} \]
Calculating this gives:
\[ \frac{1}{f} = -\frac{1}{20} - \frac{1}{40} \]
Finding a common denominator (which is 40):
\[ \frac{1}{f} = -\frac{2}{40} - \frac{1}{40} = -\frac{3}{40} \]
Thus:
\[ f = -\frac{40}{3} \text{ cm} \approx -13.33 \text{ cm} \]

### Step 6: Conclusion
The focal length of the lens is approximately:
\[ f \approx -13.33 \text{ cm} \]

---