The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is __________ cm.

Answer

To solve the problem, we need to find the focal length of a convex lens given that the distance between the object and its 3 times magnified virtual image is 20 cm.

### Step-by-Step Solution:

1. **Understanding Magnification**:
The magnification (m) produced by a lens is given by the formula:
\[ m = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance. Given that the magnification is 3 (since the image is 3 times magnified), we can write:
\[ m = 3 \implies v = 3u \]

2. **Distance Between Object and Image**:
The problem states that the distance between the object and the virtual image is 20 cm. This can be expressed as:
\[ v - u = 20 \, \text{cm} \]

3. **Substituting for \( v \)**:
We can substitute \( v \) from the magnification equation into the distance equation:
\[ 3u - u = 20 \]
Simplifying this gives:
\[ 2u = 20 \]

4. **Finding Object Distance \( u \)**:
Now, we can solve for \( u \):
\[ u = \frac{20}{2} = 10 \, \text{cm} \]

5. **Using the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
We already have \( v = 3u \) and \( u = -10 \) cm (since object distance is taken as negative in lens formula). Thus:
\[ v = 3 \times 10 = 30 \, \text{cm} \]

6. **Substituting Values into Lens Formula**:
Now substituting \( v \) and \( u \) into the lens formula:
\[ \frac{1}{f} = \frac{1}{30} - \frac{1}{-10} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{30} + \frac{1}{10} \]
Finding a common denominator (which is 30):
\[ \frac{1}{f} = \frac{1}{30} + \frac{3}{30} = \frac{4}{30} \]
Therefore:
\[ f = \frac{30}{4} = 7.5 \, \text{cm} \]

7. **Final Answer**:
The focal length of the lens used is:
\[ f = 7.5 \, \text{cm} \]