A concave lens forms the image of an object such that the distance between the object and image is 10cm and the magnification produced is 1/4. The focal length of the lens will be

Options

A. `-6.2 cm`
B. `-4.4 cm`
C. `-8.6 cm`
D. `-10 cm`

Answer

To find the focal length of a concave lens given the distance between the object and the image and the magnification, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given data**:
- The distance between the object (O) and the image (I) is given as \( |u - v| = 10 \, \text{cm} \).
- The magnification \( m = \frac{1}{4} \).

2. **Relate magnification to object and image distances**:
- The magnification \( m \) is defined as:
\[ m = \frac{v}{u} \]
- Given \( m = \frac{1}{4} \), we can write:
\[ \frac{v}{u} = \frac{1}{4} \implies v = \frac{1}{4}u \]

3. **Substitute \( v \) in the distance equation**:
- From the distance equation \( |u - v| = 10 \), we substitute \( v \):
\[ u - \frac{1}{4}u = 10 \]
- This simplifies to:
\[ \frac{3}{4}u = 10 \]
- Solving for \( u \):
\[ u = 10 \times \frac{4}{3} = \frac{40}{3} \, \text{cm} \]

4. **Calculate \( v \)**:
- Now substitute \( u \) back to find \( v \):
\[ v = \frac{1}{4}u = \frac{1}{4} \times \frac{40}{3} = \frac{10}{3} \, \text{cm} \]

5. **Apply the sign convention**:
- For a concave lens, the object distance \( u \) is taken as negative and the image distance \( v \) is also negative:
\[ u = -\frac{40}{3} \, \text{cm}, \quad v = -\frac{10}{3} \, \text{cm} \]

6. **Use the lens formula to find the focal length**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Substitute \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{-\frac{10}{3}} - \frac{1}{-\frac{40}{3}} \]
- This simplifies to:
\[ \frac{1}{f} = -\frac{3}{10} + \frac{3}{40} \]
- Finding a common denominator (which is 40):
\[ \frac{1}{f} = -\frac{12}{40} + \frac{3}{40} = -\frac{9}{40} \]
- Therefore:
\[ f = -\frac{40}{9} \, \text{cm} \approx -4.44 \, \text{cm} \]

### Final Answer:
The focal length of the concave lens is \( f \approx -4.44 \, \text{cm} \).