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

Options

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

Answer

To solve the problem step by step, we will use the information provided about the concave lens, the distance between the object and image, and the magnification.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a concave lens that forms a virtual image of an object. The distance between the object and the image is given as 10 cm. The magnification (M) produced by the lens is \( \frac{1}{4} \).

2. **Setting Up Variables**:
- Let the image distance (V) be \( -x \) (negative because the image is virtual).
- The object distance (U) can be expressed as \( -x + 10 \) (negative because it is on the same side as the object).

3. **Using the Magnification Formula**:
The magnification (M) for a lens is given by:
\[ M = \frac{V}{U} \]
Substituting the values we have:
\[ \frac{1}{4} = \frac{-x}{-x + 10} \]

4. **Cross-Multiplying**:
Cross-multiplying gives:
\[ -x + 10 = 4(-x) \]
Simplifying this:
\[ -x + 10 = -4x \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 10 = -4x + x \]
\[ 10 = -3x \]
\[ x = -\frac{10}{3} \]

6. **Finding Object Distance (U)**:
Now substituting \( x \) back to find U:
\[ U = -x + 10 = -\left(-\frac{10}{3}\right) + 10 = \frac{10}{3} + 10 = \frac{10}{3} + \frac{30}{3} = \frac{40}{3} \]

7. **Finding Image Distance (V)**:
The image distance is:
\[ V = -x = -\left(-\frac{10}{3}\right) = \frac{10}{3} \]

8. **Using the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values of V and U:
\[ \frac{1}{f} = \frac{1}{-\frac{10}{3}} - \frac{1}{-\frac{40}{3}} \]
This simplifies to:
\[ \frac{1}{f} = -\frac{3}{10} + \frac{3}{40} \]

9. **Finding a Common Denominator**:
The common denominator is 40:
\[ \frac{1}{f} = -\frac{12}{40} + \frac{3}{40} = -\frac{12 - 3}{40} = -\frac{9}{40} \]

10. **Calculating Focal Length (f)**:
Therefore, we have:
\[ f = -\frac{40}{9} \approx -4.44 \text{ cm} \]

### Conclusion:
The focal length of the concave lens is approximately \( -4.44 \) cm.