A wire mesh consisting of very small squares is viewed at a distance of `8cm` through a magnifying converging lens of focal length `10 cm`, kept close to the eye. The magnification produced by the lens is:

Options

A. 5
B. 8
C. 10
D. 20

Answer

To solve the problem of finding the magnification produced by a magnifying converging lens, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Focal length of the lens, \( f = 10 \, \text{cm} \) (positive for converging lens)
- Object distance, \( u = -8 \, \text{cm} \) (negative as per sign convention)

2. **Use the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values into the lens formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-8} \]
This simplifies to:
\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{8} \]

3. **Rearranging the Equation**:
To find \( \frac{1}{v} \), we can rearrange the equation:
\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{8} \]
To perform this subtraction, we need a common denominator. The least common multiple of 10 and 8 is 40:
\[ \frac{1}{10} = \frac{4}{40} \quad \text{and} \quad \frac{1}{8} = \frac{5}{40} \]
Therefore:
\[ \frac{1}{v} = \frac{4}{40} - \frac{5}{40} = \frac{-1}{40} \]

4. **Calculate the Image Distance \( v \)**:
Taking the reciprocal gives us:
\[ v = -40 \, \text{cm} \]
The negative sign indicates that the image is virtual and located on the same side as the object.

5. **Calculate the Magnification \( M \)**:
The magnification \( M \) is given by the formula:
\[ M = \frac{v}{u} \]
Substituting the values of \( v \) and \( u \):
\[ M = \frac{-40}{-8} = 5 \]

### Final Answer:
The magnification produced by the lens is \( M = 5 \).

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