A converging lens has a focal length of 0.12 m. To get an image of unit magnification the object should be placed at what distance from the converging lens ?

Options

A. 0.24 m
B. 0.12 m
C. 0.06 m
D. 0.4 m

Answer

To solve the problem step by step, we will use the lens formula and the concept of magnification.

### Given:
- Focal length (f) of the converging lens = 0.12 m
- Magnification (m) = 1 (unit magnification)

### Step 1: Understand the relationship between object distance (u), image distance (v), and focal length (f)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length of the lens
- \( v \) = image distance from the lens
- \( u \) = object distance from the lens

### Step 2: Use the magnification formula
The magnification (m) is defined as:
\[ m = \frac{v}{u} \]
Since we want a unit magnification (m = 1), we have:
\[ v = u \]

### Step 3: Substitute \( v \) in the lens formula
Substituting \( v = u \) in the lens formula:
\[ \frac{1}{f} = \frac{1}{u} - \frac{1}{u} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{u} - \frac{1}{u} = 0 \]
This is incorrect because we need to express \( v \) in terms of \( u \).

### Step 4: Correctly substitute \( v \) in terms of \( u \)
Since \( v = u \), we can rewrite the lens formula as:
\[ \frac{1}{f} = \frac{1}{u} - \frac{1}{u} \]
This doesn't help, so we need to express \( v \) in terms of \( u \) with the correct signs.

### Step 5: Use the sign convention
According to the sign convention:
- Object distance \( u \) is negative (since the object is placed on the left side of the lens).
- Image distance \( v \) is positive (since the image is formed on the right side of the lens).

Thus, we have:
\[ v = -u \]

### Step 6: Substitute into the lens formula
Now substituting \( v = -u \) into the lens formula:
\[ \frac{1}{f} = \frac{1}{-u} - \frac{1}{u} \]
This simplifies to:
\[ \frac{1}{f} = -\frac{1}{u} - \frac{1}{u} = -\frac{2}{u} \]
Rearranging gives:
\[ \frac{2}{u} = \frac{1}{f} \]

### Step 7: Solve for \( u \)
Now substituting \( f = 0.12 \) m:
\[ \frac{2}{u} = \frac{1}{0.12} \]
Cross-multiplying gives:
\[ 2 = \frac{u}{0.12} \]
Thus:
\[ u = 2 \times 0.12 = 0.24 \text{ m} \]

### Step 8: Apply the sign convention
Since the object distance is negative:
\[ u = -0.24 \text{ m} \]

### Final Answer:
The object should be placed at a distance of **0.24 m** from the converging lens (on the left side).

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