The magnification of an object placed in front of a convex lens is `+2` . The focal length of the lens is `2.0` metres. Find the distance by which object has to be moved to obtain a magnification of `-2` (in metres).

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
- The initial magnification \( m_1 = +2 \)
- The final magnification \( m_2 = -2 \)
- The focal length of the lens \( f = 2.0 \, \text{m} \)

### Step 2: Use the magnification formula
The magnification \( m \) for a lens is given by:
\[ m = \frac{V}{U} \]
where \( V \) is the image distance and \( U \) is the object distance.

For the first case, we have:
\[ m_1 = +2 = \frac{V_1}{U_1} \]
This implies:
\[ V_1 = 2U_1 \quad \text{(Equation 1)} \]

### Step 3: Apply the lens formula for the first case
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting \( F = 2.0 \, \text{m} \) and \( V_1 = 2U_1 \) into the lens formula:
\[ \frac{1}{2} = \frac{1}{2U_1} - \frac{1}{U_1} \]
This simplifies to:
\[ \frac{1}{2} = \frac{1 - 2}{2U_1} = \frac{-1}{2U_1} \]
Cross-multiplying gives:
\[ -U_1 = 2 \implies U_1 = -2 \, \text{m} \]

### Step 4: Calculate for the second case
For the second case, where \( m_2 = -2 \):
\[ m_2 = -2 = \frac{V_2}{U_2} \]
This implies:
\[ V_2 = -2U_2 \quad \text{(Equation 2)} \]

### Step 5: Apply the lens formula for the second case
Using the lens formula again:
\[ \frac{1}{2} = \frac{1}{V_2} - \frac{1}{U_2} \]
Substituting \( V_2 = -2U_2 \):
\[ \frac{1}{2} = \frac{1}{-2U_2} - \frac{1}{U_2} \]
This simplifies to:
\[ \frac{1}{2} = \frac{-1 - 2}{2U_2} = \frac{-3}{2U_2} \]
Cross-multiplying gives:
\[ 3 = -U_2 \implies U_2 = -3 \, \text{m} \]

### Step 6: Calculate the distance moved
The distance the object has to be moved is:
\[ \text{Distance moved} = |U_2 - U_1| = |-3 - (-2)| = |-3 + 2| = | -1 | = 1 \, \text{m} \]

### Final Answer
The distance by which the object has to be moved to obtain a magnification of \(-2\) is \(1 \, \text{m}\).

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