When an object is placed on the principal axis of a convex lens at two different positins, it produces the images with magnifications `+2` and `-4` respectively. How many times more away from the lens the image will be formed in the second position as compared to the first postion?

Options

A. 2
B. 4
C. 5
D. 10

Explanation

To solve the problem step by step, we will utilize the magnification formula and the lens formula for a convex lens.

### Step 1: Understanding the Magnification
The magnification (m) produced by a lens is given by the formula:
\[ m = \frac{V}{U} \]
where \( V \) is the image distance and \( U \) is the object distance.

### Step 2: Setting Up the First Position
For the first position, the magnification is given as \( +2 \):
\[ m_1 = 2 = \frac{V_1}{U_1} \]
Let \( U_1 = -X \) (the object distance is taken as negative in lens convention). Therefore, we can write:
\[ V_1 = 2U_1 = 2(-X) = -2X \]

### Step 3: Using the Lens Formula for the First Position
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{F} \]
Substituting \( V_1 \) and \( U_1 \):
\[ \frac{1}{-2X} - \frac{1}{-X} = \frac{1}{F} \]
This simplifies to:
\[ -\frac{1}{2X} + \frac{1}{X} = \frac{1}{F} \]
Combining the fractions:
\[ \frac{-1 + 2}{2X} = \frac{1}{F} \]
Thus:
\[ \frac{1}{2X} = \frac{1}{F} \]
From this, we find:
\[ F = 2X \]

### Step 4: Setting Up the Second Position
For the second position, the magnification is given as \( -4 \):
\[ m_2 = -4 = \frac{V_2}{U_2} \]
Let \( U_2 = -Y \). Therefore, we can write:
\[ V_2 = -4U_2 = -4(-Y) = 4Y \]

### Step 5: Using the Lens Formula for the Second Position
Using the lens formula again:
\[ \frac{1}{V_2} - \frac{1}{U_2} = \frac{1}{F} \]
Substituting \( V_2 \) and \( U_2 \):
\[ \frac{1}{4Y} - \frac{1}{-Y} = \frac{1}{F} \]
This simplifies to:
\[ \frac{1}{4Y} + \frac{1}{Y} = \frac{1}{F} \]
Combining the fractions:
\[ \frac{1 + 4}{4Y} = \frac{1}{F} \]
Thus:
\[ \frac{5}{4Y} = \frac{1}{F} \]
From this, we find:
\[ F = \frac{4Y}{5} \]

### Step 6: Equating Focal Lengths
Since the focal length \( F \) is the same for both positions, we can equate the two expressions for \( F \):
\[ 2X = \frac{4Y}{5} \]

### Step 7: Finding the Ratio of Distances
From the equation \( 2X = \frac{4Y}{5} \), we can rearrange it to find the ratio of \( Y \) to \( X \):
\[ 4Y = 10X \]
Dividing both sides by 4:
\[ Y = \frac{10X}{4} = \frac{5X}{2} \]

### Step 8: Finding the Ratio of Image Distances
Now, we want to find how many times more away from the lens the image will be formed in the second position as compared to the first position:
\[ \frac{V_2}{V_1} = \frac{4Y}{-2X} \]
Substituting \( Y = \frac{5X}{2} \):
\[ \frac{V_2}{V_1} = \frac{4 \cdot \frac{5X}{2}}{-2X} = \frac{10X}{-2X} = -5 \]

### Conclusion
The absolute value of the ratio of the distances of the images formed in the second position to that of the first position is:
\[ \frac{|V_2|}{|V_1|} = 5 \]

Thus, the image will be formed 5 times more away from the lens in the second position compared to the first position.