A convex lens of focal length 25 cm produces images of the same magnification 2, when an object is kept at two positions `x_(1)` and `x_(2) (x_(1) gt x_(2)) ` from the lens. The ratio of `x_(2)` and `x_(1)` is

Options

A. `2:1`
B. `1:3`
C. `1:2`
D. `3:1 `

Explanation

To solve the problem, we need to find the ratio of the object distances \( x_2 \) and \( x_1 \) for a convex lens with a focal length of 25 cm, given that both positions produce images of the same magnification of 2.

### Step-by-Step Solution:

1. **Understanding Magnification**:
The magnification \( m \) of a lens is given by the formula:
\[ m = -\frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance. For a magnification of 2, we can have two cases:
- Case 1: \( m = -2 \) (real and inverted image)
- Case 2: \( m = +2 \) (virtual and upright image)

2. **Using the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \( f \) is the focal length of the lens.

3. **Case 1: Magnification \( m = -2 \)**:
For \( m = -2 \):
\[ -2 = -\frac{v}{u} \implies v = 2u \]
Substituting \( v \) in the lens formula:
\[ \frac{1}{25} = \frac{1}{2u} - \frac{1}{u} \]
Simplifying:
\[ \frac{1}{25} = \frac{1 - 2}{2u} = -\frac{1}{2u} \]
Rearranging gives:
\[ 2u = -25 \implies u = -\frac{25}{2} = -12.5 \text{ cm} \]
Since \( u \) is negative, we take \( x_1 = 12.5 \) cm.

4. **Case 2: Magnification \( m = +2 \)**:
For \( m = +2 \):
\[ 2 = -\frac{v}{u} \implies v = -2u \]
Substituting \( v \) in the lens formula:
\[ \frac{1}{25} = \frac{1}{-2u} - \frac{1}{u} \]
Simplifying:
\[ \frac{1}{25} = -\frac{1}{2u} - \frac{1}{u} = -\frac{1 + 2}{2u} = -\frac{3}{2u} \]
Rearranging gives:
\[ 2u = -75 \implies u = -\frac{75}{2} = -37.5 \text{ cm} \]
Since \( u \) is negative, we take \( x_2 = 37.5 \) cm.

5. **Finding the Ratio**:
Now, we need to find the ratio \( \frac{x_2}{x_1} \):
\[ \frac{x_2}{x_1} = \frac{37.5}{12.5} = 3 \]

### Final Answer:
The ratio of \( x_2 \) to \( x_1 \) is \( 3 \).