The image of an object placed on the principal axis of a concave mirror of focal length 12 cm is formed at a point which is 10cm more distance form the mirror than the object. The magnification of the image is

Options

A. `8//3`
B. 2.5
C. 2
D. `-1.5`

Explanation

To solve the problem step by step, we will follow the principles of optics related to concave mirrors.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Focal length of the concave mirror, \( f = -12 \, \text{cm} \) (negative because it is a concave mirror).
- Let the object distance be \( u = -x \) (the object distance is negative in mirror convention).
- The image distance is given as \( v = -x + 10 \).

2. **Apply the Mirror Formula:**
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values we have:
\[ \frac{1}{-12} = \frac{1}{(-x + 10)} + \frac{1}{(-x)} \]

3. **Simplify the Equation:**
Rearranging the equation gives:
\[ \frac{1}{-12} = \frac{-x + 10 + x}{-x(-x + 10)} \]
This simplifies to:
\[ \frac{1}{-12} = \frac{10}{-x(-x + 10)} \]

4. **Cross-Multiply to Eliminate Fractions:**
Cross-multiplying gives:
\[ -x(-x + 10) = -12 \times 10 \]
Simplifying this results in:
\[ x^2 - 10x = 120 \]
Rearranging gives:
\[ x^2 - 10x - 120 = 0 \]

5. **Solve the Quadratic Equation:**
We can factor or use the quadratic formula. The quadratic can be factored as:
\[ (x - 20)(x + 6) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 20 \quad \text{or} \quad x = -6 \]
Since \( x \) represents the object distance, we take \( x = 20 \) cm (the positive value).

6. **Calculate the Image Distance:**
Using \( x = 20 \):
\[ u = -20 \, \text{cm} \quad \text{and} \quad v = -20 + 10 = -30 \, \text{cm} \]

7. **Calculate the Magnification:**
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} = -\frac{-30}{-20} = \frac{30}{20} = \frac{3}{2} = 1.5 \]
The negative sign indicates that the image is inverted.

### Final Answer:
The magnification of the image is \( -1.5 \).