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 use the mirror formula and the magnification formula for a concave mirror.

### Step 1: Understand the given information
- Focal length of the concave mirror (f) = -12 cm (negative because it's a concave mirror)
- The image is formed at a point which is 10 cm more distance from the mirror than the object.

### Step 2: Set up the variables
Let the distance of the object from the mirror be \( u \) (which will be negative in the mirror convention). Therefore, the distance of the image from the mirror will be:
\[ v = u + 10 \]

### Step 3: Apply the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Substituting the values we have:
\[ \frac{1}{-12} = \frac{1}{u} + \frac{1}{(u + 10)} \]

### Step 4: Simplify the equation
Rearranging the equation gives us:
\[ \frac{1}{u} + \frac{1}{(u + 10)} = -\frac{1}{12} \]
Finding a common denominator:
\[ \frac{(u + 10) + u}{u(u + 10)} = -\frac{1}{12} \]
This simplifies to:
\[ \frac{2u + 10}{u(u + 10)} = -\frac{1}{12} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 12(2u + 10) = -u(u + 10) \]
Expanding both sides:
\[ 24u + 120 = -u^2 - 10u \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ u^2 + 34u + 120 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 34, c = 120 \):
\[ u = \frac{-34 \pm \sqrt{34^2 - 4 \cdot 1 \cdot 120}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 34^2 - 480 = 1156 - 480 = 676 \]
Thus:
\[ u = \frac{-34 \pm 26}{2} \]
Calculating the two possible values:
1. \( u = \frac{-8}{2} = -4 \) (not valid since distance cannot be negative)
2. \( u = \frac{-60}{2} = -30 \) (valid)

### Step 8: Find the image distance \( v \)
Now substituting \( u = -30 \) into the equation for \( v \):
\[ v = u + 10 = -30 + 10 = -20 \text{ cm} \]

### Step 9: Calculate the magnification
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{-20}{-30} = \frac{20}{30} = \frac{2}{3} \]

### Final Answer
The magnification of the image is \( \frac{2}{3} \).

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