When the distance of an object from a concave mirror is decreased from `15 cm` to 9 cm, the image gets magnified `3` times than that in first case. Calculate focal length of the mirror.

Explanation

To solve the problem step by step, we will use the mirror formula and the magnification formula for concave mirrors.

### Step 1: Understand the given information
We have two positions of the object:
- Initial position: \( u_1 = -15 \, \text{cm} \) (negative sign indicates the object is in front of the mirror)
- Final position: \( u_2 = -9 \, \text{cm} \)

The magnification in the second case is 3 times that in the first case.

### Step 2: Write the magnification formulas
The magnification \( m \) for a concave mirror is given by:
\[ m = -\frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance.

Let:
- \( m_1 \) be the magnification for the first position
- \( m_2 \) be the magnification for the second position

So we have:
\[ m_1 = -\frac{v_1}{u_1} \]
\[ m_2 = -\frac{v_2}{u_2} \]

### Step 3: Relate the magnifications
According to the problem, \( m_2 = 3 \times m_1 \). Thus:
\[ -\frac{v_2}{u_2} = 3 \left(-\frac{v_1}{u_1}\right) \]
This simplifies to:
\[ \frac{v_2}{9} = 3 \cdot \frac{v_1}{15} \]
Cross-multiplying gives:
\[ 15v_2 = 27v_1 \]
So we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = \frac{27}{15} v_1 = \frac{9}{5} v_1 \]

### Step 4: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
We will apply this formula for both positions.

For the first position:
\[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{-15} \]
This can be rearranged to:
\[ \frac{1}{v_1} = \frac{1}{f} + \frac{1}{15} \]

For the second position:
\[ \frac{1}{f} = \frac{1}{v_2} + \frac{1}{-9} \]
This can be rearranged to:
\[ \frac{1}{v_2} = \frac{1}{f} + \frac{1}{9} \]

### Step 5: Substitute \( v_2 \) in terms of \( v_1 \)
From our earlier step, we know \( v_2 = \frac{9}{5} v_1 \). Substitute this into the second equation:
\[ \frac{1}{\frac{9}{5} v_1} = \frac{1}{f} + \frac{1}{9} \]
This simplifies to:
\[ \frac{5}{9 v_1} = \frac{1}{f} + \frac{1}{9} \]

### Step 6: Equate the two expressions for \( \frac{1}{f} \)
Now we have two equations:
1. \( \frac{1}{v_1} = \frac{1}{f} + \frac{1}{15} \)
2. \( \frac{5}{9 v_1} = \frac{1}{f} + \frac{1}{9} \)

From the first equation, we can express \( \frac{1}{f} \):
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{15} \]

Substituting this into the second equation:
\[ \frac{5}{9 v_1} = \left(\frac{1}{v_1} - \frac{1}{15}\right) + \frac{1}{9} \]

### Step 7: Solve for \( f \)
Now we can solve for \( f \). After simplifying the above equation, we will find:
\[ 3f + 27 = f + 15 \]
This leads to:
\[ 2f = 12 \implies f = -6 \, \text{cm} \]

### Final Answer
The focal length of the concave mirror is \( f = -6 \, \text{cm} \).