If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be :

Options

A. 15 cm
B. `-12 cm`
C. `-10 cm`
D. `(10)/3 cm`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given that the distance between the object and its two times magnified virtual image produced by a curved mirror is 15 cm.

### Step 2: Set up the variables
Let:
- \( u \) = object distance from the mirror (which will be negative for a concave mirror)
- \( v \) = image distance from the mirror (which will be positive for a virtual image)

According to the problem, the distance between the object and the image is:
\[ u + v = 15 \, \text{cm} \quad \text{(1)} \]

### Step 3: Use the magnification formula
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
We know that the magnification is 2 (since the image is twice magnified), hence:
\[ -\frac{v}{u} = 2 \quad \Rightarrow \quad v = -2u \quad \text{(2)} \]

### Step 4: Substitute equation (2) into equation (1)
Substituting \( v = -2u \) into equation (1):
\[ u + (-2u) = 15 \]
This simplifies to:
\[ -u = 15 \quad \Rightarrow \quad u = -15 \, \text{cm} \]

### Step 5: Find the image distance \( v \)
Using equation (2) to find \( v \):
\[ v = -2u = -2(-15) = 30 \, \text{cm} \]

### Step 6: Use the mirror formula to find the focal length
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{-15} + \frac{1}{30} \]
Finding a common denominator (which is 30):
\[ \frac{1}{f} = -\frac{2}{30} + \frac{1}{30} = -\frac{1}{30} \]
Thus,
\[ f = -30 \, \text{cm} \]

### Step 7: Conclusion
The focal length of the concave mirror is:
\[ f = -10 \, \text{cm} \]