If the distance between the virtual image from Its real object is 60 cm in case of a concave mirror . The focal length of the concave mirror If the image formed is 3 times magnified image is

Options

A. `+15cm`
B. `-15 cm`
C. `+22.5 cm`
D. `-22.5 cm`

Answer

To solve the problem, we need to find the focal length of a concave mirror given that the distance between the virtual image and the real object is 60 cm and the image is 3 times magnified.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a concave mirror.
- The distance between the virtual image (V) and the real object (U) is given as 60 cm.
- The magnification (M) is given as 3.

2. **Setting Up the Relationship**:
- The relationship between the object distance (U), image distance (V), and the magnification (M) is given by:
\[ M = -\frac{V}{U} \]
- Since the image is virtual, V will be positive, and U will be negative (as per the sign convention for mirrors).

3. **Using the Given Information**:
- Given that the distance between the virtual image and the real object is 60 cm:
\[ |U| + V = 60 \quad \text{(1)} \]
- Since the magnification is 3:
\[ M = 3 = -\frac{V}{U} \quad \text{(2)} \]

4. **Expressing V in terms of U**:
- From equation (2), we can express V:
\[ V = -3U \quad \text{(3)} \]

5. **Substituting V into Equation (1)**:
- Substitute equation (3) into equation (1):
\[ |U| + (-3U) = 60 \]
- Since U is negative, we can write:
\[ -U - 3U = 60 \]
\[ -4U = 60 \]
\[ U = -15 \text{ cm} \]

6. **Finding V**:
- Now substitute U back into equation (3) to find V:
\[ V = -3(-15) = 45 \text{ cm} \]

7. **Using the Mirror Formula**:
- The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{V} + \frac{1}{U} \]
- Substitute V and U into the formula:
\[ \frac{1}{f} = \frac{1}{45} + \frac{1}{-15} \]
- Finding a common denominator (which is 45):
\[ \frac{1}{f} = \frac{1}{45} - \frac{3}{45} = \frac{-2}{45} \]
- Therefore:
\[ f = -\frac{45}{2} = -22.5 \text{ cm} \]

### Final Answer:
The focal length of the concave mirror is **-22.5 cm**.