What is the distance of an object from a concave mirror of focal length 20 cm so that the size of the real image is three times the size of the object ?

Options

A. 40 cm
B. 60 cm
C. 26.67 cm
D. 6.67 cm

Explanation

To solve the problem, we need to find the distance of an object from a concave mirror such that the size of the real image is three times the size of the object. The focal length of the concave mirror is given as 20 cm.

### Step-by-Step Solution:

1. **Understand the relationship between object distance (U), image distance (V), and magnification (m)**:
- The magnification (m) is defined as the ratio of the height of the image (h') to the height of the object (h):
\[ m = \frac{h'}{h} = -\frac{V}{U} \]
- Given that the size of the real image is three times the size of the object, we can write:
\[ m = -3 \]
- Therefore, we have:
\[ -\frac{V}{U} = -3 \implies V = 3U \]

2. **Use the mirror formula**:
- The mirror formula relates the focal length (F), object distance (U), and image distance (V):
\[ \frac{1}{F} = \frac{1}{U} + \frac{1}{V} \]
- For a concave mirror, the focal length is negative, so:
\[ F = -20 \text{ cm} \]
- Substituting \(V = 3U\) into the mirror formula gives:
\[ \frac{1}{-20} = \frac{1}{U} + \frac{1}{3U} \]

3. **Combine the fractions**:
- The right side can be combined:
\[ \frac{1}{U} + \frac{1}{3U} = \frac{3 + 1}{3U} = \frac{4}{3U} \]
- Thus, we can rewrite the equation as:
\[ \frac{1}{-20} = \frac{4}{3U} \]

4. **Cross-multiply to solve for U**:
- Cross-multiplying gives:
\[ -20 \cdot 4 = 3U \implies -80 = 3U \]
- Solving for U:
\[ U = \frac{-80}{3} = -26.67 \text{ cm} \]
- Since we are interested in the magnitude, we take:
\[ U = 26.67 \text{ cm} \]

5. **Conclusion**:
- The distance of the object from the concave mirror is \(26.67 \text{ cm}\).

### Final Answer:
The distance of the object from the concave mirror is **26.67 cm**.