An object is placed at the following distances from a concave mirror of focal length 10 cm:
(a) 8 cm (b) 15 cm (c) 25 cm
Which position of the object will produce
a real image smaller than the object ?

Answer

To solve the problem of determining which position of the object will produce a real image smaller than the object when placed in front of a concave mirror with a focal length of 10 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Mirror Properties**:
- A concave mirror has a focal point (F) where the focal length is given as 10 cm. The center of curvature (C) is twice the focal length, which means C = 2F = 20 cm.

2. **Identify Object Positions**:
- The object is placed at three different distances:
- (a) 8 cm
- (b) 15 cm
- (c) 25 cm

3. **Use the Mirror Formula**:
- The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance (with the convention that \( u \) is negative for real objects).

4. **Evaluate Each Case**:
- **Case (a)**: Object at 8 cm
- Here, \( u = -8 \) cm.
- Using the mirror formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{8} \]
Solving this gives \( v \) as a negative value, indicating a virtual image, which is erect and enlarged.

- **Case (b)**: Object at 15 cm
- Here, \( u = -15 \) cm.
- Using the mirror formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{15} \]
Solving this gives \( v \) as a positive value, indicating a real image, but it is enlarged.

- **Case (c)**: Object at 25 cm
- Here, \( u = -25 \) cm.
- Using the mirror formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{25} \]
Solving this gives \( v \) as a positive value, indicating a real image that is smaller than the object.

5. **Conclusion**:
- The only case where the image is real and smaller than the object is when the object is placed at 25 cm from the mirror.

### Final Answer:
The position of the object that will produce a real image smaller than the object is at **25 cm**.