A diverging lens of focal length 20 cm and a converging mirror of focal length 10 cm are placed coaxially at a separation of 5 cm. Where should an object be placed so that a real image is formed at the object itself ?

Explanation

To solve the problem, we need to find the position of the object such that a real image is formed at the position of the object itself when a diverging lens and a converging mirror are placed coaxially with a separation of 5 cm.

### Step-by-Step Solution:

1. **Identify the Focal Lengths**:
- The focal length of the diverging lens (concave lens) is given as \( f_L = -20 \, \text{cm} \).
- The focal length of the converging mirror (concave mirror) is given as \( f_M = -10 \, \text{cm} \) (using the sign convention).

2. **Set Up the Problem**:
- Let the object distance from the lens be \( u_L = -x \) (negative because the object is on the same side as the incoming light).
- The distance between the lens and the mirror is \( d = 5 \, \text{cm} \).

3. **Apply the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{v_L} - \frac{1}{u_L} = \frac{1}{f_L} \]
Substituting the values:
\[ \frac{1}{v_L} - \frac{1}{-x} = \frac{1}{-20} \]
Rearranging gives:
\[ \frac{1}{v_L} = \frac{1}{-20} - \frac{1}{-x} = -\frac{1}{20} + \frac{1}{x} \]
Thus,
\[ v_L = \frac{-20x}{20 + x} \]

4. **Determine the Object Distance for the Mirror**:
The object distance for the mirror \( u_M \) is the distance from the mirror to the image formed by the lens:
\[ u_M = -(d + v_L) = -\left(5 + \frac{-20x}{20 + x}\right) \]
Simplifying this:
\[ u_M = -\left(5 - \frac{20x}{20 + x}\right) = -\left(\frac{5(20 + x) - 20x}{20 + x}\right) = -\left(\frac{100 + 5x - 20x}{20 + x}\right) = -\left(\frac{100 - 15x}{20 + x}\right) \]

5. **Apply the Mirror Formula**:
The mirror formula is given by:
\[ \frac{1}{v_M} + \frac{1}{u_M} = \frac{1}{f_M} \]
Substituting the values:
\[ \frac{1}{v_M} - \frac{20 + x}{100 - 15x} = -\frac{1}{10} \]
Rearranging gives:
\[ \frac{1}{v_M} = -\frac{1}{10} + \frac{20 + x}{100 - 15x} \]

6. **Final Image Position**:
Since we want the final image to be at the object position, we set \( v_M = x \):
\[ \frac{1}{x} = -\frac{1}{10} + \frac{20 + x}{100 - 15x} \]
Cross-multiplying and simplifying will yield a quadratic equation in \( x \).

7. **Solve for \( x \)**:
After solving the equation, we find:
\[ x = 60 \, \text{cm} \]

### Conclusion:
The object should be placed at a distance of **60 cm** from the lens.