An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is distance `u gt f` from the mirror. Its image will have length

Options

A. `(uf)/(u-f)`
B. `(uf)/(u+f)`
C. `(f^(2))/(u+f)`
D. `(f^(2))/(u-f)`

Answer

To solve the problem of finding the length of the image of an infinitely long rod placed along the axis of a concave mirror, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a concave mirror with a focal length \( f \).
- An infinitely long rod is placed along the axis of the mirror, with the near end of the rod at a distance \( u \) from the mirror, where \( u > f \).

2. **Draw a Diagram**:
- Sketch a concave mirror and mark the focal point \( F \) and the position of the rod. The near end of the rod is at distance \( u \) from the mirror.

3. **Apply the Mirror Formula**:
- The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
- Here, \( f \) is the focal length (negative for concave mirrors), \( u \) is the object distance (also negative), and \( v \) is the image distance.

4. **Substitute Values**:
- Since \( u \) and \( f \) are negative, we can rewrite the formula:
\[ \frac{1}{-f} = \frac{1}{-u} + \frac{1}{v} \]
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{u} - \frac{1}{f} \]

5. **Find the Image Distance**:
- Rearranging the equation for \( v \):
\[ v = \frac{uf}{u - f} \]

6. **Determine the Length of the Image**:
- The image of an infinitely long object will also be infinitely long, but we need to find the effective length of the image formed by the near end of the rod.
- The length of the image will be given by the difference in distances from the image of the near end to the image of the far end.
- The effective length of the image can be calculated as:
\[ \text{Length of image} = |v| - |f| = \frac{uf}{u - f} - |f| \]
- Simplifying this gives:
\[ \text{Length of image} = \frac{uf}{u - f} - f = \frac{uf - f(u - f)}{u - f} = \frac{f^2}{u - f} \]

7. **Final Result**:
- Therefore, the length of the image of the infinitely long rod is:
\[ \text{Length of image} = \frac{f^2}{u - f} \]