For a concave mirrorr, if real image is formed the graph between `(1)/(u)` and `(1)/(v)` is of the form

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Answer

To solve the question regarding the graph between \( \frac{1}{u} \) and \( \frac{1}{v} \) for a concave mirror when a real image is formed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Mirror Formula**:
The mirror formula for a concave mirror 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.

2. **Rearranging the Formula**:
We can rearrange this formula to express \( \frac{1}{v} \) in terms of \( \frac{1}{u} \):
\[ \frac{1}{v} = -\frac{1}{u} + \frac{1}{f} \]

3. **Identifying the Form of the Equation**:
The equation \( \frac{1}{v} = -\frac{1}{u} + \frac{1}{f} \) can be compared to the linear equation \( y = mx + c \), where:
- \( y = \frac{1}{v} \)
- \( x = \frac{1}{u} \)
- The slope \( m = -1 \)
- The y-intercept \( c = \frac{1}{f} \)

4. **Graph Characteristics**:
- The slope of -1 indicates that the graph will be a straight line with a negative slope.
- The y-intercept \( \frac{1}{f} \) is positive since the focal length \( f \) for a concave mirror is considered positive in the sign convention used.

5. **Conclusion About the Graph**:
The graph between \( \frac{1}{u} \) and \( \frac{1}{v} \) will be a straight line with:
- A negative slope of -1.
- A y-intercept at \( \frac{1}{f} \).

### Final Answer:
The graph between \( \frac{1}{u} \) and \( \frac{1}{v} \) for a concave mirror when a real image is formed is a straight line with a negative slope, intersecting the y-axis at \( \frac{1}{f} \).