An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is

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Answer

To solve the problem, we will use the mirror formula for a concave mirror and analyze the relationship between the object distance (u), image distance (v), and focal length (f).

### Step-by-Step Solution:

1. **Understand the Mirror Formula**:
The mirror formula for a concave mirror is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]

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

3. **Identifying the Equation Type**:
The equation \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \) can be rewritten as:
\[ \frac{1}{v} = -\frac{1}{u} + \frac{1}{f} \]
This is in the form of \( y = mx + c \), where \( y = \frac{1}{v} \), \( x = \frac{1}{u} \), \( m = -1 \) (the slope), and \( c = \frac{1}{f} \) (the y-intercept).

4. **Graph Characteristics**:
From the equation, we see that the slope \( m \) is negative, indicating that as \( \frac{1}{u} \) increases, \( \frac{1}{v} \) decreases. This means the graph will be a straight line with a negative slope.

5. **Conclusion**:
Therefore, the graph between \( \frac{1}{v} \) and \( \frac{1}{u} \) is a straight line with a negative slope.

### Final Answer:
The graph between \( \frac{1}{v} \) versus \( \frac{1}{u} \) is a straight line with a negative slope.

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