A student measures the focal length of a convex lens by putting an object pin at a distance `u` from the lens and measuring the distance `v` of the image pin. The graph between `u` and `v` plotted by the student should look like

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Answer

To determine the graph between the object distance \( u \) and the image distance \( v \) for a convex lens, we can start from the lens formula:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Rearranging this formula gives us:

\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

Multiplying through by \( uv \) to eliminate the fractions results in:

\[ u + v = \frac{uv}{f} \]

From this equation, we can express \( v \) in terms of \( u \):

\[ v = \frac{fu}{u - f} \]

This equation indicates that \( v \) is a function of \( u \).

### Step-by-Step Solution:

1. **Identify the lens formula**: Start with the lens formula, which relates the object distance \( u \), image distance \( v \), and focal length \( f \).

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

2. **Rearrange the lens formula**: Rearranging gives us \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \).

3. **Multiply through by \( uv \)**: To eliminate the fractions, multiply through by \( uv \):

\[ u + v = \frac{uv}{f} \]

4. **Express \( v \) in terms of \( u \)**: Rearranging the equation gives:

\[ v = \frac{fu}{u - f} \]

5. **Analyze the graph**: The equation \( v = \frac{fu}{u - f} \) indicates that as \( u \) approaches \( f \), \( v \) approaches infinity (vertical asymptote). For \( u < f \), \( v \) becomes negative, indicating that the image is virtual. For \( u > f \), \( v \) is positive, indicating a real image. Thus, the graph will be a hyperbola.

6. **Determine the quadrant**: Since \( u \) is positive and \( v \) can be both positive and negative, the graph will lie in the second quadrant (where \( u \) is positive and \( v \) can be negative).

### Conclusion:
The graph between \( u \) and \( v \) will be a hyperbola situated in the second quadrant.