Which of the following graph represents v/f vs u/f graph for a virtual erect image by a diverging lens?

Options

A.
B.
C.
D.

Explanation

To solve the problem of identifying the correct graph that represents the relationship between \( v/f \) and \( u/f \) for a virtual erect image formed by a diverging lens, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Lens Formula**: For a diverging lens, the lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Here, \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For a diverging lens, both \( v \) and \( u \) are negative.

2. **Express \( v \) in terms of \( u \)**: Rearranging the lens formula gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
This can be rewritten as:
\[ v = \frac{fu}{u - f} \]

3. **Introduce Velocity and Frequency**: If we consider the velocities \( v \) and \( u \) as proportional to the frequencies \( f \) and \( f' \) respectively, we can express:
\[ \frac{v}{f} = \frac{fu}{f(u - f)} \]
Simplifying this gives:
\[ \frac{v}{f} = \frac{u}{u - f} \]

4. **Graphical Representation**: The relationship \( \frac{v}{f} = \frac{u}{u - f} \) suggests that as \( u \) increases, \( \frac{v}{f} \) approaches 1. This indicates that the graph will start from the origin (0,0) and will approach a horizontal asymptote at \( y = 1 \) as \( x \) approaches infinity.

5. **Identify the Correct Graph**: From the options provided, we need to identify a graph that starts at the origin and approaches \( y = 1 \) as \( x \) increases. This will typically be a hyperbolic shape that lies in the first quadrant.

### Conclusion:
The correct graph that represents the relationship \( v/f \) vs \( u/f \) for a virtual erect image by a diverging lens will have the characteristics described above.