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 determining which graph represents the \( \frac{v}{f} \) vs \( \frac{u}{f} \) graph for a virtual erect image formed by a diverging lens, we can follow these steps:

### Step 1: Understand the Sign Conventions
For a diverging lens:
- The focal length \( f \) is negative.
- The object distance \( u \) is also negative because the object is placed on the same side as the incoming light.
- The image distance \( v \) is negative for virtual images, as they appear on the same side as the object.

### Step 2: Write the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]

### Step 3: Substitute the Variables
Let:
- \( y = \frac{v}{f} \)
- \( x = \frac{u}{f} \)

Then, we can rewrite the lens formula in terms of \( y \) and \( x \):
\[ \frac{1}{y} = \frac{1}{1} - \frac{1}{x} \]
This simplifies to:
\[ \frac{1}{y} = 1 + \frac{1}{x} \]

### Step 4: Rearranging the Equation
Now, we can rearrange this equation:
\[ \frac{1}{y} = 1 + \frac{1}{x} \implies y = \frac{x}{x + 1} \]

### Step 5: Analyze the Graph Behavior
1. As \( x \) approaches 0 (which corresponds to \( u \) approaching negative infinity), \( y \) approaches infinity.
2. As \( x \) approaches infinity (which corresponds to \( u \) approaching 0), \( y \) approaches 1.

### Step 6: Identify the Graph
The behavior described indicates that the graph will start from infinity when \( x \) is 0 and will approach 1 as \( x \) goes to infinity. This is characteristic of a hyperbolic curve.

### Conclusion
Based on the analysis, the correct graph representing \( \frac{v}{f} \) vs \( \frac{u}{f} \) for a virtual erect image by a diverging lens is option D.