The graph correctly representing the variation of image distance ’v’ for a convex lens of focal length ’f’ versus object distance ’u’ is_____.

Fill in the blank with the correct answer from the options given below.

The correct graph representing the variation of image distance $v$ versus object distance $u$ for a convex lens of focal length $f$ is the graph shown in the second image (where the curve lies in the upper-left region with the horizontal axis labeled pointing to the left as $u$ and the vertical axis pointing up as $v$).

Step-by-Step Explanation:

According to the lens formula, we have:
$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$
where:
- $u$ is the object distance,
- $v$ is the image distance, and
- $f$ is the focal length of the convex lens.

Rearranging the formula to express $v$ in terms of $u$ and $f$:
$$\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{u+f}{uf}$$
Taking the reciprocal, we get:
$$v=\frac{uf}{u+f}$$

By standard sign convention:
- The focal length of a convex lens is positive ($f>0$).
- For a real object, the object is placed in front of the lens, so the object distance $u$ is negative ($u<0$).

Let us analyze the behavior of the image distance $v$ for different positions of the real object:
1. When the object is at infinity ($u\to -\infty$):
$$v=\underset{u\to -\infty }{lim}\frac{f}{1+\frac{f}{u}}=f$$
The real image is formed at the focus on the other side of the lens ($v=f$).
2. When the object is at the focus ($u=-f$):
The denominator becomes zero, which means:
$$v\to \infty$$
The image is formed at infinity.

Thus, as the real object moves from infinity to the focus (i.e., $u$ varies from $-\infty$ to $-f$), the image distance $v$ is positive and increases from $f$ to $+\infty$.
This relationship forms a branch of a rectangular hyperbola in the second quadrant of the $u-v$ plane (where $u$ is negative and $v$ is positive).
Comparing this behavior with the given images:
- The graph in the second image shows the horizontal axis representing the negative object distance (indicated by the arrow pointing to the left: $\leftarrow u$) and the vertical axis representing the positive image distance (indicated by the arrow pointing up: $\uparrow v$).
- The curve asymptotically approaches the vertical line corresponding to the focus and decreases towards a constant positive value $f$ as $u$ increases in magnitude.
Therefore, this graph is the correct representation.