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Step-by-Step Explanation
Step 1: Recall the Lens Formula
For any thin lens, the lens formula is given by:
$$
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
$$
Here,
$f$ is the focal length of the lens,
$v$ is the image distance from the lens (pole), and
$u$ is the object distance from the lens (pole).
For a concave lens, $f$ is taken negative in the sign convention, but we can keep $f$ in symbolic form for the derivation.
Step 2: Express $v$ in Terms of $u$ and $f$
Rewrite the lens formula to solve for $v$:
$$
\frac{1}{v} = \frac{1}{f} + \frac{1}{u}
$$
Taking the reciprocal gives:
$$
v = \frac{1}{\frac{1}{f} + \frac{1}{u}} = \frac{uf}{u + f}
$$
This is the key relation between $u$ and $v$ for a lens (particularly relevant in representing the behavior of a concave lens).
Step 3: Analyze Special Cases
Case I: When $v = u$
Substitute $v = u$ into the derived formula:
$$
u = \frac{uf}{u + f}
$$
This implies
$$
u + f = f \quad \Longrightarrow \quad u = 0 \quad \Longrightarrow \quad v = 0.
$$
So the line $v = u$ is satisfied only at the origin (the point where both $u = 0$ and $v = 0$).
Case II: When $u \to \infty$
If the object is placed very far from the lens ($u \to \infty$), from
$$
v = \frac{uf}{u + f},
$$
as $u \to \infty,$ the term $\frac{u}{u+f}$ approaches 1, so
$$
v \to f.
$$
Hence, for a very distant object, the image distance approaches the focal length $f$, consistent with the behavior of a concave lens.
Step 4: Identify the Correct Graphical Representation
From these boundary conditions and the form $v = \frac{uf}{u + f}$, the graph that correctly matches these properties is:
This curve meets at the origin when $v = u$ and asymptotically approaches $v = f$ when $u \to \infty$, thereby confirming that it is the correct representation for a concave lens with focal length $f$.
