For a concave lens of focal length f, the relation between object and image distances u and v, respe

Q: Why do we use magnitudes for $u$ and $v$ in the graph?

The graphs are plotted in the first quadrant, which implies we are looking at the absolute distances of the object and image from the optical center, ignoring their negative signs as per the Cartesian sign convention.

Q: Why is the curve tangent to $u=v$ at the origin?

For very small object distances ($u \to 0$), the lens formula $v = \frac{uf}{u+f}$ simplifies to $v \approx \frac{uf}{f} = u$. This means the slope of the $v$ vs $u$ curve at the origin is exactly 1, matching the line $u=v$.

Q: Why does the image distance $v$ never exceed the focal length $f$?

For a concave lens, the virtual image is always formed between the optical center and the principal focus. Mathematically, as $u \to \infty$, $v \to f$. Since $u$ is positive, the denominator $u+f$ is always greater than $u$, making $v < f$ for all finite $u$.

Q: How do we know the curve is concave downwards?

By taking the second derivative of $v$ with respect to $u$, we get $\frac{d^2v}{du^2} = -\frac{2f^2}{(u+f)^3}$. Since $u$ and $f$ are positive magnitudes, the second derivative is negative, indicating a concave downward shape.

Q: What happens if we place the object exactly at the focus ($u=f$)?

Substituting $u=f$ into our magnitude relation $v = \frac{uf}{u+f}$, we get $v = \frac{f^2}{2f} = \frac{f}{2}$. The image is formed exactly halfway between the optical center and the focus.

Question

For a concave lens of focal length $f$, the relation between object and image distances $u$ and $v$, respectively, from its pole can best be represented by ($u = v$ is the reference line)

Sigma Percentile · Accepted Answer

### Understanding the Lens Formula

To determine the correct graphical representation of the relationship between the object distance $u$ and the image distance $v$ for a concave lens, we must first establish the mathematical equation governing these variables. We begin with the standard lens formula:

$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$

However, the graphs provided in the options are plotted in the first quadrant, which implies we are dealing strictly with the **magnitudes** of these distances. To derive the relation for magnitudes, we must apply the Cartesian sign convention. For a concave lens, the object is placed in front of the lens, the virtual image is formed on the same side, and the principal focus is also on the same side. Therefore, all three quantities—object distance, image distance, and focal length—are negative.

Let's substitute $-u$, $-v$, and $-f$ into the lens formula to represent their magnitudes:

$$ \frac{1}{-v} - \frac{1}{-u} = \frac{1}{-f} $$

### Deriving the Magnitude Relation

Simplifying the equation by multiplying the entire expression by $-1$, we get:

$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{u} + \frac{1}{f} $$

Now, we can solve for $v$ by taking the least common multiple (LCM) on the right-hand side:

$$ \frac{1}{v} = \frac{u + f}{uf} $$

Inverting both sides yields the explicit function for the image distance $v$ in terms of the object distance $u$:

$$ v = \frac{uf}{u + f} $$

This is the master equation we need to plot. Let's analyze its behavior at key points to identify the correct graph.

### Analyzing the Graph's Behavior

**1. Behavior near the Origin ($u 	o 0$):**
Imagine placing the object extremely close to the optical center of the lens. As $u$ approaches $0$, the denominator $u + f$ is dominated by $f$. Thus, the equation simplifies to:

$$ v \approx \frac{uf}{f} = u $$

This tells us that near the origin, the curve $v(u)$ behaves like the straight line $v = u$. Geometrically, this means the curve must start at the origin $(0,0)$ and be **perfectly tangent** to the reference line $u = v$.

**2. Behavior at Infinity ($u 	o \infty$):**
What happens if we move the object infinitely far away? We can evaluate the limit of our function as $u$ tends to infinity. Dividing the numerator and the denominator by $u$, we get:

$$ v = \lim_{u 	o \infty} \frac{f}{1 + \frac{f}{u}} = \frac{f}{1 + 0} = f $$

This indicates that as the object distance increases indefinitely, the image distance approaches the focal length $f$ but never exceeds it. Therefore, the graph must have a **horizontal asymptote** at $v = f$.

**3. Checking an Intermediate Point ($u = f$):**
To further distinguish the correct curve, let's evaluate the function exactly at the focal point, $u = f$:

$$ v = \frac{f \cdot f}{f + f} = \frac{f^2}{2f} = \frac{f}{2} $$

This means the curve must pass through the coordinate $(f, \frac{f}{2})$. Notice that the reference line $u = v$ passes through $(f, f)$. Since $\frac{f}{2} < f$, our curve lies strictly below the reference line.

### Conclusion

By synthesizing our findings, we are looking for a curve that:
1. Starts at the origin and is tangent to the line $u = v$.
2. Stays entirely below the line $u = v$.
3. Passes through the point $(f, \frac{f}{2})$.
4. Asymptotically approaches the horizontal line $v = f$ as $u$ becomes very large.

Furthermore, taking the second derivative of $v$ with respect to $u$ yields a negative value, confirming that the curve is **concave downwards** everywhere. Matching these precise mathematical characteristics with the given visual options, we find that option (b) is the only graph that flawlessly represents this physical reality.

Animated Solution for Physics - Optics: For a concave lens of focal length $f$ , the relation between object and image distances $u$ and $v$ , respectively, from its pole can best be represented by ( $u = v$ is the reference line)

Select Answer:

Visualized Solution

Lens Formula

Sign Convention

Magnitude Relation

Expressing $v$

Behavior near Origin

Behavior at Infinity

Intermediate Point

Final Graph

The Sigma Insight: Lens

Understanding the Lens Formula

Deriving the Magnitude Relation

Analyzing the Graph's Behavior

Conclusion

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