A student measures the focal length of a convex lens by putting an object pin at a distance u from t

Q: Why is the graph in the second quadrant?

By Cartesian sign convention, the object is placed in front of the lens, making $u$ negative. A convex lens forms a real image on the opposite side, making $v$ positive. Hence, the coordinates $(u, v)$ fall in the second quadrant.

Q: Why isn't the graph a straight line?

The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ is an inverse relationship. When you solve for $v$, you get $v = \frac{uf}{u+f}$, which is the equation of a rectangular hyperbola, not a linear equation like $y = mx + c$.

Q: What does the vertical asymptote represent?

The vertical asymptote at $u = -f$ represents the physical situation where the object is placed exactly at the principal focus of the lens. In this case, the refracted rays become parallel, and the image is formed at infinity.

Q: What does the horizontal asymptote represent?

The horizontal asymptote at $v = f$ represents the scenario where the object is infinitely far away ($u \to -\infty$). The incident rays are parallel to the principal axis, and they converge exactly at the focal point.

Q: Can the graph ever cross the axes?

No. If $v = 0$, the image would be at the optical center, which is impossible for a real object. If $u = 0$, the object is at the optical center, which is a trivial case where no refraction occurs. The curve approaches the axes but never crosses them in this physical context.

Question

A student measures the focal length of a convex lens by putting an object pin at a distance $u$ from the lens and measuring the distance $v$ of the image pin. The graph between $u$ and $v$ plotted by the student should look like

Sigma Percentile · Accepted Answer

### The Setup: Real Objects and Real Images

Imagine you are in a dark room, sliding a candle along an optical bench towards a convex lens. As you move the candle (the object), the sharp image projected on the screen on the other side also moves. We want to capture this dynamic relationship mathematically by plotting the image distance $v$ against the object distance $u$.

Before we draw any curves, we must establish our coordinate system using the Cartesian sign convention. For a real object placed in front of the lens, the incident light travels from left to right. Since we measure the object distance against the direction of incident light, $u$ is strictly negative ($u < 0$). Conversely, the real image is formed on the other side of the lens, measured in the direction of light, making $v$ strictly positive ($v > 0$).

Because $u$ is negative and $v$ is positive, our entire graph must reside in the **second quadrant** of the Cartesian plane.

### The Master Equation: The Lens Formula

The relationship between the object distance, image distance, and the focal length $f$ is governed by the thin lens formula:

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

To plot $v$ as a function of $u$, we need to isolate $v$. Let's rearrange the terms:

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

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

Inverting both sides gives us our master function:

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

This is not a linear equation; it is the equation of a **rectangular hyperbola**.

### Unveiling the Asymptotes

To sketch this hyperbola accurately, we need to find its boundaries, known as asymptotes. We do this by pushing our object to extreme physical limits.

**The Vertical Asymptote:** What happens if we place the object exactly at the principal focus? Mathematically, we take the limit as $u 	o -f$. The denominator $(u + f)$ approaches zero, causing the image distance $v$ to blow up to infinity.

$$\lim_{u 	o -f} \frac{uf}{u + f} = \infty$$

This gives us a vertical asymptote at $u = -f$. Physically, this means the refracted rays become perfectly parallel and never converge on a screen.

**The Horizontal Asymptote:** Now, let's pull the object infinitely far away from the lens. We take the limit as $u 	o -\infty$. In the denominator, the constant $f$ becomes negligible compared to the massive value of $u$.

$$\lim_{u 	o -\infty} \frac{uf}{u + f} = \lim_{u 	o -\infty} \frac{uf}{u} = f$$

This gives us a horizontal asymptote at $v = f$. Physically, parallel rays from a distant star will converge exactly at the focal point.

### The Final Shape: A Rectangular Hyperbola

By connecting these two extreme limits, we trace a smooth curve in the second quadrant. The curve starts near the horizontal line $v = f$ when $u$ is very large and negative, and it sweeps upwards, approaching the vertical line $u = -f$ as the object gets closer to the focus.

Looking at the given options, only the graph in option (c) correctly depicts a hyperbolic curve residing in the second quadrant, perfectly bounded by these physical asymptotes.

Animated Solution for Physics - Optics: A student measures the focal length of a convex lens by putting an object pin at a distance $u$ from the lens and measuring the distance $v$ of the image pin. The graph between $u$ and $v$ plotted by the student should look like

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Visualized Solution

\text{Graph of } v \text{ vs } u

\text{Lens Formula}

\text{Sign Convention}

\text{Rearranging the Formula}

\text{Vertical Asymptote}

\text{Horizontal Asymptote}

\text{Tracing the Curve}

The Sigma Insight: Lens

The Setup

The Master Equation

Unveiling the Asymptotes

The Final Shape

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