The graph shows how the magnification m produced by a thin lens varies with image distance v. What i

Q: Why did we multiply the lens formula by $v$?

We multiplied by $v$ to introduce the magnification term $m = v/u$ directly into the equation. This allows us to create a direct relationship between $m$ and $v$, matching the axes of the given graph.

Q: Why is the slope negative in the formula but positive in the graph?

The theoretical slope is $-1/f$. The graph shows a line going upwards, meaning the slope $c/b$ is positive. For $-1/f$ to be positive, the focal length $f$ must be negative. This tells us that the lens used is actually a concave (diverging) lens!

Q: What does the y-intercept of this graph represent?

The y-intercept is 1. This means that when the image distance $v$ approaches 0, the magnification $m$ approaches 1. This is a theoretical limit representing the optical center of the lens.

Q: Can we use the object distance $u$ instead of $v$?

You could derive a relationship between $m$ and $u$, but the given graph is specifically plotted with $v$ on the x-axis. We must tailor our mathematical derivation to match the visual data provided.

Q: Why do we take the magnitude of $f$ at the end?

The options provided in the question are all positive expressions. They represent the absolute value or magnitude of the focal length. The negative sign we derived simply indicates the diverging nature of the lens.

Question

The graph shows how the magnification $m$ produced by a thin lens varies with image distance $v$. What is the focal length of the lens used?

Options:

(A) $\frac{b^2c}{a}$

(B) $\frac{b^2}{ac}$

(C) $\frac{a}{c}$

(D) $\frac{b}{c}$

Sigma Percentile · Accepted Answer

### Decoding the Magnification Graph: Finding the Focal Length

Graphs in physics are not just lines on a grid; they are visual stories of mathematical relationships. In this problem, we are given a graph showing how the magnification $m$ of a thin lens varies with the image distance $v$. Our goal is to extract the focal length $f$ from this visual data.

### The Math Behind the Graph

To understand the graph, we first need to establish the mathematical relationship between $m$ and $v$. We start with the fundamental thin lens formula:

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

We know that magnification $m$ is defined as the ratio of image distance to object distance, $m = \frac{v}{u}$. To introduce $m$ into our equation, we can multiply the entire lens formula by $v$:

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

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

Rearranging this to solve for $m$, we get:

$$ m = -\frac{1}{f}v + 1 $$

This equation is beautifully simple. It is in the form of a straight line, $y = mx + c$, where our y-axis is $m$ and our x-axis is $v$. The slope of this line is exactly $-\frac{1}{f}$.

### Extracting Data from the Graph

Now, let's turn our attention to the provided graph. We need to find the slope of the line using the given points.

The graph highlights two distinct points. The horizontal distance between these two points on the $v$-axis is given as $b$. This represents our change in $x$, or $\Delta v = b$.

The vertical distance between these same two points on the $m$-axis is given as $c$. This represents our change in $y$, or $\Delta m = c$.

The slope of any straight line is the 'rise over run', or the change in $y$ divided by the change in $x$:

$$ 	ext{Slope} = \frac{\Delta m}{\Delta v} = \frac{c}{b} $$

### The Grand Finale

We now have two expressions for the slope of the line: one from our theoretical derivation and one from the graphical data. By equating them, we can solve for the focal length $f$:

$$ -\frac{1}{f} = \frac{c}{b} $$

$$ f = -\frac{b}{c} $$

The negative sign indicates the nature of the lens (in this case, a concave lens, since the slope $c/b$ is positive, making $f$ negative). However, the question asks for the magnitude of the focal length, which is simply:

$$ |f| = \frac{b}{c} $$

And there we have it! By bridging the gap between algebraic formulas and graphical geometry, we've successfully decoded the focal length.

Animated Solution for Physics - Optics: The graph shows how the magnification $m$ produced by a thin lens varies with image distance $v$ . What is the focal length of the lens used?

Select Answer:

Visualized Solution

Lens Formula

Magnification Relation

Equation of Straight Line

Analyzing the Graph

Graphical Slope

Equating Slopes

The Sigma Insight: Lens

Decoding the Magnification Graph

The Math Behind the Graph

Extracting Data from the Graph

The Grand Finale

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