An object is gradually moving away from the focal point of a concave mirror along the axis of the mi

Q: Why do we substitute $u = -x$?

According to the Cartesian sign convention, the pole of the mirror is the origin. Since the object is placed in front of the mirror (against the direction of incident light), its coordinate is negative.

Q: Why is the focal length taken as $-f$?

For a concave mirror, the focus lies in front of the reflecting surface. Therefore, measuring from the pole against the incident light gives a negative focal length.

Q: What does $m = 1$ physically mean at $x = 2f$?

At $x = 2f$, the object is at the center of curvature. The mirror forms a real, inverted image at the exact same location with the exact same size as the object, hence the magnitude of magnification is 1.

Q: Why doesn't the graph start from $x = 0$?

The problem explicitly states that the object is moving *away* from the focal point. Therefore, our domain of interest starts from $x = f$ and goes towards infinity.

Q: How do we know the curve is a hyperbola and not a straight line?

The derived relation is $m = \frac{f}{x-f}$. Since the variable $x$ is in the denominator with a power of 1, it represents a rectangular hyperbola, which curves downwards, unlike a linear equation $y = mx + c$.

Question

An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification ($m$) versus distance of the object from the mirror ($x$) is correctly given by (graphs are drawn schematically and are not to scale)

Sigma Percentile · Accepted Answer

Have you ever wondered how the size of an image changes as you walk away from a curved mirror? It’s not just a random blur; it follows a beautiful, predictable mathematical rhythm. Let’s dive into the fascinating world of a concave mirror and uncover the exact relationship between the object's distance and its magnification.

### Setting the Stage

Imagine a concave mirror. You are holding an object exactly at its focal point, $f$. Now, you start moving the object slowly away from the mirror, along the principal axis. The question asks us to visualize how the magnitude of linear magnification, $m$, changes as the distance $x$ increases.

To solve this, we need to translate our physical setup into a mathematical equation.

### The Master Equation

The heart of this problem lies in the magnification formula that directly connects focal length $f$ and object distance $u$:
$$m = \frac{f}{f - u}$$

Now, we must be extremely careful with our sign conventions. For a concave mirror, the focal length is always negative, so we replace $f$ with $-f$. The object is placed in front of the mirror at a distance $x$, so the object distance $u$ becomes $-x$.

Let's substitute these into our master equation:
$$m = \frac{-f}{-f - (-x)}$$

Simplifying the denominator, we get:
$$m = \frac{-f}{x - f}$$

### The Mathematical Translation

The question specifically asks for the **magnitude** of the magnification, which we will denote as $m$ for the graph. Since the object is moving *away* from the focal point, we know that $x > f$. This means the term $(x - f)$ is always positive.

Taking the absolute value, our equation beautifully simplifies to:
$$m = \frac{f}{x - f}$$

This is the equation of a rectangular hyperbola! But to confidently pick the right graph, we need to test a few critical anchor points.

### Analyzing the Extremes

**1. Close to the Focus ($x 	o f$):**
What happens when the object is just barely past the focal point? As $x$ approaches $f$, the denominator $(x - f)$ becomes incredibly small, approaching zero. Dividing by a tiny number gives a massive result. Therefore, $m 	o \infty$. The graph must start from infinity, acting as an asymptote at $x = f$.

**2. Far Away ($x 	o \infty$):**
Now, imagine moving the object miles away from the mirror. As $x$ becomes infinitely large, the denominator $(x - f)$ explodes. Dividing by a massive number yields almost zero. So, as $x 	o \infty$, $m 	o 0$. The curve must gently approach the x-axis.

### The Anchor Point

To be absolutely certain, let's check a famous landmark: the center of curvature. We know this occurs at $x = 2f$.

Let's plug this into our equation:
$$m = \frac{f}{2f - f} = \frac{f}{f} = 1$$

This is a perfect physical match! When an object is at the center of curvature, the image is formed at the exact same spot, real and inverted, meaning the size of the image equals the size of the object ($m = 1$).

### The Grand Finale

Let's piece the puzzle together. Our graph must:
1. Start from infinity at $x = f$.
2. Pass exactly through the point $(2f, 1)$.
3. Taper off towards zero as $x$ increases.

Looking at the given options, only **Option (d)** perfectly captures this elegant hyperbolic descent. The math and the physics align flawlessly!

Select Answer:

Visualized Solution

Problem Setup

Magnification Formula

Sign Convention

Substitution

Simplification

Magnitude of Magnification

Boundary Condition 1

Boundary Condition 2

Boundary Condition 3

Conclusion

The Sigma Insight: Spherical Mirror

Setting the Stage

The Master Equation

The Mathematical Translation

Analyzing the Extremes

The Anchor Point

The Grand Finale

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