An object and a concave mirror of focal length f = 10 cm both move along the principal axis of the m

Q: Why is the focal length taken as negative?

According to the Cartesian sign convention, the direction of incident light is taken as positive. For a concave mirror, the focus lies in front of the mirror, opposite to the direction of incident light, making it negative.

Q: Why do we use relative velocity in the differentiated formula?

The standard mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ assumes the pole of the mirror is the origin $(0,0)$. If the mirror itself is moving, the coordinate system is moving. Therefore, differentiating $u$ and $v$ gives the velocities relative to the mirror, not the ground.

Q: What does it mean physically for the image to be at rest?

It means that to an observer standing still on the ground, the image appears completely stationary at that specific instant, even though both the object and the mirror are moving.

Q: Why is the magnification squared in the velocity formula?

When you differentiate $\frac{1}{v}$ with respect to time, you get $-\frac{1}{v^2} \frac{dv}{dt}$. Similarly for $u$. Rearranging these terms naturally produces the factor $\frac{v^2}{u^2}$, which is the square of the transverse magnification $m$.

Q: The diagram in the question shows the mirror moving to the left, but our answer is positive. Why?

The arrows in physics problems are often just generic labels indicating that a variable exists. By setting up a strict coordinate system (right is positive) and letting the math guide us, we discovered that the mirror must actually move to the right ($+3\text{ cm/s}$) to keep the image stationary.

Question

An object and a concave mirror of focal length $f = 10	ext{ cm}$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_0 = 15	ext{ cm s}^{-1}$ towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$. When $u = 30	ext{ cm}$, the speed of the mirror $V_m$ is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_m$ is ________ $	ext{cm s}^{-1}$.

Sigma Percentile · Accepted Answer

### The Illusion of Rest: A Dance of Mirrors and Objects

Imagine you are standing on a moving walkway, looking at a mirror that is also moving, while trying to observe the reflection of a friend walking towards you. It sounds chaotic, right? But what if, amidst all this motion, the reflection of your friend appears perfectly still to you? This is the beautiful paradox we are about to unravel in this problem.

We are given a concave mirror and an object, both moving along the principal axis. The object is heading towards the mirror at a brisk $15	ext{ cm/s}$. Our mission is to find the exact speed of the mirror such that the image appears completely at rest to a stationary observer in the laboratory frame.

### Analyzing the Setup

Before we dive into the kinematics, we must freeze time and figure out exactly where the image is located at the given instant. We know the focal length of the concave mirror is $f = -10	ext{ cm}$, and the object is currently at a distance $u = -30	ext{ cm}$.

We deploy our trusty mirror formula:

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

Substituting our known values with strict adherence to the sign convention:

$$\frac{1}{v} + \frac{1}{-30} = \frac{1}{-10}$$

Solving for $v$, we find:

$$\frac{1}{v} = \frac{1}{30} - \frac{1}{10} = -\frac{2}{30}$$

$$v = -15	ext{ cm}$$

The negative sign confirms that a real image is formed $15	ext{ cm}$ in front of the mirror.

### The Master Equation: Kinematics of the Image

Now comes the thrilling part—bringing motion into the picture. The mirror formula is valid in the rest frame of the mirror. To find the velocities, we differentiate the mirror formula with respect to time. This yields a powerful relationship between the relative velocities of the image and the object with respect to the mirror:

$$\vec{v}_{I/m} = -\left(\frac{v}{u}ight)^2 \vec{v}_{O/m}$$

Expanding the relative velocity terms, we get:

$$\vec{v}_I - \vec{v}_m = -\left(\frac{v}{u}ight)^2 (\vec{v}_O - \vec{v}_m)$$

### Final Calculation

We are given that the image is instantaneously at rest in the laboratory frame, meaning $\vec{v}_I = 0$. The object is moving towards the mirror, so we take its velocity as $\vec{v}_O = +15	ext{ cm/s}$ (assuming the positive direction is towards the mirror). Let's substitute everything we know:

$$0 - \vec{v}_m = -\left(\frac{-15}{-30}ight)^2 (15 - \vec{v}_m)$$

The magnification ratio $\frac{v}{u}$ is $\frac{1}{2}$, and squaring it gives $\frac{1}{4}$. The equation simplifies beautifully:

$$-\vec{v}_m = -\frac{1}{4} (15 - \vec{v}_m)$$

Multiplying both sides by 4 to clear the fraction:

$$4\vec{v}_m = 15 - \vec{v}_m$$

Bringing the $\vec{v}_m$ terms together:

$$5\vec{v}_m = 15$$

**$$\vec{v}_m = 3	ext{ cm/s}$$**

The positive sign is a revelation! It tells us that the mirror is actually moving in the same direction as the object (to the right), despite what a generic diagram might suggest. The magnitude of the mirror's velocity is exactly $3	ext{ cm/s}$.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Mirror Formula$

$Substituting Values$

$Calculating Image Position$

$Velocity Relation$

$Substituting Velocities$

$Simplification$

$Final Answer$

The Sigma Insight: Spherical Mirror

The Illusion of Rest

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mirror Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Image Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Velocities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer