An object and a concave mirror of focal length 𝑓 = 10 𝑐𝑚 both move along the principal axis of the mirror with constant speeds. The object moves with speed 𝑉� = 15 𝑐𝑚 𝑠�� 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 𝑐𝑚, the speed of the mirror 𝑉� 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� is _____ 𝑐𝑚 𝑠��.

 Alright, let's tackle this problem step by step. I'm going to approach it methodically to ensure I understand all the components before arriving at the final answer.

Understanding the Problem

We have:

• A concave mirror with a focal length $f=10$ cm.

• An object moving towards the mirror with a speed $V_o=15$ cm/s relative to the laboratory frame.

• The mirror is also moving along the principal axis with some speed $V_m$ (this is what we need to find).

• At a particular moment, the distance between the object and the mirror is $u=30$ cm.

• At this instant, the image is instantaneously at rest with respect to the laboratory frame.

• The image formed is real.

We need to find the magnitude of $V_m$, the speed of the mirror, under these conditions.

Sign Conventions

First, let's recall the standard sign conventions for mirrors:

• The object distance (u) is measured from the mirror to the object. It's negative if the object is on the same side as the incoming light (real object) and positive if on the opposite side (virtual object). Typically, for real objects, $u$ is negative, but sometimes conventions vary. Here, since the object is in front of the mirror, we'll take $u$ as negative.

However, the problem states "the distance between the object and the mirror is denoted by $u$", and doesn't specify the sign. Given that the object is approaching the mirror and forms a real image, it's conventional to take $u$ as negative in mirror formulas. But to avoid confusion, let's proceed carefully.

Given $u=30$ cm, and the object is in front of the mirror, we'll take $u=-30$ cm (assuming the mirror is at the origin and the object is on the negative side).

The focal length (f) for a concave mirror is negative by the standard convention (since the focus is on the same side as the incoming light). So, $f=-10$ cm.

But sometimes, in problems where distances are given as positive magnitudes, the formula is adjusted accordingly. Given the ambiguity, let's proceed with the mirror formula and see.

Mirror Formula

The mirror formula is:

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

Where:

• $f$ = focal length

• $u$ = object distance

• $v$ = image distance

Given $f=-10$ cm (concave mirror, focus is in front), $u=-30$ cm:

$$\begin{gathered} \frac{1}{-10}=\frac{1}{v}+\frac{1}{-30} \\ -\frac{1}{10}=\frac{1}{v}-\frac{1}{30} \\ \frac{1}{v}=-\frac{1}{10}+\frac{1}{30}=-\frac{3}{30}+\frac{1}{30}=-\frac{2}{30}=-\frac{1}{15} \\ v=-15\text{ cm} \end{gathered}$$

A negative $v$ means the image is formed on the same side as the object, which is correct for a real image formed by a concave mirror.

Velocities and Relative Motion

Now, let's consider the velocities:

• Object speed towards the mirror: $V_o=15$ cm/s (relative to lab frame).

• Mirror speed: $V_m$ (we need to find; direction isn't specified, but since the object is moving towards the mirror, and we need the image to be at rest, the mirror must be moving in a certain way).

At the instant when $u=30$ cm, the image is at rest in the lab frame. That means the velocity of the image relative to the lab frame is zero.

First, let's find the velocity of the image relative to the mirror, and then relate it to the lab frame.

Velocity of Image Relative to Mirror

The transverse magnification $m$ is given by:

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

Differentiating both sides with respect to time (to relate velocities), and noting that $f$ is constant:

From the mirror formula:

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

Differentiating:

$$\begin{gathered} -\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}=0 \\ \frac{dv}{dt}=-{\left(\frac{v}{u}\right)}^2\frac{du}{dt} \end{gathered}$$

Here:

• $\frac{du}{dt}$ is the rate of change of object distance with respect to time, i.e., the velocity of the object relative to the mirror.

• $\frac{dv}{dt}$ is the velocity of the image relative to the mirror.

Let’s denote:

• $V_{o,rel}$ = velocity of object relative to mirror.

• $V_{i,rel}$ = velocity of image relative to mirror.

From the differentiation:

$$V_{i,rel}=-{\left(\frac{v}{u}\right)}^2{V}_{o,rel}$$

Now, let's find $V_{o,rel}$.

Relative Velocity of Object to Mirror

In the lab frame:

• Object velocity: $V_o=15$ cm/s towards the mirror (let's take towards mirror as positive).

• Mirror velocity: $V_m$ (unknown; let's assume it's moving towards the object as well, but we'll see).

The relative velocity of the object with respect to the mirror is:

$$V_{o,rel}=V_o-V_m$$

Similarly, the velocity of the image in the lab frame is:

$$V_{i,lab}=V_{i,rel}+V_m$$

We're given that $V_{i,lab}=0$, so:

$$\begin{gathered} V_{i,rel}+V_m=0 \\ {V}_{i,rel}=-V_m \end{gathered}$$

From earlier:

$$\begin{gathered} V_{i,rel}=-{\left(\frac{v}{u}\right)}^2{V}_{o,rel} \\ -V_m=-{\left(\frac{v}{u}\right)}^2(V_o-V_m) \\ {V}_m={\left(\frac{v}{u}\right)}^2(V_o-V_m) \end{gathered}$$

Now, plug in $v=-15$ cm, $u=-30$ cm:

$$\begin{gathered} \frac{v}{u}=\frac{-15}{-30}=\frac{1}{2} \\ {\left(\frac{v}{u}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4} \\ {V}_m=\frac{1}{4}(15-V_m) \\ 4V_m=15-V_m \\ 5V_m=15 \\ {V}_m=3\text{ cm/s} \end{gathered}$$

Verifying Directions

We assumed both object and mirror are moving towards each other (positive direction towards mirror). The calculation gives $V_m=3$ cm/s.

If the mirror is moving towards the object at 3 cm/s, and the object is moving towards the mirror at 15 cm/s, their relative speed is $15-3=12$ cm/s.

The image speed relative to the mirror is:

$$V_{i,rel}=-{\left(\frac{1}{2}\right)}^2\times 12=-\frac{1}{4}\times 12=-3\text{ cm/s}$$

Then, the image speed in the lab frame:

$$V_{i,lab}=V_{i,rel}+V_m=-3+3=0\text{ cm/s}$$

This matches the given condition.

Conclusion

The magnitude of the mirror's speed $V_m$ is $3$ cm/s.

Final Answer

The magnitude of $V_m$ is $\boxed{{\displaystyle 3}}$ cm/s.