An object and a concave mirror of focal length $f=10\text{ cm}$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_0=15{\text{ cm s}}^{-1}$ 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 $u=30\text{ cm}$, the speed of the mirror $V_m$ is such that the image is instantaneously at rest with respect to the laboratory frame, and object forms a real image. The magnitude of $V_m$ is _____cm s⁻¹.

The correct answer is 3.

Step-by-step Explanation:

As shown in the given diagram, an object moves to the right towards a concave mirror with a speed $V_0=15{\text{ cm s}}^{-1}$ relative to the laboratory frame. The distance between the object and the mirror is given as $u=30\text{ cm}$, and the focal length of the concave mirror is $f=10\text{ cm}$.

1. Using the Mirror Formula to find the image distance (v):

According to the Cartesian sign convention, with the direction of incident light taken as positive (to the right):

Object distance, $u=-30\text{ cm}$

Focal length of concave mirror, $f=-10\text{ cm}$

The mirror formula is:

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

Substituting the values into the formula:

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

$\frac{1}{v}=-\frac{1}{10}+\frac{1}{30}=\frac{-3+1}{30}=-\frac{2}{30}=-\frac{1}{15}$

Thus, $v=-15\text{ cm}$.

2. Relation between velocities:

Differentiating the mirror formula with respect to time $t$:

$-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}=0$

Here, $\frac{dv}{dt}=V_{i,m}$ is the velocity of the image with respect to the mirror, and $\frac{du}{dt}=V_{o,m}$ is the velocity of the object with respect to the mirror.

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

Let $V_I$, ${\mathrm{Vmo}}_{}=V_0$, and $V_m$ be the velocities of the image, object, and mirror relative to the laboratory frame, respectively:

$V_I-V_m=-{\left(\frac{v}{u}\right)}^2\left(V_0-V_m\right)$

3. Calculating the magnitude of $V_m$:

Given that the image is instantaneously at rest in the laboratory frame, $V_I=0$.

The object is moving to the right, so $V_0=+15{\text{ cm s}}^{-1}$.

Also, ${\left(\frac{v}{u}\right)}^2={\left(\frac{-15}{-30}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$.

Substitute these values into the velocity equation:

$0-V_m=-\frac{1}{4}\left(15-V_m\right)$

$-V_m=-\frac{15}{4}+\frac{1}{4}{V}_m$

$\frac{15}{4}=V_m+\frac{1}{4}{V}_m$

$\frac{15}{4}=\frac{5}{4}{V}_m$

$V_m=3{\text{ cm s}}^{-1}$

Thus, the magnitude of $V_m$ is 3 cm s⁻¹.