Question

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 $V_0=15\, cm s ^{-1}$ 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 $u=30\, cm$, 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 _______ $cm\, s ^{-1}$

Answer 1

Correct Answer: 3

We are given a concave mirror with focal length \( f = 10 \, \text{cm} \) and an object moving towards the mirror with speed \( V_0 = 15 \, \text{cm/s} \) relative to the laboratory frame. We are tasked with finding the speed of the mirror, \( V_m \), such that the image is instantaneously at rest relative to the laboratory frame when the object is at a distance \( u = 30 \, \text{cm} \) from the mirror. The object forms a real image at this position. We will use the mirror equation and apply the principle of relative motion to solve this problem.

1. Mirror Equation:
The mirror equation relates the object distance \( u \), the image distance \( v \), and the focal length \( f \) of the concave mirror:

\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] where: - \( u \) is the object distance from the mirror, - \( v \) is the image distance from the mirror, - \( f \) is the focal length of the mirror.

2. Setting up the Scenario:
We are told that when \( u = 30 \, \text{cm} \), the image is at rest with respect to the laboratory frame. This means that the relative velocity between the image and the mirror is zero at this moment. Therefore, the rate of change of the image distance \( v \) with respect to time must be zero. This condition allows us to find the speed of the mirror \( V_m \).

3. Relating Speeds Using Differentiation:
To find the relationship between the speeds, we differentiate the mirror equation with respect to time. Using the chain rule, we have:

\[ 0 = \frac{d}{dt} \left( \frac{1}{u} + \frac{1}{v} \right) = -\frac{1}{u^2} \frac{du}{dt} - \frac{1}{v^2} \frac{dv}{dt} \] where \( \frac{du}{dt} \) is the velocity of the object relative to the mirror, and \( \frac{dv}{dt} \) is the velocity of the image relative to the mirror. Since \( \frac{dv}{dt} = 0 \) (because the image is at rest with respect to the laboratory frame), we get: \[ 0 = -\frac{1}{u^2} V_0 - \frac{1}{v^2} V_m \] where \( V_0 = 15 \, \text{cm/s} \) is the velocity of the object towards the mirror and \( V_m \) is the velocity of the mirror.

4. Applying the Mirror Equation at \( u = 30 \, \text{cm} \):
At \( u = 30 \, \text{cm} \), we can find \( v \) from the mirror equation. Since the mirror forms a real image, the image distance \( v \) will be positive. Substituting \( u = 30 \, \text{cm} \) and \( f = 10 \, \text{cm} \) into the mirror equation:

\[ \frac{1}{10} = \frac{1}{30} + \frac{1}{v} \] Solving for \( v \), we get: \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} \Rightarrow v = 15 \, \text{cm} \]

5. Substituting Values to Find \( V_m \):
Now, we substitute \( u = 30 \, \text{cm} \), \( v = 15 \, \text{cm} \), and \( V_0 = 15 \, \text{cm/s} \) into the equation:

\[ 0 = -\frac{1}{30^2} \times 15 - \frac{1}{15^2} \times V_m \] Simplifying this equation: \[ 0 = -\frac{1}{900} \times 15 - \frac{1}{225} \times V_m \] \[ 0 = -\frac{15}{900} - \frac{V_m}{225} \] \[ \frac{V_m}{225} = -\frac{15}{900} \] \[ V_m = -\frac{15}{900} \times 225 = -\frac{15 \times 225}{900} = -\frac{3375}{900} = -3.75 \, \text{cm/s} \]

Final Answer:
The magnitude of the mirror's speed is \( |V_m| = 3.75 \, \text{cm/s} \), which corresponds to option 3.