When a girl moves towards a plane mirror at a speed of 4.0m/s, the distance between the girl & her image reduces at a speed of

Analysis

This question tests your understanding of the properties of plane mirrors, specifically the relationship between the motion of an object and the motion of its image.

Concept

In a plane mirror, the image is always formed at the same distance behind the mirror as the object is in front of it.

• If the object moves towards the mirror at a speed $v$, the image also moves towards the mirror at the same speed $v$.
• However, the question asks for the speed at which the distance between the girl and her image reduces. Since both the girl and the image are moving towards each other, their relative speed is the sum of their individual speeds.

Step-by-Step Solution

1. Identify the speed of the girl ($v_g$): The girl moves towards the mirror at $4.0\text{ m/s}$.
2. Identify the speed of the image ($v_i$): Because the image moves at the same speed as the object relative to the mirror, the image also moves towards the mirror at $4.0\text{ m/s}$.
3. Calculate the relative speed: Since the girl and the image are moving towards each other, the rate at which the gap between them closes is the sum of their speeds: $$\text{Relative speed} = v_g + v_i$$ $$\text{Relative speed} = 4.0\text{ m/s} + 4.0\text{ m/s}$$ $$\text{Relative speed} = 8.0\text{ m/s}$$

Conclusion

The distance between the girl and her image reduces at a speed of $8.0\text{ m/s}$.

Exam Tip

Always remember the "Double Rule" for plane mirrors:

• If an object moves towards a mirror at speed $v$, the image moves towards the object at $2v$.
• If you rotate a plane mirror by an angle $\theta$, the reflected ray rotates by $2\theta$. Keeping these "doubling" effects in mind will help you solve similar mirror problems quickly in the exam hall!