Physics optics問題

When Tommy moves at a speed of 2 m/s, its image moves at the same speed towards him. This result is apparent as according to the Law of Reflection, the distance of the object from the mirror equals to the distance of image from the mirror. Hence, when both object and image move at the same speed towards the mirror, the same distances from the mirror will be maintained.

When a mirror moves towards an object, the image moves at twice the speed of the mirror towards the object (you could prove this result easily or make reference to standard physics textbooks). Thus, when the mirror moves at 1 m/s towards Tommy, his image moves at 2 m/s towards him.

Combining the effect of the two movements, the image of Tommy will then moves at a speed of (2 + 2) m/s = 4 m/s towards him.

I will use mathematics to solve this problem.

Assume that at the beginning, the mirror is placed at the origin O and Tommy is 10m(x = +10) on the right of the mirror. Then his image is 10m on the left of the mirror (x = -10)

After 1 second, Tommy goes to x = +8 and the mirror goes to x = +1

The distance between Tommy and mirror is 7m. So, his image will be appeared at x = 1 - 7 = -6.

This means that the speed of the mirror is -6 - (-10) = 4m/s

You can let the object distance and image diatance be u and v respectively at a moment.

After 1s, object distance becomes u-3 and so image distance becomes v-3

In this way, Tommy moved 2m in 1s and the distance between Tommy and his image shortened by 6m, so the image moved 4m in 1s

Therefore the speed of image is 4m/s