# Homework Help: Problem with light

1. Jul 11, 2007

### Mekiel

Hello everyone, I'm knew to this forum. I was wondering if someone could help me solve this physics problem. I've searched my text book, the internet, and asked tons of people, but I still can't find an answer. Here's the question:

What is the speed of the image, relative to Dubbie, if Dubbie walk away from the mirror surface at 3.5 m/s at an angle of 30 degrees to the mirror surface?

Obviously theres a formula I need to use, but which formula is it?
:uhh:

2. Jul 11, 2007

### ice109

why wouldn't it be -3.5m/s

3. Jul 11, 2007

### Mekiel

My apologies, I didn't notice the sticky about homework questions. I'll post this in the appropriate forum.

My apologies again.

4. Jul 11, 2007

### Mekiel

Wouldn't the 30 degree angle make a difference in the speed? I mean, if it was just a flat/plane mirror, then I could see the speed being equal.

5. Jul 11, 2007

### Danger

Welcome to PF, Mekiel. The 'speed' of the image will be the speed of light through whatever refractive indices are provided by the atmosphere and the material of the mirror.

6. Jul 11, 2007

### Mekiel

Okay, so if the material is made of glass, and the atmosphere was oxygen, I would need to find the index of refraction for O2?

7. Jul 12, 2007

### ice109

i really think that has nothing to do with the problem

upon rereading the problem i think by symmetry alone you could reason that it's -7m/s

8. Jul 12, 2007

### Danger

So you have somehow managed to alter the speed of light from 300,000Km/sec to -7m/s? How'd you manage that? There might be a paper in the future.

9. Jul 12, 2007

### ice109

the question asks what is the speed of the image's recession relative to the object not how fast does the movement's information propagate. obviously if i pull something along the normal away from a mirror at 3.5m/s its image is not receding from at c.

10. Jul 12, 2007

### Danger

edit: Somehow, my post pointing out that only 23% or so of the atmosphere is oxygen, and it's primarily nitrogen, disappeared.

edit #2: Ice, I think that I just realized the difference in our approach to the problem. I was considering how fast the image responds to movement of the original; you were dealing with how fast the image moves in relation to its environment. A subtle, yet critical, difference. By the way that you're approaching the problem, I agree with you.

Last edited: Jul 12, 2007
11. Jul 12, 2007

### Mekiel

3x10^5km/second eh? I reckon my car can go faster =p. It's a '92 honda accord by the way.

So, I've been further researching and have come across various formulae such as snell's law, critical angle, lens-marker's formula, but I dont know how to implement them or if they are even useful for solving this.

Could someone shed some "light" on this.... Pardon the pun.

12. Jul 12, 2007

### Danger

We tried to, from 2 different perspectives. Depending upon how you meant the queston, one of them answered it.
And a '92 Honda is not a car; it's a roller skate with delusions of grandeur.

13. Jul 12, 2007

### Integral

Staff Emeritus
Danger, Danger :rofl: But please remember this is now in the Homework help forum.

Ok, now for the problem. First start by drawing a picture.

Decompose D's velocity into components. One parallel to the mirror the other normal to the mirror. From the diagram you should see that the image of D will have the same velocity parallel to the mirror but will be receding "into" the mirror at the same rate but in the opposite direction that D is receding from the mirror.

I have given to much already.

14. Jul 12, 2007

### andrevdh

If two objects are receeding from each other with a speed v their relative speed of separation would be 2v.

15. Jul 13, 2007

### belliott4488

That's where the 30 deg. angle comes in; v is a vector, not a speed (scalar). You need to consider vector arithmetic, as someone already hinted at above.

Last edited: Jul 13, 2007
16. Jul 13, 2007

### ice109

no you don't

edit

maybe you do, the question is very vague

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook