- #1

FAS1998

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Would any other answer of the form (13.4 + 2npi) also be correct, assuming n is a non-zero integer? Or is 13.4 the only correct answer because it tells you how many wavelengths one wave is ahead of the other?

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- Thread starter FAS1998
- Start date

- #1

FAS1998

- 50

- 1

Would any other answer of the form (13.4 + 2npi) also be correct, assuming n is a non-zero integer? Or is 13.4 the only correct answer because it tells you how many wavelengths one wave is ahead of the other?

- #2

Merlin3189

Homework Helper

Gold Member

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But we would say for all such paths that the phase difference was ## \frac {π}{3} ##

In your question you know the actual paths and the path difference. So I'd think you are being asked to say what phase difference an observer at the given point will see. They will not be able to tell how many whole waves difference there is between the paths. All they will see is the difference in phase within one cycle.

So you're probably near enough there. But your answer does not tell me the phase difference! What phase is 13.4 ?

If it's 13.4 radians, that's odd, because that's more than 2π.

Maybe it's 13.4 degrees? That's less than 2π radians, so it could be a phase difference. But you need to say.

Maybe it's a distance? If so you need to convert it into a phase. If you tell me the path difference is 13.4 cm and the wavelength is 53.6cm, then I could work out that the phase is ## \frac{2π}{4} = \frac {π}{2} or 90^o ## but so could you, and you should.

Maybe it's 13.4 msec ? Then the same applies. finish the calculation and give a phase.

Edit: Sorry, I'm assuming in the first paragraph that I'm working in radians. I should have said (just as I tell you to do!)

- #3

FAS1998

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The answer was in radians. For context, the homework problem was online, had automatic grading, and said my answer of 13.4 radians was correct. The equation I used was P = 2(pi)L/λ, where L is the difference in path lengths and P is the phase difference. Plugging in gave me an answer of 13.4. I’m not sure if the automatic grading would’ve accepted other answers.

But we would say for all such paths that the phase difference was ## \frac {π}{3} ##

In your question you know the actual paths and the path difference. So I'd think you are being asked to say what phase difference an observer at the given point will see. They will not be able to tell how many whole waves difference there is between the paths. All they will see is the difference in phase within one cycle.

So you're probably near enough there. But your answer does not tell me the phase difference! What phase is 13.4 ?

If it's 13.4 radians, that's odd, because that's more than 2π.

Maybe it's 13.4 degrees? That's less than 2π radians, so it could be a phase difference. But you need to say.

Maybe it's a distance? If so you need to convert it into a phase. If you tell me the path difference is 13.4 cm and the wavelength is 53.6cm, then I could work out that the phase is ## \frac{2π}{4} = \frac {π}{2} or 90^o ## but so could you, and you should.

Maybe it's 13.4 msec ? Then the same applies. finish the calculation and give a phase.

Edit: Sorry, I'm assuming in the first paragraph that I'm working in radians. I should have said (just as I tell you to do!)

Would it be incorrect to write a phase difference of pi/3 as 7pi/3, or just unconventional?

- #4

DaveE

Science Advisor

Gold Member

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Just unconventional, IMO; not technically wrong. However, it is unconventional enough to confuse people, I think.Would it be incorrect to write a phase difference of pi/3 as 7pi/3, or just unconventional?

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