Path difference between the waves

AI Thread Summary
The discussion revolves around calculating the path difference between waves using trigonometric relationships. The initial approach involved forming a right triangle, but confusion arose regarding the correct allocation of angles and sides. Participants clarified that the path difference should be expressed using cosine rather than sine, as it relates to the leg along the wave's path. There was also a consensus that the problem did not require expressing the path difference in terms of wavelengths. The conversation concluded with an acknowledgment of the correct approach and understanding of the trigonometric relationships involved.
toforfiltum
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Homework Statement


upload_2015-11-2_20-59-1.png


Homework Equations

The Attempt at a Solution


I formed a right angle triangle between the arrow downwards,d, and perpendicular line x from arrow. As such, sin θ = x/d.

x= d sinθ
path difference: d sinθ/λ

My answer is C, but answer is B. Where did cos θ come from?
 
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Looking at the figure, in which case would you expect it to be no difference in path length?
 
Your description of how you formed your triangle is not very rigorous, so I may be misinterpreting what you've described. That said, I think you've allocated angle ##\theta## to the wrong corner of your triangle.

Perhaps you could sketch your construction on the image and post the result?
 
DrClaude said:
Looking at the figure, in which case would you expect it to be no difference in path length?
Is it when the waves are received at angle perpendicular to the horizontal?
 
toforfiltum said:
Is it when the waves are received at angle perpendicular to the horizontal?
Which corresponds to which value of θ?
 
DrClaude said:
Which corresponds to which value of θ?
90°?
 
gneill said:
Your description of how you formed your triangle is not very rigorous, so I may be misinterpreting what you've described. That said, I think you've allocated angle ##\theta## to the wrong corner of your triangle.

Perhaps you could sketch your construction on the image and post the result?
upload_2015-11-2_21-26-47.png

I did it this way, but I think now I'm wrong to assume that x is parallel to the wavelength, right?
 
You've drawn the correct triangle but as you say, you've chosen the wrong "leg" for x. You want the leg that lies along the wave's path.
Fig1.png
 
gneill said:
You've drawn the correct triangle but as you say, you've chosen the wrong "leg" for x. You want the leg that lies along the wave's path.
View attachment 91213
Ah, I see now why it's cos θ. But, why it is not D now? Shouldn't path difference be expressed in terms of fraction of wavelengths?
 
  • #10
toforfiltum said:
Ah, I see now why it's cos θ. But, why it is not D now? Shouldn't path difference be expressed in terms of fraction of wavelengths?
In this case the path difference is just the length, otherwise the problem would have specified to express it in terms of wavelengths.
 
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  • #11
gneill said:
In this case the path difference is just the length, otherwise the problem would have specified to express it in terms of wavelengths.
Ok. Thanks!
 
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