Giancoli: Wave Nature of Light: Parallel Crests 2.5cm Apart

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Homework Help Overview

This discussion revolves around a problem related to the wave nature of light, specifically involving water waves with parallel crests and their behavior as they pass through two openings. The original poster expresses difficulty in visualizing the concept of "parallel crest water waves" and seeks clarification on the problem's setup.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of wavelength and its relation to the distance between crests. There are inquiries about how to calculate the wavelength and the angle of wave action, with some participants attempting to apply relevant equations.

Discussion Status

Some participants have provided calculations for the angle of wave action, while others question the relevance of the distance provided in the problem. There is an ongoing exploration of the assumptions made in the calculations and the implications of the distance on the results.

Contextual Notes

Participants note the importance of the distance being significantly greater than the wavelength for the approximations used in their calculations to hold true. There is also a reference to visual aids that may help in understanding the wave behavior.

leolaw
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This question is related to the wave nature of light from Giancoli:

Water waves having parallel crests 2.5cm apart pass through two openings 5cm apart in a board. At a point 2.0m beyond the board, at what angle relative to the "straight-through" direction would there be little or no wave action?

The reason I am having a problem with this question is because I cannot visualize what a "parallel crest watever wave" is. Is it like the one in this picture (http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/sinslitwid.gif) ?
 
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Yes, those are paralllel crests.
 
how can i find the wavelength of the water wave?
Is it simply just 1/2.5cm or 40m?
 
Last edited:
what's the definition of wavelength?
 
distance from crust to crust, gottcha
 
I find the angle to be 14.48 degree, but I didn't use the 2 meters that the question provdied. Am I missing something?

I use [tex]d sin (\theta) = (m + \frac {1}{2} )(0.025m)[/tex] where d = 0.05m, and m = 0 to find [tex]\theta[/tex].
 
Last edited:
leolaw said:
I find the angle to be 14.48 degree, but I didn't use the 2 meters that the question provdied. Am I missing something?

I use [tex]d sin (\theta) = (m + \frac {1}{2} )(0.025m)[/tex] where d = 0.05m, and m = 0 to find [tex]\theta[/tex].

Since the problem asked you for the angle, and not a distance from the midpoint, you do not need the 2m as long as that distance is much greater than a wavelength. If the distance were only a couple of wavelengths, the approximations used to derive the equation you used would not be justified. Look carefully at this diagram and you will see that there are two angles [itex]\theta[/itex] and [itex]\theta^\prime[/itex] that are only approximately equal.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html
 

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