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

AI Thread Summary
The discussion centers on a physics problem regarding the wave nature of light, specifically water waves with parallel crests 2.5cm apart passing through two openings 5cm apart. Participants clarify that the wavelength is defined as the distance between crests, confirming it is 2.5cm. The calculation for the angle of little or no wave action is discussed, with one participant finding an angle of 14.48 degrees using the formula d sin(θ) = (m + 1/2)(wavelength). It is noted that the 2-meter distance does not need to be included in the angle calculation, as long as it is significantly larger than the wavelength. The conversation emphasizes the importance of understanding the relationship between distance, wavelength, and angle in wave phenomena.
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?
 
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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 d sin (\theta) = (m + \frac {1}{2} )(0.025m) where d = 0.05m, and m = 0 to find \theta.
 
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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 d sin (\theta) = (m + \frac {1}{2} )(0.025m) where d = 0.05m, and m = 0 to find \theta.

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 \theta and \theta^\prime that are only approximately equal.

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