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GR9677 #57
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Alternate Solutions |
scottopoly
2006-11-03 19:39:18 | Use Raleigh's criterion! |  |
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Comments |
mianghazanfar786
2010-07-20 18:17:20 | There is a confusion that "one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.)"
We should use directly for diffraction pattern,
1 for minima and 1/2 for maxima so we use m=1
to solve the question by Bragg's Diffraction law |  | archard
2010-05-30 13:20:12 | If all you knew was that minima occur at half integers, you could make a wild guess by noticing that C and D differ by a factor of a half, and since ETS is always trying to trick you it's probably one of the two, and most likely C because 1 is half of 2. | | sullx
2009-11-02 18:39:02 | Careful! The value 'm' corresponds to Minimum for single slit diffraction, and maximum for double slit interference.
Yosun: "The single slit diffraction formula is d \sin \theta = \lambda m, where one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.) "
niux
2009-11-05 14:15:02 |
Gee. Seems Yosun using different semantics?. I agree with RootMeanSquare explanation, and for single slit, m should tell the number of minima you are dealing with.
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| | tan
2009-10-23 23:09:13 | Central maximum occurs at  = 0
First minimum occurs at d sin =  /2
Answer should be 5 x  .
What's wrong? | | boundforthefloor
2006-11-24 23:43:14 | Wolfram explains this well giving minima at  = m\lambda/d) and the primary maxima at 0.
see:
http://scienceworld.wolfram.com/physics/FraunhoferDiffractionSingleSlit.html
physicsisgod
2008-10-29 22:12:18 |
Yea, you can think of it like the center max is actually at  , and then the optical path difference between a ray diffracted at an angle at the center (), and one at 0, is  if there's a minimum (destructive inteference), and that has to be equal to  .
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| | RootMeanSquare
2006-11-24 14:31:54 | Okay, I'll never remember that I don't have to log in before posting...
Anyway: It seems here is a little confusion about the factor 2.  gives you the optical retardation between waves originating a distance d apart within the slit. For a full minimum, you need to find a pair of beams that cancel each other out for every point within the slit, i.e. you actually need a optical retardation of  for point  apart. Hence you end up with  as the width of the slit.
mhas035
2007-03-25 21:54:27 |
Thanks! Some clarity!
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| | RootMeanSquare
2006-11-24 14:25:50 | | | scottopoly
2006-11-03 19:39:18 | Use Raleigh's criterion!
physicsisgod
2008-10-29 22:04:35 |
For those of you who are not as enlightened as our fortunate friend scottopoly, Raleigh's Criterion is a formula for deriving the distance between two point sources necessary to resolve them, given their angular separation,  , and the wavelength of emitted light, . The formula is
But Raleigh's Criterion only applies for a circular slit, thus the factor of 1.22. If you just use the regular old equation for the distance between the center maximum and the first minimum given a single slit diffraction,
 ,
and the small angle approximation, you'll derive the same formula without the 1.22 factor
Source: http://www.ux1.eiu.edu/~cfadd/3050/Ch20WO/OpRes.html
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apr2010
2010-04-06 14:20:00 |
Very nice, also makes me remembering the raileigh criterion!
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| | buddy.epson
2006-10-13 16:01:42 | Re: Previous User Comment
For single slit diffraction, m corresponds to the center of the dark band (minima). For multiple slits and x-ray diffraction, it corresponds to the center of the bright bands (maxima). The posted solution is correct.
pablojm
2006-10-28 14:39:50 |
In fact, nahmad is right and the problem is incorrectly worded. You can check in any optics book that the formula for minima is d(theta)=m(lambda), where m is a half-integer. Plugging in m=0 and the given values for (lambda) and (theta), you get answer B. The mistake in the original solution is that you stated that m should be a half-integer for minima, and then you plugged in m=1.
Cheers,
Pablo
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f4hy
2009-04-03 17:37:10 |
The problem states angle between the first minimum and the central max. Not the angle between two mins.
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| | nahmad
2006-03-31 22:40:22 | I don't understand. The problems gives the angle between the first minimum and the central max. Doesn't that correspond to m=1/2 which leads to d = 5 x 10^-5 which is B? |  |
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