GR8677 #100



Alternate Solutions 
Simplicio 20090331 18:51:18  I would agree with hungrychemist and consider the order of magnitude.
The reasonably values I used are
,
(A) ≈
(B) : Good luck making a camera pinhole the size of the wavelength ...
(C) : This is even worse!
(D) : LOL!
(E) : Can this even be called a "pin"hole?
So by elimination, (A) gives a small yet reasonable number!  

Comments 
calvin_physics 20140327 22:39:05  Just like others have suggested,
answers from B to E are simply silly if we apply some logic to it.
Visible lights has wavelength 400~700nm
Distance for an image is usually few meters.
Only (A) gives few millimeter as answer, which fits the description of pinhole.   syreen 20130923 18:45:05  I used rayleigh criterion and small angle approx:
Rayleigh
sin(theta)=~theta=lambda/d
small angle
theta~=arclength/distance~="blur"/D
Set the two theta equations equal, solve for blur, then
derivative of blur with respect to d, set to 0 to minimize blur. Solve for d. gives A.   mpdude8 20120419 17:18:46  Yeah, bag all the optics and look at it logically. If you know the wavelength of light is on the order of nanometers, a pinhole, while small, is probably on the order of millimeters, and the screen, probably on the order of centimeters, you can use orders of magnitude to figure out which answers are possible and which are nonsense.   rizkibizniz 20111007 02:10:00  tho im quite sure Yosun's answer is more accurate, but all this PGRE stuff is about approximation isn it? can someone point out why its a big mistake to approximate using Rayleigh's criterion?   rizkibizniz 20111007 02:01:41  couldn' we approximate all this just using Rayleigh's criterion? for sharp images, (to be able to distinguish two separate objects with small angle ) :
,
,
Setareh 20111029 01:07:31 
Nice!

rizkibizniz 20111106 21:28:13 
Hmm. I come to realize that I have mistakenly defined which doesn seem to be the correct reasoning to get the angle subtended by the objects through the pinhole on the screen which is what is.rnrnWhat I had (mistakenly) defined was the angle subtended by lines from the edges of the pinhole to a point on the center of the screen. rnrnI think it might be a coincidence that the answers came out to be similar if not the same. Or can they both be in fact the same to some certain approximations? Anyone have any thoughts on this?

livieratos 20111108 07:30:56 
i think that this is the case:
somebody correct me if i'm wrong please...

ewcikewqikd 20140719 12:41:00 
The problem with ur solution is that the "d" in the single slit equation is the width of the slit (pinhole), not the width of the screen. The width of the pinhole is never given in the problem.

  Simplicio 20090331 18:51:18  I would agree with hungrychemist and consider the order of magnitude.
The reasonably values I used are
,
(A) ≈
(B) : Good luck making a camera pinhole the size of the wavelength ...
(C) : This is even worse!
(D) : LOL!
(E) : Can this even be called a "pin"hole?
So by elimination, (A) gives a small yet reasonable number!
flyboy621 20101115 20:48:43 
Nice!

  thebigshow500 20081014 09:44:48  Not sure if this make sense: Can we simply apply single slit diffraction's equation?
1.
Let m=1 to let the sharp image reach its 1st order minima, and then we have
2.
For the Fraunhoffer diffraction, so that we can assume a small angle.
Plug 2. back to 1., and we eventually get
medellin 20081104 13:01:09 
I think you got the answer just by chance, but there is not really physics there. I think your procedure got the answer just by the orders of magnitude involve but no more.

medellin 20081104 13:02:53 
You need FRESNEL DIFFRACTION as mentioned below.

tinytoon 20081106 00:04:52 
Actually, your solution makes perfect sense, physically also.

h.fei10 20121107 22:14:23 
I think, in your second equation, you should use the radius of the blur, not just the diameter of the pinhole. It doesn't make sense.

ewcikewqikd 20140719 12:38:12 
The problem with ur solution is that the "d" in the single slit equation is the width of the slit (pinhole), not the width of the screen. The width of the pinhole is never given in the problem.

  Richard 20071101 18:55:36  I did this one by relying on my intuition, namely, that if we increase the distance the screen is from the hole, the optimum diameter of the hole should increase, and also that if you let the wavelength become very large
the diameter of the hole should increase to get a resolved image. The only solution that fits this is (A).
For those wondering what is meant by "blur," it may be useful to look at the site Yosun quoted/pirated. "Blur" is basically a quantification of the interference on the image, how much "blur" there is. There is interference due to incident light entering the hole and interference due to diffraction around the hole's edges. These combine as Yosun quotes and the sum's minimum is obtained by a differentiation.
  hungrychemist 20070917 08:00:32  Use order of magnitude approach. Say, if we set = 1 unit of length. D is about 10^8 of that same unit. (Size of camera should be in order of m where the size of the visible light wave length is order of nm).rnrnNow, you want d (pin hole size) at the order of 10^2 or so. (About size of mm). rnrnChoise E is way too big(bigger than the camera)rnChoice D is way too smal(smaller than the wavelength of light)rnChoice C is still too small(still smaller than the wavelength of light)rnChoice B is same as wavelength of light(still quiet small, and we know from double slit experiment, this magnitude of hole will create dispersion of wave, which blurs things). rnChoice A is only reasonable magnitude.   yosun 20051031 01:14:19  Fresnel Diffraction would complicate things. Franhouffer should be a good approximation, especially when one has only about a minute to answer the question.
alpha 20051031 12:47:49 
Shouldn't Fresnel Diffraction be used instead, as ?

ee7klt 20051111 04:44:44 
Hi Alpha,
I think the question says , thus justifying the Fraunhofer regime.
I have one question, why is reasoning behind defining the blur equation as written?

ee7klt 20051111 04:45:27 
oops..I meant "what is the reasoning.."

student2008 20081014 07:38:54 
the condition is needed to neglect the second term under the square root in the Fresnel zones formula:

 




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