GREPhysics.NET: GR8677 #100

GR8677 #100

Problem

GREPhysics.NET Official Solution

Optics $\Rightarrow$ }Diffraction

The key equation involved is the (Fraunhoffer) diffraction equation, $\sin \theta = \lambda/d$ , where the obvious quantities are used.

For small angle, the equation simplifies to $\theta \approx \lambda/d$ .

The blur would be due to both the size of the hole and the diffraction. The arclength of the diffraction is approximated by $d_2=D(2\theta) = 2D\lambda/d$ , where the factor of 2 comes from the fact that the arclength angle is symmetrical about the diffraction axis, thus twice the diffraction angle. Define a blur equation $D'=d+d_2=d+2D\lambda/d$ . Take the first derivative, with respect to the pinhole diameter, set it to 0, to get $dD'/dd = 1-2D\lambda/d^2=0\Rightarrow d=\sqrt{2D\lambda}\approx \sqrt{D\lambda}$ , as in choice (A).
\par

This solution is due to: \begin{verbatim} http://www.exo.net/~pauld/summer_institute
/summer_day3eye_and_brain/pinhole_optimum_size.html
\end{verbatim}

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Comments

thebigshow500
2008-10-14 09:44:48

Not sure if this make sense: Can we simply apply single slit diffraction's equation?
1. $d\sin \theta = m\lambda$

Let m=1 to let the sharp image reach its 1st order minima, and then we have
2. $\sin \theta \approx \tan\theta = d/D$

For the Fraunhoffer diffraction, $\lambda \ll D$ so that we can assume a small angle.

Plug 2. back to 1., and we eventually get
$d (d/D) = \lambda$
$d=\sqrt{D\lambda}$

medellin
2008-11-04 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
2008-11-04 13:02:53

You need FRESNEL DIFFRACTION as mentioned below.

tinytoon
2008-11-06 00:04:52

Actually, your solution makes perfect sense, physically also.

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Richard
2007-11-01 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.

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hungrychemist
2007-09-17 08:00:32

Use order of magnitude approach. Say, if we set $\lambda$ = 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.

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yosun
2005-10-31 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
2005-10-31 12:47:49

Shouldn't Fresnel Diffraction be used instead, as $D<<1$ ?

ee7klt
2005-11-11 04:44:44

Hi Alpha,

I think the question says $\\lambda << D$ , thus justifying the Fraunhofer regime.

I have one question, why is reasoning behind defining the blur equation as written?

ee7klt
2005-11-11 04:45:27

oops..I meant "what is the reasoning.."

student2008
2008-10-14 07:38:54

the condition $\lambda \ll D$ is needed to neglect the second term under the square root in the Fresnel zones formula: $R_n=\sqrt{n\lambda D + n^2 \frac{\lambda^2}{4}}$

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Fresnel Diffraction would complicate things. Franhouffer should be a good approximation, especially when one has only about a minute to answer the question.

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