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An aperture of size $a$ illuminated by a parallel beam sends diffracted beam (the central maximum) in angular width approximately $\lambda/a$. Travelling a distance $z$, it acquires the width $z \lambda / a$ due to diffraction. The distance at which this width equals the size of the aperture is called the Fresnel distance.

As far as I know, the angular width of central maximum is given by $2 \lambda / a$ and not $\lambda/a$.

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    $\begingroup$ There's no question here. $\endgroup$
    – DanielSank
    Commented Jun 2, 2016 at 6:10
  • $\begingroup$ @DanielSank I think OP wants to ask , why don't we equate (2(lambda)z/a with a so that we will get z=a^2/2lamda and not a^2/lambda which is correct equation for fresnel distance. HOPE YOU UNDERSTAND PLEASE! $\endgroup$
    – JM97
    Commented Jun 2, 2016 at 14:55

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This is really just a geometry problem since we can not use Fraunhofer approximation in general. The solution depends on what you mean by "width" of the central maximum. If you take it to be the distance between the two adjacent minima (full width) then the solution is:

$$z_{\text{Fresnel}} = \frac{a^2}{4 \lambda} - \frac{\lambda}{4}$$

Alternatively the full width half maximum would probably have a factor of 2 in front or something like that. Note that for $\lambda \ll a$ the result becomes what LC7 wrote in his answer (who calculated full width half maximum), but it is not the most general form.

Here is my working. The crucial step is that at the minimum the interference is fully destructive, so for every point in the aperture exists another point in the aperture that is $\pi$ out of phase. Hence the phase difference at the minimum between edge of the aperture and the centre of the aperture is $\pi$ (see picture). The rest is then just doing the geometry.

enter image description here

EDIT in response to comments

I will try to give an explanation of how this fits with the question.

The distance at which this width equals the size of the aperture is called the Fresnel distance.

I calculated this distance above, fixing the ambiguous "width" to be the full width, as I explained (as opposed to full width half maximum). My derivation does not make any approximation other than scalar diffraction theory.

An aperture of size $a$ illuminated by a parallel beam sends diffracted beam (the central maximum) in angular width approximately $\lambda/a$. Travelling a distance $z$, it acquires the width $z \lambda / a$ due to diffraction.

The OP talks about "beams" here. This is only valid in the Fraunhofer approximation, i.e. the far field limit. Why? Interference/diffraction causes a complicated wave field behind the aperture. Far away from it it looks a bit simpler, since only the linear phase terms will be significant. Now linear phase terms give you plane waves. These are the "beams" you are talking about. However the Fresnel distance does not lie in the Frauhofer regime region!! So unless you make further approximations this is not a valid approach.

As far as I know, the angular width of central maximum is given by $2 \lambda / a$ and not $\lambda/a$.

Again this is a property of the diffraction pattern in the Fraunhofer limit and not true in general, which is where the whole misconception arises.

Comment about the comments

From what I gather from the comments people are just trying to find a derivation of some particular formula they learnt in some course. If that is the case I think the question should be closed. I think I have explained all conceptual aspects of the problem and am open to further suggestions about how to improve that, but I won't go into throwing around formulae without concepts.

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  • $\begingroup$ why the downvote? is the answer incorrect? $\endgroup$ Commented Jun 1, 2016 at 11:40
  • $\begingroup$ @Numork Please have a look on JM97's comment below the question. $\endgroup$
    – user102917
    Commented Jun 4, 2016 at 16:29
  • $\begingroup$ @physics In my answer I defined all the quantities I computed in a precise manner, so I'm pretty sure there is nothing wrong in my answer. Of course the word "Fresnel distance", which is really just a quantification of the geometric limit distance (e.g. see this question: physics.stackexchange.com/questions/90854/…) may be defined differently, but the OP didn't state that in his question. Statements like JM97's comment there and also under the answer by LC7 don't make much sense. They would if he did define the Fresnel distance precisely. $\endgroup$ Commented Jun 4, 2016 at 16:58
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    $\begingroup$ @Vilvanesh Good point! The answer is simple though: If you consider any one point in the aperture, there is always a second point that fulfils the destructive interference condition in the answer. On the screen, this gives an offset. Since we are calculating the width, the calculation therefore gives at least a first order estimate. $\endgroup$ Commented Aug 11, 2020 at 9:06
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    $\begingroup$ @Vilvanesh I suggest you try doing the integral to fin the intensity at that point. Hope that gives some insight. $\endgroup$ Commented Aug 14, 2020 at 13:04
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"As far as I know, the angular width of central maximum is given by 2λ/a and not λ/a."

The half-power angular width of the main beam is λ/a radians. That's why it is preferred.

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It is $\frac{\lambda}a$ because we are talking about only two rays, so the formula of double slit fringe width is more appropriate. If there were a vast no. of rays, $\frac{2\lambda}d$ would be correct. As told by Feynman the only significant difference between diffraction and interference is the number of light sources at slit.

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$I(\theta)= I_{max} \left[ \dfrac{\sin\left(\frac{\pi a \sin \theta}{\lambda}\right)}{\frac{\pi a \sin \theta}{\lambda}}\right]^2$

where $a$ is the size of aperture. This is the diffraction pattern, now you can easily see that the minimum of diffraction are where $\sin \theta = m \lambda /a$. So as you say the angular width of central maxima is $2 \lambda /a$, in fact from the relation $\Delta x = z \Delta \theta$ with $z$ travelling distance and $\Delta x$ width of the central maximum on the detector, in approximation $a\gg \lambda \Rightarrow \Delta \sin \theta \simeq \Delta \theta = 2 \lambda /a$.

We obtain that $\Delta x=z 2 \lambda /a$. Now the Fresnel distance is where $\Delta x = a$ i.e.:

$a=2 z \lambda /a \Rightarrow z_{fresnel}= \frac {a^2}{2 \lambda}$

That is the final result.

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  • $\begingroup$ Fresnel distance is (a^2)/lamda and not half of it $\endgroup$
    – JM97
    Commented Jun 1, 2016 at 7:42
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    $\begingroup$ @LC7 -1 this is incorrect, the diffraction pattern you quote is only valid in the far-distance approximation, i.e. the Fraunhofer regime. The Fraunhofer approximation is valid when $L>>a^2/\lambda$ (L being the screen aperture distance) which the $z_fresnel$ clearly does NOT fulfill at all. Hence the Fraunhofer approximation is not appropriate here. $\endgroup$ Commented Jun 1, 2016 at 8:36

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