# Fraunhofer diffraction pattern

• says
In summary: There is a way to get a better approximation than ## \Delta x=\frac{L \, \lambda}{b} ## for ## L<45 ## cm. But you would have to do a little bit of work to get it. ## \\ ## If you were to derive the actual form of the equations for ## \Delta \theta ## and ## \Delta x ## that are valid in the near field, you would find that the equations are no longer the same. The equations for ## \Delta \theta ## and ## \Delta x ## are not the same for the near field as for the far field. ## \\ ## The equations are derived from the Huygens-F
says

## Homework Statement

A square aperture with a side of length 0.5 mm is illuminated with light of wavelength 550 nm. At what distance from the aperture would the Fraunhofer diffraction pattern have a central maximum with a width also equal to 0.5 mm? What can you say about the Fraunhofer condition under these circumstances?

## Homework Equations

W2 / Lλ << 1

where
W : aperture or slit size
L : distance from aperture
λ : wavelength

3. The Attempt at a Solution

0.5mm = 0.0005m
550 nm = 5.5*10-7m

L = W2 / λ
= (0.0005m)2 / 5.5*10-7m
= 45 cm - distance at which Fraunhofer diffraction pattern is observed

I'm a bit lost with this question / Fraunhofer diffraction. I think I may have used the wrong equation maybe. Any help would be much appreciated.

What you did is ok. If you notice in your calculation, since ## \Delta x=L \, \Delta \theta=\frac{L \lambda}{b}=b ##, the conditions for this problem are ## \frac{b^2}{L \lambda}=1 ##. What this is saying is that although you computed ## L= 45 ## cm, this answer is unreliable, because the formula that you used for this case, that ## \Delta x=\frac{L \lambda}{b} ## only works for ## \frac{b^2}{L \lambda}<<1 ##, which means we must have ## L>>\frac{b^2}{\lambda} ##. If not, the answer is unreliable. (And we don't have ## L>> \frac{b^2}{\lambda} ##, because our calculated ## L=\frac{b^2}{\lambda} ##). ## \\ ## Clearly, the bright spot produced by this pattern must be at least as big as the aperture. (The Fraunhofer equation, ## \Delta x=\frac{L \, \lambda}{b} ##, would tell us for ## L< 45 ## cm, that the bright spot is smaller than the aperture, i.e. ## \Delta x <b ## , which is clearly wrong). The Fraunhofer formulas do not work in this case for ## L \lesssim 45 ## cm. The Fraunhofer equations work in the far-field. In this case, ## L=45 ## cm is in the near field. The far field would begin somewhere around ## L \approx 10 ## meters, or thereabouts for this problem. ##\\ ## Note: I'm using ## b ## for aperture width, in place of ## w ##.

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The only other equation I have found is:
df = 2λz / W
where
df=width of central band in diffraction pattern
λ=wavelength
z=viewed at a distance of...
W=slit width

z = (0.5mm)(0.5mm)2 / 2(550nm) = 22.72 cm

This equation is for a slit of infinite depth. I'm not totally sure how to convert it to a square aperture...

says said:
The only other equation I have found is:
df = 2λz / W
where
df=width of central band in diffraction pattern
λ=wavelength
z=viewed at a distance of...
W=slit width

z = (0.5mm)(0.5mm)2 / 2(550nm) = 22.72 cm

This equation is for a slit of infinite depth. I'm not totally sure how to convert it to a square aperture...
Please read my post above. The factor of 2 that you have is of little consequence. Depending how you choose to measure the width, you can get an extra factor of two. That's of little concern in what is going on in the problem that you have.

What is Δx in your working?

says said:
What is Δx in your working?
says said:
What is Δx in your working?
For a single slit diffraction pattern in the far field, ## m \lambda =b \sin{\theta} ##, with integer ## m \neq 0 ##, gives the positions of zero intensity. ## \\ ##For ## \sin{\theta} \approx \theta ##, the angular spread between ## m=-1 ## and ## m=+1 ##, which is the (approximate) width of the central maximum, is ## \Delta \theta=\frac{2 \lambda}{b} ##. This corresponds to a distance ## \Delta x=\frac{2 L \, \lambda}{b} ##. ## \\ ## When estimating this width ## \Delta x ##, sometimes the factor of 2 in the numerator is omitted. ## \\ ## If you use the formula with a factor of 2, it complicates the analysis a little bit, but the same conclusions are reached in that the formula for ## \Delta x ## is a far-field formula that doesn't work in the near field.

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so Δx= Lλ / b will give us the distance between m=0 (central maximum) and m=±1

says said:
so Δx= Lλ / b will give us the distance between m=0 (central maximum) and m=±1
The central maximum is a bright blob that peaks in the center and falls off to zero at ## m=+1 ## and ## m=-1 ##. ## \\ ## It is somewhat arbitrary how you estimate the width of this bright blob. Picking it as half the width of the zero points should seem reasonable.

I still get an answer of 45.45cm when I solve for L...

says said:
I still get an answer of 45.45cm when I solve for L...
That's ok. What your equations say though is if you let ## L=10 ## cm, that the bright blob will be smaller than the aperture, and that is impossible. ## \\ ## The purpose of the problem is to figure out why the equations are unreliable. The reason is that the equations you are using are only accurate for computing the size of he blob if the screen is far away from the source. ## \\ ## There is actually a very special case with a precisely circular aperture where I think you can create a Fresnel zone plate where the central peak might be smaller than the aperture, but that doesn't normally occur. See also: http://vanderbei.princeton.edu/images/Questar/PoissonSpot.html

So to answer the question I would say the distance would be approximately 45 cm, however, the Fraunhofer condition under these circumstances are unreliable, because if let the distance be shorter the bright blob will be smaller than the aperture, which is impossible

says said:
So to answer the question I would say the distance would be approximately 45 cm, however, the Fraunhofer condition under these circumstances are unreliable, because if let the distance be shorter the bright blob will be smaller than the aperture, which is impossible
Yes. The Fraunhofer condition, for where these diffraction equations are accurate is ## \frac{b^2}{L \lambda} << 1 ##. This means we are required to have ## L>> 45 ## cm for the equations to be reliable. ## \\ ## ## L ## really needs to be ## L \gtrsim 10 ## meters for your equation for ## \Delta x ## to be reliable.

says

## 1. What is Fraunhofer diffraction pattern?

Fraunhofer diffraction pattern is a phenomenon that occurs when a coherent light wave passes through a narrow slit and produces a pattern of bright and dark fringes on a screen placed at a distance from the slit.

## 2. How is Fraunhofer diffraction pattern different from Fresnel diffraction pattern?

The main difference between Fraunhofer diffraction pattern and Fresnel diffraction pattern is the location of the light source in relation to the diffracting object. Fraunhofer diffraction occurs when the light source is far away from the diffracting object, while Fresnel diffraction occurs when the light source is close to the diffracting object.

## 3. What factors affect the shape and size of Fraunhofer diffraction pattern?

The shape and size of Fraunhofer diffraction pattern are affected by the wavelength of the light, the size of the diffracting object, and the distance between the diffracting object and the screen where the pattern is observed.

## 4. What is the importance of Fraunhofer diffraction pattern in scientific research?

Fraunhofer diffraction pattern is important in scientific research as it allows us to study the properties of light, such as wavelength and intensity, and can also provide information about the size and shape of diffracting objects.

## 5. Can Fraunhofer diffraction pattern be observed with other types of waves besides light waves?

Yes, Fraunhofer diffraction pattern can also be observed with other types of waves, such as sound waves and water waves. This phenomenon occurs whenever waves pass through a narrow aperture or around an obstacle.

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