- #1

PFuser1232

- 479

- 20

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter PFuser1232
- Start date

In summary, when the width of a gap is smaller than the wavelength, diffraction effects become more pronounced and are at their peak when the gap width is equal to the wavelength of the diffracted waves. This is explained in the Cambridge International AS and A Level Physics Coursebook by David Sang, Graham Jones, Richard Woodside and Gurinder Chadha. The diffraction pattern formed in this scenario can be described using the equation for intensity of the single-slit diffraction pattern, and there will be zeros (dark bands) at angles where the ratio of the gap width to the wavelength is equal to 1. The central peak of the intensity distribution fills the "field of view" looking outwards from the slit, with the zero-intensity

- #1

PFuser1232

- 479

- 20

Science news on Phys.org

- #2

olivermsun

Science Advisor

- 1,268

- 136

MohammedRady97 said:I know that a smaller width of a gap would lead to "more diffraction", anddiffraction effects are at their peak when the wavelength of the diffracted waves is equal to the gap width.

Are you sure about that?

- #3

PFuser1232

- 479

- 20

olivermsun said:Are you sure about that?

Well, this is what my A level Physics book says.

- #4

olivermsun

Science Advisor

- 1,268

- 136

Would you please quote a bit of the passage or context from the Physics book?

- #5

jtbell

Staff Emeritus

Science Advisor

Homework Helper

- 15,959

- 6,156

Does the book tell you the equation that gives the angles at which minimum intensity occurs?

- #6

PFuser1232

- 479

- 20

jtbell said:Does the book tell you the equation that gives the angles at which minimum intensity occurs?

Not at all.

- #7

PFuser1232

- 479

- 20

PHP:

olivermsun said:Would you please quote a bit of the passage or context from the Physics book?

"As the gap becomes narrower, the diffraction effect becomes more pronounced. It is greatest when the width of the gap is equal to the wavelength of the ripples."

Cambridge International AS and A Level Physics Coursebook by David Sang, Graham Jones, Richard Woodside and Gurinder Chadha.

- #8

Malverin

- 139

- 7

MohammedRady97 said:PHP:

"As the gap becomes narrower, the diffraction effect becomes more pronounced. It is greatest when the width of the gap is equal to the wavelength of the ripples."

Cambridge International AS and A Level Physics Coursebook by David Sang, Graham Jones, Richard Woodside and Gurinder Chadha.

Here it is explained well

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1

- #9

olivermsun

Science Advisor

- 1,268

- 136

MohammedRady97 said:PHP:

"As the gap becomes narrower, the diffraction effect becomes more pronounced.It is greatest when the width of the gap is equal to the wavelength of the ripples."

Cambridge International AS and A Level Physics Coursebook by David Sang, Graham Jones, Richard Woodside and Gurinder Chadha.

What I'm questioning is the bolded part above. Think about the shape of the outgoing wavefront in the limit of (gap width) « wavelength.

Alternatively, examine the equation for the intensity of the single-slit diffraction pattern, given at Hyperphysics and also at Wikipedia:

[itex]I(\theta) = I_0 \, \mathrm{sinc}^2 \left(\dfrac{d\pi}{\lambda} \sin\theta \right),[/itex]

where

But why doesn't the spreading continue to increase for [itex]d < \lambda[/itex]?

- #10

olivermsun

Science Advisor

- 1,268

- 136

MohammedRady97 said:PHP:

"As the gap becomes narrower, the diffraction effect becomes more pronounced.It is greatest when the width of the gap is equal to the wavelength of the ripples."

Cambridge International AS and A Level Physics Coursebook by David Sang, Graham Jones, Richard Woodside and Gurinder Chadha.

What I'm questioning is the bolded part above. Think about the shape of the outgoing wavefront in the limit of (gap width) « wavelength.

Alternatively, examine the equation for the intensity of the single-slit diffraction pattern, given at Hyperphysics and also at Wikipedia:

[itex]I(\theta) = I_0 \, \mathrm{sinc}^2 \left(\dfrac{d\pi}{\lambda} \sin\theta \right),[/itex] where

But why doesn't the spreading continue to increase for [itex]d/\lambda < 1[/itex]?

- #11

jtbell

Staff Emeritus

Science Advisor

Homework Helper

- 15,959

- 6,156

olivermsun said:There will be zeros (dark bands) at at angles such that [itex]\sin\theta_m = \lambda / d[/itex], where [itex]\theta_m[/itex] getsbiggeras [itex]d/\lambda[/itex] gets smaller (as stated in the first line from the course book).

But why doesn't the spreading continue to increase for [itex]d/\lambda < 1[/itex]?

It does continue to increase.

At ##d/\lambda = 1## i.e. ##d = \lambda##, ##\theta_1 = 90^{\circ}##. The central peak of the intensity distribution exactly fills the "field of view" looking outwards from the slit, with the zero-intensity points at ±90°.

For ##d/\lambda < 1##, the intensity decreases as you "look" outwards at larger angles from the slit, but it never reaches zero. In effect, you can "see" only a central "slice" of the central peak.

- #12

craigi

- 615

- 36

MohammedRady97 said:"As the gap becomes narrower, the diffraction effect becomes more pronounced. It is greatest when the width of the gap is equal to the wavelength of the ripples."

olivermsun said:What I'm questioning is the bolded part above.

It's a little vague, but he's referring to the width of the central peak and the ratio of intensities of the maxima and minima.

You might want to play with that function a little to convince yourself of it.

You can do it here: "rechneronline.de/function-graphs/"

Try:

sinc(0.5*pi*sin(x))*sinc(0.5*pi*sin(x))

sinc(1.0*pi*sin(x))*sinc(1.0*pi*sin(x))

sinc(2.0*pi*sin(x))*sinc(2.0*pi*sin(x))

sinc(9.9*pi*sin(x))*sinc(9.9*pi*sin(x))

As you increase the factor, a value of 1.0 forms a sort of tipping point for that function, where zero intensity is present, but the secondary maxima aren't present yet.

Last edited by a moderator:

- #13

olivermsun

Science Advisor

- 1,268

- 136

That's what I think too, which is why I found the passage from the course book surprising.jtbell said:It does continue to increase.

- #14

olivermsun

Science Advisor

- 1,268

- 136

Why would those be the metrics? In particular, why the ratio of intensities between maxima and minima? A ratio of 1:1 would just mean the beam has become fully cylindrical, right?craigi said:It's a little vague, but he's referring to the width of the central peak and the ratio of intensities of the maxima and minima.

I'd expect something like the deviation from ray behavior to be a more obvious indication of how "pronounced" the diffraction effect has become.

- #15

craigi

- 615

- 36

olivermsun said:Why would those be the metrics? In particular, why the ratio of intensities between maxima and minima? A ratio of 1:1 would just mean the beam has become fully cylindrical, right?

I'd expect something like the deviation from ray behavior to be a more obvious indication of how "pronounced" the diffraction effect has become.

It's not a technical term. If you understand the importance of the slit width to wavelength ratio, then you understand at least as much as he's trying to illustrate. My advice is to ignore the word "pronounced".

The beam isn't cylindrical, even with a circular hole instead of a slit. There is never a sharp cut off. This is true regardless of the slit width to wavelength ratio. If you follow the link that I gave you should be able to see that by playing with the function.

This video might help if you haven't seen this phenomena yet:

Last edited by a moderator:

- #16

sophiecentaur

Science Advisor

Gold Member

- 29,358

- 7,074

The authors of that textbook have managed to demonstrate just how inadequate ordinary language can be as a substitute for some very simple Maths. Let's face it, there is absolutely no non-mathematical explanation for the way a diffraction pattern is formed in detail. (There's a challenge for someone.)

- #17

olivermsun

Science Advisor

- 1,268

- 136

The argument of the sinc function scales as 1/craigi said:It's not a technical term. If you understand the importance of the slit width to wavelength ratio, then you understand at least as much as he's trying to illustrate. My advice is to ignore the word "pronounced".

The beam exhibits cylindrical spreading in the limit of a narrow vertical slit. The spreading would be spherical with the limit of a small circular aperture.The beam isn't cylindrical, even with a circular hole instead of a slit.

There is never a sharp cut off. This is true regardless of the slit width to wavelength ratio. If you follow the link that I gave you should be able to see that by playing with the function.

This video might help if you haven't seen this phenomena yet:

Thanks, very nice video. I'm quite familiar with the phenomenon however.

Last edited by a moderator:

- #18

olivermsun

Science Advisor

- 1,268

- 136

Even in a professional paper on the topic, it would be perfectly normal for there to be a (relatively) conversational description in the text to accompanying the "formal Maths." I wouldn't expect the discussion to describesophiecentaur said:The sinc function is clearly the best way to express the diffraction pattern of a single slit. What point is there in putting it in a conversational way and then discussing how well those words fit the sinc description? For people who can't cope with the details of the formal Maths for this sort of problem then why not just accept the diagrams corresponding to a few different slit widths? What relevance has the word "pronounced" and why is it worth losing any sleep over? Is one circle any more 'round' than another circle or a rectangle any more rectangular than another rectangle?

Here what I'm saying is that the conversational description doesn't seem to make sense and in fact may be misleading. Worse yet, the formal Maths don't seem to have been included in the course book.

I think it's useful to say, in ordinary language, that the diffraction effect becomes more pronounced with decreasing slit widthThe authors of that textbook have managed to demonstrate just how inadequate ordinary language can be as a substitute for some very simple Maths. Let's face it, there is absolutely no non-mathematical explanation for the way a diffraction pattern is formed in detail. (There's a challenge for someone.)

- #19

sophiecentaur

Science Advisor

Gold Member

- 29,358

- 7,074

olivermsun said:Even in a professional paper on the topic, it would be perfectly normal for there to be a (relatively) conversational description in the text to accompanying the "formal Maths." I wouldn't expect the discussion to describeeverythingcontained within the maths, but at the least it should be usefulandtechnically correct.

Here what I'm saying is that the conversational description doesn't seem to make sense and in fact may be misleading. Worse yet, the formal Maths don't seem to have been included in the course book.

I think it's useful to say, in ordinary language, that the diffraction effect becomes more pronounced with decreasing slit widthd. If however I say the effect ismost pronouncedwhend= λ, then I think that deserves some further explanation (or at least definition).

I would agree. In this case, More was Less, as it added confusion. Text book authors should always realize that many of their readers are coming to what is in the book for the first time. 'You and I' can look at statements and see them in the light of experience. A student, on his or her own, when presented with some 'helpful' embellishment can easily be thrown into panic.

Words need to be very carefully selected in a one way statement, whereas, in a conversation, the two parties can arrive at an understanding much more reliably. That's where PF comes in handy.

Single slit diffraction is a phenomenon that occurs when light waves pass through a narrow slit. The light waves spread out and interfere with each other, creating a pattern of bright and dark bands on a screen behind the slit. This occurs because the slit acts as a point source of light, causing the waves to diffract or bend around the edges of the slit.

The width of the slit has a direct effect on the diffraction pattern. A wider slit will produce a narrower pattern with more intense light bands, while a narrower slit will produce a wider pattern with less intense bands. This is because a wider slit allows more light to pass through, resulting in a larger central maximum and smaller secondary maxima.

The wavelength of light also affects the diffraction pattern. As the wavelength increases, the distance between the bright and dark bands in the pattern also increases. This is because longer wavelengths diffract more, causing the bands to spread out further. Thus, the diffraction pattern can be used to determine the wavelength of light.

The distance between the slit and the screen, also known as the distance of observation, affects the diffraction pattern by changing the spacing between the bright and dark bands. As the distance increases, the bands become more spread out and the central maximum becomes wider. This is because the light waves have more time to diffract as they travel a longer distance from the slit to the screen.

Yes, single slit diffraction can occur with other types of waves, such as sound waves and water waves. Just like with light waves, the waves spread out and interfere with each other, creating a diffraction pattern. The only difference is that the spacing between the bands may vary depending on the wavelength and properties of the specific type of wave.

- Replies
- 5

- Views
- 1K

- Replies
- 17

- Views
- 2K

- Replies
- 17

- Views
- 3K

- Replies
- 20

- Views
- 2K

- Replies
- 13

- Views
- 2K

- Replies
- 5

- Views
- 2K

- Replies
- 2

- Views
- 1K

- Replies
- 2

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Replies
- 9

- Views
- 2K

Share: