Is the diffraction grating formula wrong?

[ChrisXenon]

So, two wave fronts leave adjacent "slits" in a grating and head for a screem When they get to a point of local maximum, they will have interfered constructively, and so they must be in phase. Therefore the path different will be a whole multiple of the wavelenth. People routinely draw a right angled triangle, pointing out that the path difference between these wavers is d sin (theta), hence d sin (theta) = N x lamba.
Makes perfect sense. Except that right angled triangle. If we chop off the path difference part (d sin (theta) then we should be back to equal path lengths, right? Wrong - because the remaining triangle is not isosceles - it's right angled, and the two long sides cannot be the same length.

So where did I go wrong? I can't see it.

Thanks!

 

[kuruman]

Please post a drawing to support your argument, specifically what this "remaining" triangle is about. Thanks

 

[Merlin3189]

You are quite right of course.
The thing about the large triangle is that it is so large.
The width of the short side is about the line spacing of the grating. So for 250 lines/mm it is about 3.94 μm
If the screen is 1m away, then my calculator gives the difference between the two long sides as 0.0000000 m. Edited as I put one too many 0 & edit 2, its m. So 0.0 μm.
That is much less than the wavelength of red light at about 0.68 μm
Edit 3 ! so no guaruntee that it is that much less than 0.68 μm. It could be nearly0.05 μm
Edit 4 But it isn't. A better calculator shows about 8 pm difference

 

[ZapperZ]

Summary:: Shouldn't the right angled triangle should be isosceles?

I can't see it.

Neither can we. Put yourself in our shoes and see if your post makes any sense.

Zz.

 

[ChrisXenon]

You are quite right of course.
The thing about the large triangle is that it is so large.
The width of the short side is about the line spacing of the grating. So for 250 lines/mm it is about 3.94 μm
If the screen is 1m away, then my calculator gives the difference between the two long sides as 0.0000000 m. Edited as I put one too many 0 & edit 2, its m. So 0.0 μm.
That is much less than the wavelength of red light at about 0.68 μm
Edit 3 ! so no guaruntee that it is that much less than 0.68 μm. It could be nearly0.05 μm
Edit 4 But it isn't. A better calculator shows about 8 pm difference

Thanks Merlin! Much appreciated.

 

[ChrisXenon]

Neither can we. Put yourself in our shoes and see if your post makes any sense.

Zz.

... and yet, miraculously, Merlin understood exactly what I meant.
I guess we're telepathically linked... or could it be some defficiency on your part... it's a toughy.

 

[ChrisXenon]

Please post a drawing to support your argument, specifically what this "remaining" triangle is about. Thanks

Thanks for making time to reply Kurman. I felt that anyone who knew about diffraction would know about "the triangle". In fact there are very many wrongly-drawn diagram, but figure 6 in http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter037.htm is what I'm talking about.

 

[mjc123]

It is explained in the panel to the left of the diagram (though not very well, since the arc is always perpendicular to both lines). What it means is that if you drop a perpendicular from S1 to BS2, hitting it at point A', then A' is almost exactly coincident with A, and the error involved in assuming it is the same is very small. This leads to the simple formula for sin θ.

 

[Merlin3189]

Interestingly, I actually thought of the diagram you mention (or similar) which is for Young's double slit.
My reply was based on consideration of just such a pair of slits very close together.

When I searched for a suitable diagram to add to my post, I realized that the grating is not exactly the same. And I even found an interesting derivation, which I didn't know, or had forgotten, that does take account of the different distances from each line. This, and other treatments considering the multiple lines, account not only for the location of the primary peaks, but also for peak narrowing and small intermediate peaks.

So in fact it's not telepathy that puts me on your wavelength, rather our similar level of ignorance (if you'll excuse my putting it that way.) It's one of my few advantages on PF, that from my lower level, I can sometimes see what simple questioners are asking for, while the experts realize it's all much more complicated and dive into explanations which don't satisfy the questionner.

For me, answering PF queries to help people is not so much altruism, as a way of refreshing my own knowledge and stimulating myself to look into things more deeply.

 

[nasu]

I think it make more sense to say that you draw the line S1A so that the two paths are equal after they cross S1A. So make the triangle isosceles, as you said initially. Then there is the problem to calculate the path difference from the triangle S1-S2-A. This is the exact path but the angle A is not exactly 90 degrees. However the difference is of the order of $10^{-5} - 10^{-6}$ radians. So the angle in degrees will be something like 89.9999. You can use the sine theorem to calculate the path difference (it works even if the triangle is not right-angle). But the difference between sin(90) and sin(89.9999) is of the order of $10^{-12}$ so it's safe to assume a right angle triangle. Unfortunately some authors, knowing that the angle is very close to 90 degrees, will say that you draw S1A as a perpendicular which even though is very close to reality may make students wonder why.

 

[hutchphd]

I guess we're telepathically linked... or could it be some defficiency on your part... it's a toughy.

If someone from whom one is soliciting enlightenment says your question doesn't make sense to him, it would seem impudent to call him out. Or perhaps I too am deficient... It's a toughy.

 

[jbriggs444]

or could it be some defficiency on your part.

Or perhaps I too am deficient

Deffinitely.

 

[ChrisXenon]

If someone from whom one is soliciting enlightenment says your question doesn't make sense to him, it would seem impudent to call him out. Or perhaps I too am deficient... It's a toughy.

Hutch - you're absolutely right, but there is an IF at the start of your comment. The IF condition fails and with it - your point.

 

[ChrisXenon]

It is explained in the panel to the left of the diagram (though not very well, since the arc is always perpendicular to both lines). What it means is that if you drop a perpendicular from S1 to BS2, hitting it at point A', then A' is almost exactly coincident with A, and the error involved in assuming it is the same is very small. This leads to the simple formula for sin θ.

Thanks

 

[ChrisXenon]

Interestingly, I actually thought of the diagram you mention (or similar) which is for Young's double slit.
My reply was based on consideration of just such a pair of slits very close together.

When I searched for a suitable diagram to add to my post, I realized that the grating is not exactly the same. And I even found an interesting derivation, which I didn't know, or had forgotten, that does take account of the different distances from each line. This, and other treatments considering the multiple lines, account not only for the location of the primary peaks, but also for peak narrowing and small intermediate peaks.

So in fact it's not telepathy that puts me on your wavelength, rather our similar level of ignorance (if you'll excuse my putting it that way.) It's one of my few advantages on PF, that from my lower level, I can sometimes see what simple questioners are asking for, while the experts realize it's all much more complicated and dive into explanations which don't satisfy the questionner.

For me, answering PF queries to help people is not so much altruism, as a way of refreshing my own knowledge and stimulating myself to look into things more deeply.

Impressive Merlin. Thank you again.

 

[sophiecentaur]

It always amazes me that, in this age of brilliantly useable computer drawing apps, people come to PF without supplying simple diagrams. When I read the OP, I assumed that I had the same diagram in my head as was being discussed bit could I be sure? No, of course not and any response I could make would risk being the wrong one.
You only have to learn ONCE how to draw lines, circles, etc. and the intellectual content of that operation is so much lower than the content of a Physics question so why not be bothered to learn that simple skill? Alternatively, a photo of a textbook page or a copied and pasted diagram straight off a Google hit would be useful and a courtesy to helpers.
Grumpy of git, perhaps but I do have a serious point.

 

[Mister T]

Alternatively, a photo of a textbook page or a copied and pasted diagram straight off a Google hit would be useful and a courtesy to helpers.

Courtesy being the main point here. Just look at the responses given by the OP.

Maybe I'm just being an old git, too, but there seems to be a lack of common courtesy among many young learners. They seek an education but blame the educator when they're being taught something that requires the effort of a back-and-forth discussion.

I got a nasty email the other day from a student whose complaint seemed to be that it's my fault that he doesn't know how to follow the instructions given for writing an essay. We're half way through the term now. He could have asked for clarification earlier, but instead waited until six essays (all with the same instructions) have already been graded. I always grade the essays before the next one is due, and unless the student gets a perfect score I include an explanation of which directions were not followed.

 

[sophiecentaur]

Maybe I'm just being an old git, too, but there seems to be a lack of common courtesy among many young learners. They seek an education but blame the educator when they're being taught something that requires the effort of a back-and-forth discussion.

I get the impression that no one has ever shouted at them or even dared to criticised for fear of 'complaints'. You reap what you sow.

 

[Merlin3189]

I agree with Sophie about providing diagrams.
I always have a pad and pencil to hand when I browse PF and for many questions, the first thing I do is sketch a diagram of the situation. I often think that many HW questioners would help themselves a lot if they did the same, even if they don't show it to us - though obviously it's much better if they do.

But what prompted me to respond here, was the thought that maybe diagrams could also be used more when responding. It is more work and many of you are much busier than I, responding to many more questions than I, so I understand that it might not be feasible. I often spend so much time producing diagrams, that by the time I'm ready to post, the question has already been answered (I don't find it quite as easy as Sophie suggests.)

I expect I'm unusual here in needing diagrams, most people seem to think in much more abstract ways, but I would suggest that people asking the low level questions may tend to be more like me than like the experts.

I get the impression that no one has ever shouted at them or even dared to criticised ...

Well people on PF certainly do their best to correct that error