# Diffraction grating help (Physics AS level)

Watch
Announcements
#1
I'm answering questions in the Nelson Thornes book and it doesnt really give you an explanation and my teacher isnt allowed to respond so that's great. If anyone can help me understand this, it would be much appreciated.

4) A diffraction grating is designed with a slit width of 0.83 micrometres. When used in a spectrometer to view light of wavelength 430nm, diffracted beams are observed at angles 14 degrees 55' and 50 degrees 40' to the zero-order beam.
a) Assuming the low-angle diffracted beams is the first order beam, calculate the number of lines per mm on the grating.
b) Explain why there is no diffracted beam between the two observed beams. What is the order number for the beams at 50 degrees 40'.
(Reminder: 1 degree = 60 mins)

A) 599mm^-1 ( I didn't get this and I dont understand why it's to the power of -1)
B) 3 (It doesnt explain the first part so I dont understand it but i do know how to get 3 for the second part )

My working for a:
d x sin(15.92) = 1 x (430x10^-9)
d = 1.567655759 x 10^-6
N = (0.83 x 10^-6)/(1.567655759 x 10^-6) = 0.5294529717 lines per m
0.5294529717/10^-3 = 529.4529717
= 529 lines per mm
Last edited by PetitePanda; 1 year ago
0
1 year ago
#2
this question has stumped so many people (including me, still does) - i think it has been answered in a past thread. look it up
0
1 year ago
#3
A quick look at this shows that you have used 15.92 as the angle given as 15o 40' in the question.
40' is 40/60 mins (60 mins in a degree) which would make this 15.67
Does this give you the correct answer now?
0
#4
(Original post by Stonebridge)
A quick look at this shows that you have used 15.92 as the angle given as 15o 40' in the question.
40' is 40/60 mins (60 mins in a degree) which would make this 15.67
Does this give you the correct answer now?
I meant 55' I typed it wrong
0
#5
(Original post by forbearne)
this question has stumped so many people (including me, still does) - i think it has been answered in a past thread. look it up
That's no help for me because I want to know where I went wrong in my calculations. I have looked up previous threads but I only found two and I still dont understand
0
1 year ago
#6
Firstly, could you check the original question, in particular the line
'a) Assuming the low-angle diffracted beams are observed at order beam'
it doesn't make sense. There is a word or something missing there.

Secondly
The number of lines per mm of a grating is simply 1/d where d is the separation between the slits. It does not involve the actual slit width, given as 0.83 micrometres here.
Having said that, it still does not produce the answer given.
The reason why the answer is 'to the power of -1' is because the unit of number of lines per mm is 'per mm' which is written as mm-1 . Just like 'per second' is s-1
Maybe if we can sort out what is wrong here, we can move on to the reason why there is no 2nd order beam. The answer to which is actually a little more complicated and is a consequence of the slit width mentioned earlier.
0
#7
(Original post by Stonebridge)
Firstly, could you check the original question, in particular the line
'a) Assuming the low-angle diffracted beams are observed at order beam'
it doesn't make sense. There is a word or something missing there.

Secondly
The number of lines per mm of a grating is simply 1/d where d is the separation between the slits. It does not involve the actual slit width, given as 0.83 micrometres here.
Having said that, it still does not produce the answer given.
The reason why the answer is 'to the power of -1' is because the unit of number of lines per mm is 'per mm' which is written as mm-1 . Just like 'per second' is s-1
Maybe if we can sort out what is wrong here, we can move on to the reason why there is no 2nd order beam. The answer to which is actually a little more complicated and is a consequence of the slit width mentioned earlier.
You're defintely right I typed this wrong but then again I shouldnt have done this late at night. "Assuming the low-angle diffracted beam is the first order beam..." - I'll change that now. Ohhh I thought it was length of slit width over the diffraction - I dont know why I did that. I still dont know what I did else I did wrong because it still doesnt give me the answe like you said. Ohhh I got the power of -1 now.
0
1 year ago
#8
Well, the formula you have for the diffraction grating is correct

d sin θ = nλ

and in the question we are given
λ = 430 nm
n = 1
θ = 15o 55' = 15.92 deg
I get this to give d = 1.57 μm
which is the slit spacing.
Do you get that too?
The number of slits per m is 1/d
and I get 1 / 1.57 μm to be 636843, which is
637 slits per mm or 637mm-1
Do you get that?
Unless I am missing something really obvious, or just being a bit slow tonight, or there is another typo in the question data somewhere (can you check?) that must be the answer.
1
#9
(Original post by Stonebridge)
Well, the formula you have for the diffraction grating is correct

d sin θ = nλ

and in the question we are given
λ = 430 nm
n = 1
θ = 15o 55' = 15.92 deg
I get this to give d = 1.57 μm
which is the slit spacing.
Do you get that too?
The number of slits per m is 1/d
and I get 1 / 1.57 μm to be 636843, which is
637 slits per mm or 637mm-1
Do you get that?
Unless I am missing something really obvious, or just being a bit slow tonight, or there is another typo in the question data somewhere (can you check?) that must be the answer.
I'm so so sorry I just realise my whole mistake because I cant read. It's supposed to be 14 instead of 15. I've worked it out and it did give me the answer. Thank you so much and again I'm genuinely sorry. I should've proof read it.
0
1 year ago
#10
Well thank heavens we got there in the end.
0
#11
(Original post by Stonebridge)
Well thank heavens we got there in the end.
Yayyyyyy If you don't mind, could you explain why there is diffracted beams between the two observed beams?
0
1 year ago
#12
I will get you something for that tomorrow, as I'm logging off now for the night.
It will help me in my answer if you can say if you have also studied single slit diffraction. It is usually studied before you do the diffraction grating, which is of course a multi-slit diffraction.
0
#13
(Original post by Stonebridge)
I will get you something for that tomorrow, as I'm logging off now for the night.
It will help me in my answer if you can say if you have also studied single slit diffraction. It is usually studied before you do the diffraction grating, which is of course a multi-slit diffraction.
Thank you so much. I did but I dont think I understood it that well (tbh that's with anything optics). I did diffraction grating because I checked my process tracker and diffraction grating was a red area. I hope you have a nice night
Last edited by PetitePanda; 1 year ago
0
1 year ago
#14
Ok. The explanation for why we don't see the n=2 beam.

This is the beam that we would expect to lie between the 2 that we have in the question. The n=1 and the n=3.
So far we have the n=1 beam at 14o 55' and the n=3 beam at 50o 40'
So where should the n=2 beam be. At what angle? Somewhere between those two angle values, of course.

If you apply the formula for the grating again we can find out exactly where it should be.

d sin θ = nλ will do it, putting n=2 in the formula
we know λ is 430nm from the question
we know d from the calculation we did before with the same formula for n=1 and angle 14.92deg

So, work out what the angle is for n=2

Now the theory for why it's missing.
In a single slit diffraction, as opposed to this multi slit diffraction grating diffraction, there are no sharp beams, but rather a series of so called fringes. A light and dark alternating pattern. The theory for single slits (I will leave you to look this up) says that the dark parts (where there is no light) of the light/dark fringe pattern are separated by a distance on the screen such that they form an angle where

sin θ = λ/a where a is the actual width of the single slit. (We are given this in the question. This is why. )

When you do a multi-slit diffraction grating experiment, the bright beams you see, are actually contained in an 'envelope' determined by the single slit pattern of the individual slits. The diagram here shows this:

The dotted line shows the diffraction pattern due to a single slit, and the vertical bars show the bright beams you get with the grating.
The bright beams have to fall inside the dotted curve. The vertical height of the graph shows the brightness of the light at any angle.
In this diagram, the n=1 and n=2 beams fall inside the curve and would be visible. In our question it is similar but we need to do a calculation first to see where these points are in our example.

Now we know where the n=2 beam should be from the calculation above.
We need to look at where that 'minimum' is on the dotted line, to the right of centre, at the position given by that other formula, sin θ = λ/a
At that point, if there is a bright beam expected, it will not appear, as it occurs at a minimum of the envelope curve, due to the diffraction pattern of the single slits that make up the grating.

So sub the values we know into this second formula to find the angle theta at which this minimum occurs.
λ is 430nm and a = 0.83 μm as given in the question.

What do we find? I think you may already have guessed.

If the two coincide to within a degree or 2, we have our explanation.
Last edited by Stonebridge; 1 year ago
0
#15
(Original post by Stonebridge)
Ok. The explanation for why we don't see the n=2 beam.

This is the beam that we would expect to lie between the 2 that we have in the question. The n=1 and the n=3.
So far we have the n=1 beam at 14o 55' and the n=3 beam at 50o 40'
So where should the n=2 beam be. At what angle? Somewhere between those two angle values, of course.

If you apply the formula for the grating again we can find out exactly where it should be.

d sin θ = nλ will do it, putting n=2 in the formula
we know λ is 430nm from the question
we know d from the calculation we did before with the same formula for n=1 and angle 14.92deg

So, work out what the angle is for n=2

Now the theory for why it's missing.
In a single slit diffraction, as opposed to this multi slit diffraction grating diffraction, there are no sharp beams, but rather a series of so called fringes. A light and dark alternating pattern. The theory for single slits (I will leave you to look this up) says that the dark parts (where there is no light) of the light/dark fringe pattern are separated by a distance on the screen such that they form an angle where

sin θ = λ/a where a is the actual width of the single slit. (We are given this in the question. This is why. )

When you do a multi-slit diffraction grating experiment, the bright beams you see, are actually contained in an 'envelope' determined by the single slit pattern of the individual slits. The diagram here shows this:

The dotted line shows the diffraction pattern due to a single slit, and the vertical bars show the bright beams you get with the grating.
The bright beams have to fall inside the dotted curve. The vertical height of the graph shows the brightness of the light at any angle.
In this diagram, the n=1 and n=2 beams fall inside the curve and would be visible. In our question it is similar but we need to do a calculation first to see where these points are in our example.

Now we know where the n=2 beam should be from the calculation above.
We need to look at where that 'minimum' is on the dotted line, to the right of centre, at the position given by that other formula, sin θ = λ/a
At that point, if there is a bright beam expected, it will not appear, as it occurs at a minimum of the envelope curve, due to the diffraction pattern of the single slits that make up the grating.

So sub the values we know into this second formula to find the angle theta at which this minimum occurs.
λ is 430nm and a = 0.83 μm as given in the question.

What do we find? I think you may already have guessed.

If the two coincide to within a degree or 2, we have our explanation.
The angle is 31 degrees. So it doesn't appear because of the angle isn't under the envelope curve, where the minimum occurs at 31.2 degrees?
0
1 year ago
#16
Yes. The bright beam for n=2, just happens to be very close to the point where the dotted line is a minimum. This means it will not be visible.
0
#17
(Original post by Stonebridge)
Yes. The bright beam for n=2, just happens to be very close to the point where the dotted line is a minimum. This means it will not be visible.
Thank you so much once again PRSOM
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Would you give consent for uni's to contact your parent/trusted person in a mental health crisis?

Yes - my parent/carer (117)
34.21%
Yes - a trusted person (92)
26.9%
No (91)
26.61%
I'm not sure (42)
12.28%