Turn on thread page Beta
    • Thread Starter
    Offline

    15
    ReputationRep:
    They're long questions but any help would be much appreciated


    1) Light directed normally at a diffraction grating contains wavelengths of 580 and 586 nm only. Grating has 600 lines per mm. There are 2 diffracted orders.

    For the highest order, calculate the angle between the two diffracted beams.

    I'm not even clear on what this question is asking, the difference between the 2 angles of diffraction for the 1st and 2nd order beam? And should I just take 583 nm as the wavelength in the calculation?


    2) A diffraction grating is designed with a slit width of 0.83 um. When used in a spectrometer to view light of wavelength 430 nm, diffracted beams are observed at angles of 14 degrees 55' and 50 degrees 40' to the zero order beam.
    Assuming the low-angle diffracted beam is the first order beam, calculate the number of lines per mm on the grating. (Done this part)

    Explain why there is no diffracted beam between the two observed beams. What is the order number for the beam at 50 degrees 40'?

    Worked out the order is 3 but that means order 2 is missing and I have no idea why
    Offline

    17
    ReputationRep:
    (Original post by G.Y)
    They're long questions but any help would be much appreciated


    1) Light directed normally at a diffraction grating contains wavelengths of 580 and 586 nm only. Grating has 600 lines per mm. There are 2 diffracted orders.

    For the highest order, calculate the angle between the two diffracted beams.

    I'm not even clear on what this question is asking, the difference between the 2 angles of diffraction for the 1st and 2nd order beam? And should I just take 583 nm as the wavelength in the calculation?


    2) A diffraction grating is designed with a slit width of 0.83 um. When used in a spectrometer to view light of wavelength 430 nm, diffracted beams are observed at angles of 14 degrees 55' and 50 degrees 40' to the zero order beam.
    Assuming the low-angle diffracted beam is the first order beam, calculate the number of lines per mm on the grating. (Done this part)

    Explain why there is no diffracted beam between the two observed beams. What is the order number for the beam at 50 degrees 40'?

    Worked out the order is 3 but that means order 2 is missing and I have no idea why
    For question 1 there are two distinct wavelengths of light. Each will produce different patterns when passed through the grating.

    The question is saying work out the maximum order for each wavelength. Then find the angle between these two maxima.
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    17
    ReputationRep:
    (Original post by G.Y)
    They're long questions but any help would be much appreciated


    1) Light directed normally at a diffraction grating contains wavelengths of 580 and 586 nm only. Grating has 600 lines per mm. There are 2 diffracted orders.

    For the highest order, calculate the angle between the two diffracted beams.

    I'm not even clear on what this question is asking, the difference between the 2 angles of diffraction for the 1st and 2nd order beam? And should I just take 583 nm as the wavelength in the calculation?


    2) A diffraction grating is designed with a slit width of 0.83 um. When used in a spectrometer to view light of wavelength 430 nm, diffracted beams are observed at angles of 14 degrees 55' and 50 degrees 40' to the zero order beam.
    Assuming the low-angle diffracted beam is the first order beam, calculate the number of lines per mm on the grating. (Done this part)

    Explain why there is no diffracted beam between the two observed beams. What is the order number for the beam at 50 degrees 40'?

    Worked out the order is 3 but that means order 2 is missing and I have no idea why
    Im a bit confused as to why u said the slit width in question 2 is 0.83um, from the first part of that question i get a different value for d.

    So im not entirely sure on the second question.
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Shaanv)
    Im a bit confused as to why u said the slit width in question 2 is 0.83um, from the first part of that question i get a different value for d.

    So im not entirely sure on the second question.
    I've just typed the question as it is. I think they included the slit width to throw you off. d is not the same as the slit width.

    And wow I completely misinterpreted the first question, thank you
    Offline

    17
    ReputationRep:
    (Original post by G.Y)
    I've just typed the question as it is. I think they included the slit width to throw you off. d is not the same as the slit width.

    And wow I completely misinterpreted the first question, thank you
    Ahh i missed the part that slit width is irrelevant in the question. I agree with u about the other beam being 3rd order.

    Im not sure what happened to the second order beam. It could have something to do with the slit width and the single slit diffraction occurring at each slit as the slit width is much greater than the wavelength. This could cause an interference pattern which destructively interferes where the second order beam should be.

    Im just rambling tho. Im curious as to what the answer is let me know when and if u find out.
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Shaanv)
    Ahh i missed the part that slit width is irrelevant in the question. I agree with u about the other beam being 3rd order.

    Im not sure what happened to the second order beam. It could have something to do with the slit width and the single slit diffraction occurring at each slit as the slit width is much greater than the wavelength. This could cause an interference pattern which destructively interferes where the second order beam should be.

    Im just rambling tho. Im curious as to what the answer is let me know when and if u find out.
    Yeah that sounds like it should be right, hate when textbooks don't include answers with the questions ugh
    Hoping a similar question will be on a past paper so I can check the mark scheme for a definitive conclusion

    Thank you
    • Community Assistant
    Offline

    12
    ReputationRep:
    Community Assistant
    (Original post by G.Y)
    ......

    2) A diffraction grating is designed with a slit width of 0.83 um. When used in a spectrometer to view light of wavelength 430 nm, diffracted beams are observed at angles of 14 degrees 55' and 50 degrees 40' to the zero order beam.
    Assuming the low-angle diffracted beam is the first order beam, calculate the number of lines per mm on the grating. (Done this part)

    Explain why there is no diffracted beam between the two observed beams. What is the order number for the beam at 50 degrees 40'?

    Worked out the order is 3 but that means order 2 is missing and I have no idea why
    (Original post by Shaanv)
    Ahh i missed the part that slit width is irrelevant in the question. I agree with u about the other beam being 3rd order.

    Im not sure what happened to the second order beam. It could have something to do with the slit width and the single slit diffraction occurring at each slit as the slit width is much greater than the wavelength. This could cause an interference pattern which destructively interferes where the second order beam should be.

    ....

    I would address the missing order explanation.

    If you look at the double-slit interference pattern with finite width slits, you can see that there is a diffraction pattern. See the link below which shows the combined effects of two-slit and single-slit interference.

    https://books.google.com.sg/books?id...page&q&f=false

    From the diagram, you can see that the first order of diffraction minimum coincides with a particular interference maximum order.

    The missing order in the interference maximum order can be computed using


     \dfrac{d}{a} = n

    where d is the separation between the two slits and a is the width of the slit.

    Apply the similar reasoning to the diffraction grating, you would realize the info on “a slit width of 0.83 μm” is relevant and important.

    As you work out the separation between the slits in the diffraction grating to be 1.67 μm,


     \dfrac{d}{a} = \dfrac{1.67}{0.83} \approx 2

    Which implies the second order of the interference pattern in the diffraction grating coincides with the first order of the diffraction minimum. This explains the missing second order.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Eimmanuel)
    I would address the missing order explanation.

    If you look at the double-slit interference pattern with finite width slits, you can see that there is a diffraction pattern. See the link below which shows the combined effects of two-slit and single-slit interference.

    https://books.google.com.sg/books?id...page&q&f=false

    From the diagram, you can see that the first order of diffraction minimum coincides with a particular interference maximum order.

    The missing order in the interference maximum order can be computed using



     \dfrac{d}{a} = n



    where d is the separation between the two slits and a is the width of the slit.

    Apply the similar reasoning to the diffraction grating, you would realize the info on “a slit width of 0.83 μm” is relevant and important.

    As you work out the separation between the slits in the diffraction grating to be 1.67 μm,



     \dfrac{d}{a} = \dfrac{1.67}{0.83} \approx 2



    Which implies the second order of the interference pattern in the diffraction grating coincides with the first order of the diffraction minimum. This explains the missing second order.
    So the second order of the diffraction grating pattern is cancelled out by the first min of single slit interference? But how did you work out it's the first min?
    • Community Assistant
    Offline

    12
    ReputationRep:
    Community Assistant
    (Original post by G.Y)
    So the second order of the diffraction grating pattern is cancelled out by the first min of single slit interference? …
    You can say so.

    (Original post by G.Y)
    …But how did you work out it's the first min?
    Not sure why you ask this question. :confused: I thought it is pretty obvious. Just apply the slit diffraction pattern formula
    a sin θ =
    where a is the width of the slit.

    Note that n is the order of minimum.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Eimmanuel)
    You can say so.



    Not sure why you ask this question. :confused: I thought it is pretty obvious. Just apply the slit diffraction pattern formula

    a sin θ =
    where a is the width of the slit.

    Note that n is the order of minimum.
    What do you mean you can say so? Is that not exactly what's happening?

    Wasn't obvious to me thanks
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Eimmanuel)
    I would address the missing order explanation.

    If you look at the double-slit interference pattern with finite width slits, you can see that there is a diffraction pattern. See the link below which shows the combined effects of two-slit and single-slit interference.

    https://books.google.com.sg/books?id...page&q&f=false

    From the diagram, you can see that the first order of diffraction minimum coincides with a particular interference maximum order.

    The missing order in the interference maximum order can be computed using



     \dfrac{d}{a} = n



    where d is the separation between the two slits and a is the width of the slit.

    Apply the similar reasoning to the diffraction grating, you would realize the info on “a slit width of 0.83 μm” is relevant and important.

    As you work out the separation between the slits in the diffraction grating to be 1.67 μm,



     \dfrac{d}{a} = \dfrac{1.67}{0.83} \approx 2



    Which implies the second order of the interference pattern in the diffraction grating coincides with the first order of the diffraction minimum. This explains the missing second order.
    Also, does this graph show that single slit interference combined with double slit diffraction causes maxima to be split up into smaller fringes?
    • Community Assistant
    Offline

    12
    ReputationRep:
    Community Assistant
    (Original post by G.Y)
    So the second order of the diffraction grating pattern is cancelled out by the first min of single slit interference? …
    (Original post by G.Y)
    What do you mean you can say so? Is that not exactly what's happening? …
    I mean you can “interpret” at your level if you have not gone through the maths to derive the intensity curve formula. The intensity curve formula tends to scare off a lot of A level students I would not usually advocate such interpretation. For some students, such interpretation is easy for them to understand the solution. However, it can give rise to some issues like the question you asked below.

    I would prefer to say “The interference pattern is modulated by the diffraction pattern. At the place of the diffraction minimum, the intensity is zero.”

    A better version (IMO) would be the following: (I cannot remember where this is from)
    Sharp maxima occur due to constructive interference of light emerging from the two slits.
    Their intensity is modulated by the envelope due to diffraction by each individual slit.
    • Community Assistant
    Offline

    12
    ReputationRep:
    Community Assistant
    (Original post by G.Y)
    Also, does this graph show that single slit interference combined with double slit diffraction causes maxima to be split up into smaller fringes?

    NO.

    The intensity curve formula is given as equation 38.3. If you look at the formula, the square bracket term is due to the single slit diffraction while the cosine squared term is the double-slit interference pattern.

    So the “complicated pattern” of the two slits of width a that has a separation of d is the interference pattern of two point sources separated by d modulated by the diffraction pattern of single-slit of width a.

    To make thing simple. Think in this way in the “ideal” situation separately. (Take the following as a way to see the result.)
    Single-slit gives rise to a diffraction pattern.
    Two-slits gives rise to an interference pattern.

    But we are not living in the ideal world, so we need to combine (or superimpose) them together and this gives a complicated pattern.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Eimmanuel)
    NO.

    The intensity curve formula is given as equation 38.3. If you look at the formula, the square bracket term is due to the single slit diffraction while the cosine squared term is the double-slit interference pattern.

    So the “complicated pattern” of the two slits of width a that has a separation of d is the interference pattern of two point sources separated by d modulated by the diffraction pattern of single-slit of width a.

    To make thing simple. Think in this way in the “ideal” situation separately. (Take the following as a way to see the result.)
    Single-slit gives rise to a diffraction pattern.
    Two-slits gives rise to an interference pattern.

    But we are not living in the ideal world, so we need to combine (or superimpose) them together and this gives a complicated pattern.
    So, whenever double slit interference occurs, realistically single slit diffraction is also occurring?
    • Community Assistant
    Offline

    12
    ReputationRep:
    Community Assistant
    (Original post by G.Y)
    So, whenever double slit interference occurs, realistically single slit diffraction is also occurring?
    Yes.
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by Eimmanuel)
    Yes.
    I definitely need more practice with this topic

    Thank you
    • Community Assistant
    Offline

    12
    ReputationRep:
    Community Assistant
    (Original post by G.Y)
    I definitely need more practice with this topic

    Thank you
    It is more like you need to have a better reference on this topic.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 9, 2018

3,379

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should universities take a stronger line on drugs?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.