The light from a `special LED' consists of two colours of light with wavelengths of 530nm and 630nm respectively. The light is shone through a diffraction grating with 500lines/mm, and the two colours need to be separated by at least 5.0∘. What is the minimum order of interference needed in order to do this?
How do I do this one?
-
joyoustele
- Follow
- 3 followers
- 17 badges
- Send a private message to joyoustele
- Thread Starter
Offline17ReputationRep:- Follow
- 1
- 07-03-2018 16:22
-
- Follow
- 2
- 07-03-2018 20:21
For the diffraction grating (not youngs slits) you ahve an equation
nλ = d sin theta
The n has to be the same for both, d is the diffraction grating spacing and you have two wavelengths.
At the moment I can't see any other way to do it that start from n=1 (n=0 will give angle zero for both) and solve for theta until you find 5º of difference for the two wavelengths. (ie, start with n=1 and if the difference isn't enough, try n=2 and so on)
Someone else more inspired might come up with a shortcut. -
Eimmanuel
- Follow
- 4 followers
- 11 badges
- Send a private message to Eimmanuel
Offline11ReputationRep:- Follow
- 3
- 08-03-2018 15:31
(Original post by joyoustele)
The light from a `special LED' consists of two colours of light with wavelengths of 530nm and 630nm respectively. The light is shone through a diffraction grating with 500lines/mm, and the two colours need to be separated by at least 5.0∘. What is the minimum order of interference needed in order to do this?
How do I do this one?
(Original post by phys981)
For the diffraction grating (not youngs slits) you have an equation
nλ = d sin θ
The n has to be the same for both, d is the diffraction grating spacing and you have two wavelengths.
At the moment I can't see any other way to do it that start from n=1 (n=0 will give angle zero for both) and solve for theta until you find 5º of difference for the two wavelengths. (ie, start with n=1 and if the difference isn't enough, try n=2 and so on)
Someone else more inspired might come up with a shortcut.
I would not say the following is a shortcut but some learning moments for the students after they have successfully answered the question.
For such question, I would usually have some extension questions or discussions depending on the level of students.
If this is the first time that the students are attempting such question, I would get them to use the brute force method outlined by phys981.
Then I would have a discussion using a spreadsheet calculation based on the brute force method or guiding questions to explore for a “shortcut”.
Let θSn and θLn be the angles for the wavelength of 530nm and 630nm respectively based on the diffraction grating formula (d sin θ = nλ)
Compute the ratio λ /d for the wavelength of 530nm and 630nm respectively and then find the difference (which is 0.05).
Using the brute force method, the following can be “deduced” for "small angle":
Note that the angle in radians NOT degree.
So for 5°, convert it to radian which is about 0.087 rad. Substitute 0.087 rad to the above approximated equation to solve for n.
-
joyoustele
- Follow
- 3 followers
- 17 badges
- Send a private message to joyoustele
- Thread Starter
Offline17ReputationRep:- Follow
- 4
- 08-03-2018 17:20
(Original post by Eimmanuel)
I would not say the following is a shortcut but some learning moments for the students after they have successfully answered the question.
For such question, I would usually have some extension questions or discussions depending on the level of students.
If this is the first time that the students are attempting such question, I would get them to use the brute force method outlined by phys981.
Then I would have a discussion using a spreadsheet calculation based on the brute force method or guiding questions to explore for a “shortcut”.
Let θSn and θLn be the angles for the wavelength of 530nm and 630nm respectively based on the diffraction grating formula (d sin θ = nλ)
Compute the ratio λ /d for the wavelength of 530nm and 630nm respectively and then find the difference (which is 0.05).
Using the brute force method, the following can be “deduced” for "small angle":
Note that the angle in radians NOT degree.
So for 5°, convert it to radian which is about 0.087 rad. Substitute 0.087 rad to the above approximated equation to solve for n.
-
Eimmanuel
- Follow
- 4 followers
- 11 badges
- Send a private message to Eimmanuel
Offline11ReputationRep:- Follow
- 5
- 08-03-2018 17:57
(Original post by joyoustele)
I just didnt understand the question well. Thanks a lot, Very much appreciate it. -
joyoustele
- Follow
- 3 followers
- 17 badges
- Send a private message to joyoustele
- Thread Starter
Offline17ReputationRep:- Follow
- 6
- 08-03-2018 18:04
-
UCL
-
University of Manchester
-
Physics, Astrophysics and Cosmology
Lancaster University
-
Physics with Quantum Technologies
University of Surrey
-
University of East Anglia (UEA)
-
Keele University
-
PGDE Secondary Teaching (English Medium) Physics with Science
University of the Highlands and Islands
-
University of Oxford
-
Physics with Astrophysics with a Year in Industry
University of Kent
-
Physics with Astrophysics with Study Year Abroad.
University of Bath
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
- charco
- Mr M
- Changing Skies
- F1's Finest
- rayquaza17
- Notnek
- RDKGames
- davros
- Gingerbread101
- Kvothe the Arcane
- TeeEff
- The Empire Odyssey
- Protostar
- TheConfusedMedic
- nisha.sri
- claireestelle
- Doonesbury
- furryface12
- Amefish
- harryleavey
- Lemur14
- brainzistheword
- Rexar
- Sonechka
- TheAnxiousSloth
- EstelOfTheEyrie
- CoffeeAndPolitics
- an_atheist
- Labrador99
- EmilySarah00