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Rayleigh's criterion with two wavelengths...? Watch

1. Hi everyone,

A question we've been set is incredibly vague and I have no idea where to start. It says, "A source emits light with two monochromatic components of wave-lengths λ1 = 510.50 nm and λ2 = 510.90 nm. Using the Rayleigh criterion, find the minimum number of slits of a grating that must be illuminated by a beam from the source in order to resolve these components."

So it says to use Rayleigh, but there are two wavelengths...? I tried setting the equation with λ1 and λ2 on each side equal to each other by assuming D was the same in both, but now there are still two unknowns in terms of angles - . Then I worked out the resolving power using the formula for diffraction gratings, but I don't know the order the light is received in (m) or N, so again there's two unknowns.

2. (Original post by velkyr)
Hi everyone,

A question we've been set is incredibly vague and I have no idea where to start. It says, "A source emits light with two monochromatic components of wave-lengths λ1 = 510.50 nm and λ2 = 510.90 nm. Using the Rayleigh criterion, find the minimum number of slits of a grating that must be illuminated by a beam from the source in order to resolve these components."

So it says to use Rayleigh, but there are two wavelengths...? I tried setting the equation with λ1 and λ2 on each side equal to each other by assuming D was the same in both, but now there are still two unknowns in terms of angles - . Then I worked out the resolving power using the formula for diffraction gratings, but I don't know the order the light is received in (m) or N, so again there's two unknowns.

Have you tried simultaneous equations?

Like:

1.22lambda1/theta1 = 1.22lambda2/theta

and 1.22lambda1theta2/1.22lambda2theta1 = 1

Would that work?
3. (Original post by Kyx)
Have you tried simultaneous equations?

Like:

1.22lambda1/theta1 = 1.22lambda2/theta

and 1.22lambda1theta2/1.22lambda2theta1 = 1

Would that work?
But... those two equations are exactly equivalent anyway. There would have to be a separate equation with different multiples of the same unknown variables to solve simultaneously?
4. (Original post by velkyr)
But... those two equations are exactly equivalent anyway. There would have to be a separate equation with different multiples of the same unknown variables to solve simultaneously?
I was thinking that

Then IDK
5. (Original post by velkyr)
Hi everyone,

A question we've been set is incredibly vague and I have no idea where to start. It says, "A source emits light with two monochromatic components of wave-lengths λ1 = 510.50 nm and λ2 = 510.90 nm. Using the Rayleigh criterion, find the minimum number of slits of a grating that must be illuminated by a beam from the source in order to resolve these components."

So it says to use Rayleigh, but there are two wavelengths...? I tried setting the equation with λ1 and λ2 on each side equal to each other by assuming D was the same in both, but now there are still two unknowns in terms of angles - . Then I worked out the resolving power using the formula for diffraction gratings, but I don't know the order the light is received in (m) or N, so again there's two unknowns.

Well if you just considered it for each individual wavelength, you could presumably get the minimum number of slits to resolve that wavelength. Then i would assume you would just pick the bigger number of slits of those 2 as you need to be able to resolve both wavelengths
6. (Original post by samb1234)
Well if you just considered it for each individual wavelength, you could presumably get the minimum number of slits to resolve that wavelength. Then i would assume you would just pick the bigger number of slits of those 2 as you need to be able to resolve both wavelengths
Okay that seems to be the way forward, but I still have a lot of unknowns. In Rayleigh, if I were to consider each wavelength separately, I don't know the corresponding angle or the diameter of aperture (nor do I know what the relevance of an aperture is to diffraction grating...) In the resolution equation I can actually use to find N, I still don't know the order of light. So while I would like to follow through on this idea I have no clue how to.
7. We have the wavelengths. We need to use the same diffraction grating to make the angles of the different wavelengths considerably different. Am I correct?
8. Maybe simultaneous equations with the Rayleigh and the lambda = dsintheta/n equation?
9. (Original post by Kyx)
Maybe simultaneous equations with the Rayleigh and the lambda = dsintheta/n equation?
Sounds like a good idea. I'll give it a go for one of the wavelengths and see if it works.

EDIT: never mind... trying to solve with there being a sin term (and only one sin term, in one equation) is making them impossible to solve from my standpoint.
10. (Original post by velkyr)
Sounds like a good idea. I'll give it a go for one of the wavelengths and see if it works.

EDIT: never mind... trying to solve with there being a sin term (and only one sin term, in one equation) is making them impossible to solve from my standpoint.
ah well...

gl
11. (Original post by Kyx)
ah well...

gl
I asked around at Physicsforums and it turned out as simple as using the resolution equation and assuming that m must be 1 because that's the brightest light you can receive. The answer turned out to be 1277.

Thank you for your help anyway ^^
12. (Original post by velkyr)
I asked around at Physicsforums and it turned out as simple as using the resolution equation and assuming that m must be 1 because that's the brightest light you can receive. The answer turned out to be 1277.

Thank you for your help anyway ^^
OK

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