Rayleigh criterion (two wavelengths) + diffraction grating

[velkyr]

Homework Statement

"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."

Homework Equations

\theta = \frac{1.22\lambda}{D}

m\lambda = dsin\theta

R = \frac{\Delta\lambda}{\lambda} = mN

The Attempt at a Solution

First, I substituted in λ₁ and λ₂ into Rayleigh and assumed they would have an equal aperture diameter, giving an equation that looks like \frac{1.22\lambda_1}{\theta_1} = \frac{1.22\lambda_2}{\theta_2}. Of course this left two unknown values of θ. So, I then attempted to use the resolving power formula using the two wavelengths. I calculated R to be 1276.25. However, not knowing the order of light being received, I couldn't then use this to calculate N.

My current thinking is to assume m = 1 and use the diffraction maximum condition to find values for θ in each case. However if I were to then use those angles in Rayleigh to calculate D, I'm not sure what relevance the aperture diameter actually has to finding the number of slits... I would really appreciate a nudge in the right direction because I'm running out of ideas. Thank you.

 

[blue_leaf77]

$R = \frac{\Delta\lambda}{\lambda} = mN$

It should be $R = \frac{\lambda}{\Delta\lambda} = mN$. Use this equation to find $N$ with $m=1$ (as this order contains the brightest light aside from the zeroth order).

 

[velkyr]

It should be $R = \frac{\lambda}{\Delta\lambda} = mN$. Use this equation to find $N$ with $m=1$ (as this order contains the brightest light aside from the zeroth order).

Oops sorry... I do have the equation written down the right way around, I just mistyped it.

Is it simply just a matter of finding N for each wavelength and taking the largest answer, then? It almost feels too simple after all the time I've spent thinking about it.

 

[blue_leaf77]

Oops sorry... I do have the equation written down the right way around, I just mistyped it.

Is it simply just a matter of finding N for each wavelength and taking the largest answer, then? It almost feels too simple after all the time I've spent thinking about it.

You are using this equation once. $\lambda$ can be taken as the average between the two wavelengths. It only works if the two wavelengths are very close though, which is true in this case. It becomes simple because you already have the right equation.

 

[velkyr]

You are using this equation once. $\lambda$ can be taken as the average between the two wavelengths. It only works if the two wavelengths are very close though, which is true in this case. It becomes simple because you already have the right equation.

I see. Thank you very much, I think I've got the right answer now (1277 is the minimum). :)