0
$\begingroup$

Today I read that if you have two light beams with a wavelength difference equal to $\Delta$$\lambda$ to resolve the two into two disjoint spots, the following must be true:

enter image description here

Where N is the number of slits in a diffraction grating. I've been drawing the situation and looking online but I can't quite figure out why this is the condition. Could anyone give some reasoning why this condition is true. Also does this condition have a name?

$\endgroup$
1
  • $\begingroup$ Are you trying to derive the Rayleigh criterion for resolving the 2 wavelengths using the grating equation? The Rayleigh criterion is where the maximum of one wavelength coincides with the first minimum of the of the second wavelength.and and vice versa? $\endgroup$ Apr 4, 2019 at 0:09

2 Answers 2

1
$\begingroup$

You should first of all read the answers to Fringe width and spacing and number of slits in diffraction experiments and Intensity of subsidiary maxima in a diffraction grating pattern? where it is explained that as the number of slits $N$ increases the width of the principal maxima decreases.
For a grating with $N$ slits there are $N-1$ subsidiary minima and $N-2$ subsidiary maxima between principal maxima.

The condition for the $n^{\rm th}$ order principal maximum is $n\lambda = d \sin \theta_{\rm n}$ where $\lambda$ is the wavelength and $d$ is the adjacent slit separation.

If there is a grating with $N$ slits then the path difference between the first slit and the $N^{\rm th}$ slit is approximately $Nn\lambda$ remembering that $N\gg 1$.
The first subsidiary minimum occurs when the path difference between the two extreme slits is $Nn\lambda\pm \lambda$.

The Rayleigh criterion for just being able to resolve two wavelengths is that the principal maximum for light of wavelength $\lambda + \Delta \lambda$ occurs at an adjacent subsidiary minimum to the principal maximum of wavelength $\lambda$.

This means that $Nn\lambda+\lambda = Nn(\lambda +\Delta \lambda) \Rightarrow \dfrac{\lambda }{\Delta \lambda} = Nn$ which is the resolving power of a diffraction grating with $N$ slits in the $n^{\rm th}$ order.

$\endgroup$
3
  • $\begingroup$ Thanks, great response, that answered a lot of my questions. $\endgroup$ Apr 4, 2019 at 0:17
  • $\begingroup$ Let $\lambda+\Delta\lambda=\lambda_2$ and $\lambda=\lambda_1$. Then max($\lambda_2$) $\rightarrow$ first min($\lambda_1$). Okay, that's necessary. But if the first min($\lambda_2$) is beyond max($\lambda_1$), then the 2 different wavelengths haven't been resolved. $\endgroup$ Apr 4, 2019 at 1:51
  • $\begingroup$ @CinaedSimson The intensity of the principal maxima is very much greater than that of the subsidiary maxima so if the principal maximum of one wavelength is located at the same position as the subsidiary maximum of another wavelength one observes just the principal maximum. $\endgroup$
    – Farcher
    Apr 4, 2019 at 5:44
1
$\begingroup$

The key to this is that the number of slits (or grooves, or whatever) in the grating, multiplied by the spacing between slits, is equal to the aperture width of the diffraction grating. See Single-Slit Diffraction. The beam diffracted by the grating is also diffracted by the aperture (which acts like a single wide slit), so even if the beam is initially collimated, the diffracted beam spreads by an angle that depends on the aperture width and the wavelength. If the angular spread of the beams corresponding to $\lambda$ and $\delta \lambda$ is greater than the separation of the beam centers, then instead of two distinct spots you have two partially overlapped spots.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.