The basic equation of the intensity for a grating of ## N ## equally spaced lines separated by a distance ## d ## is ## I(\theta)=I_o \frac{\sin^2(N \phi)/2)}{\sin^2(\phi/2)} ##where ## \phi =\frac{ 2 \pi d \, \sin(\theta)}{\lambda} ##. The primary maxima are found when both numerator and denominator are equal to zero, and in the limit as the denominator approaches ## 0 ##, the intensity of the maxima ## I=I_o N^2 ## at these locations. The resolving power is given by ## R=\frac{\lambda}{\Delta \lambda}=N m ## where ## N ## is the number of illuminated lines and ## m ## is the order. This can be derived from the intensity equation, but for someone not specializing in spectroscopy, simply using this resolving power formula is sufficient. ## \\ ## For the second question, (the width of the ruling), I'm not sure what they are asking. Maybe it has a simple answer, but they didn't seem to give sufficient information for that part. ## \\ ## See also this previous homework posting:
https://www.physicsforums.com/threads/diffraction-grating-question-dispersion-and-resolving.897090/ ## \\ ## Editing: I think I figured out what part (ii) is asking. This higher order will occur at a fairly wide angle ## \theta ## on the spectrometer. In order to have the intensity be appreciable at that angle, the single-slit diffraction intensity factor ## i(\theta)= \frac{\sin^2(\phi'/2)}{(\phi'/2)^2} ## where ## \phi'=\frac{2 \pi b \, \sin(\theta)}{\lambda} ## that is multiplied by the above interference result to get the complete intensity needs to be greater than ## .1 ## or thereabouts. (For this formula, with a reflection grating, you can assume ## b ## is the width of the ruling). ## \\ ## [Edit: "In principle", the order ## m ## maximum from above needs to occur inside of, really well inside of, the first zero intensity of the single slit diffraction pattern. That first zero occurs at ## \theta ## such that ## \lambda=b \sin(\theta) ##. In addition, might it be a good answer to say ## b ## should be such that ## b<\lambda ##? I'll let you decide that, based on the approximate value that you get for ## \sin(\theta) ## at the maximum of order ## m ## ]. ## \\ ## I think the second answer I gave here, ( the part [Edit: "In principle"..., etc.], is simpler, and probably better than trying to mathematically analyze the single-slit diffraction intensity function.
