Diffraction Grating - Calculating Ruling Width

[Martin89]

Homework Statement

The Attempt at a Solution

Hi All!

This is probably going to have a very simple answer. I'm stuck on question (b)(ii). The only equation I can find with the ruling width in it is the one shown and I can't find a way to calculate is as I don't know Beta.

Thanks

 

[Charles Link]

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.

 

[Martin89]

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.

Thanks a lot! I managed to briefly see my professor this morning and he said exactly the same thing as you