# Diffraction Grating Question, Dispersion and Resolving

## Homework Statement

One diffraction grating has 9600 lines uniformly spaced over a width of 3cm, it is illuminated by light from a mercury vapour discharge.

The other has 5000 lines/cm and is a 3.5cm grating, this one is used in the second order to resolve spectral lines close to 587.8002 nm spectral line of iron

part 1 to 2 concern first grating, part 3 to 5 concern 2nd grating

i) What is the expected dispersion, in the third order, in the vicinity of the intense green line whose wavelength = 546nm

ii) What is the resolving power of this grating in the 5th order?

For second grating

iii) For wavelengths > the iron line, whats the shortest wavelength you can distringuish from the iron line

iv) for wavelengths < the iron line, whats the longest you could distringuish?

v) What range of λ is indistinguishable from the iron line?

## Homework Equations

Resolving power = Nm = λ/Δλ
Δλ = λ/m = free spectral range

## The Attempt at a Solution

i) 9600/3 = 3200 lines per cm => 1/3200 is the seperation = d

θ = mλ/d = 3*546*10^-9 * 3200? Is that what they want?

ii) Would that just be 3200*5? = Nm?

iii) sinθ=mλ/d unsure, find where sinθ = 1? Very unsure to be honest

I'm really sketchy on this part of the subject, and my notes have no examples and when I google I'm not sure what formulas I'm supposed to be using.

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## Homework Statement

One diffraction grating has 9600 lines uniformly spaced over a width of 3cm, it is illuminated by light from a mercury vapour discharge.

The other has 5000 lines/cm and is a 3.5cm grating, this one is used in the second order to resolve spectral lines close to 587.8002 nm spectral line of iron

part 1 to 2 concern first grating, part 3 to 5 concern 2nd grating

i) What is the expected dispersion, in the third order, in the vicinity of the intense green line whose wavelength = 546nm

ii) What is the resolving power of this grating in the 5th order?

For second grating

iii) For wavelengths > the iron line, whats the shortest wavelength you can distringuish from the iron line

iv) for wavelengths < the iron line, whats the longest you could distringuish?

v) What range of λ is indistinguishable from the iron line?

## Homework Equations

Resolving power = Nm = λ/Δλ
Δλ = λ/m = free spectral range

## The Attempt at a Solution

i) 9600/3 = 3200 lines per cm => 1/3200 is the seperation = d

θ = mλ/d = 3*546*10^-9 * 3200? Is that what they want?

ii) Would that just be 3200*5? = Nm?

iii) sinθ=mλ/d unsure, find where sinθ = 1? Very unsure to be honest

I'm really sketchy on this part of the subject, and my notes have no examples and when I google I'm not sure what formulas I'm supposed to be using.
From the resolving power equation, you obtain the resolution ## \Delta \lambda ##. Because a diffraction grating spectrometer works with diffraction principles, it means that a perfectly monochromatic spectral line will not occur precisely at ## m \lambda=d \sin(\theta) ##, but will have a spread of ## \Delta \theta ## which also means that two monochromatic lines that are closely spaced in wavelength could blur across each other and be indistinguishable. The minimum spread ## \Delta \lambda ## that can be resolved is given is what this ## \Delta \lambda ## is referring to. That's what questions 3 and 4 are about. Compute ## \Delta \lambda ##. In this case you have other iron lines besides the 587.8 line. They want to know e.g. can you separate 587.9 or would it blend together? They want to know the closest to 587.8002 nm that you can get and be able to tell you have a different wavelength. ## \\ ## For additional info on the diffraction pattern for a grating with ## N ## grooves, you can use the equation ## I(\theta)=I_o \frac{sin^2(N \phi/2)}{sin^2(\phi/2)} ## where ## \phi=\frac{2 \pi}{\lambda}d \sin(\theta) ##. The intensity pattern can be computed from this formula, and the primary maxima occur when the numerator and denominator are both zero. You take the limit at the angle the denominator goes to zero (at these locations, ## m \lambda=d \sin(\theta_{max}) ## and ##limit \, I(\theta_{max})=N^2 I_o ##), and the spread ## \Delta \lambda ## is found by taking the location of the first zero that occurs in the numerator after the denominator is no longer zero. (You compute the wavelength ## \lambda +\Delta \lambda ## that is centered at this position.) This formula for ## I(\theta) ## is exact, but a little complicated to derive. (If you had an advanced Modern Physics course, perhaps they derived it for you when they treated diffraction theory=two-slit and multi-slit, etc.) The formula is also a little tricky to analyze, but once you figure out how to use it, it's not overly complex. You might find this extra info of interest if you want to know the details behind the above formulas. ## \\ ## One additional item about diffraction gratings: Normally, the dffraction theory is shown using slits and Huygens sources at each of the slits. Diffraction gratings are normally reflective rather than transmissive, but each of the lines on a diffraction grating can be considered to be a "Huygens mirror". The same interference equations apply for the reflective case that apply for the transmissive case.

Last edited:
• PhilJones
From the resolving power equation, you obtain the resolution ## \Delta \lambda ##. Because a diffraction grating spectrometer works with diffraction principles, it means that a perfectly monochromatic spectral line will not occur precisely at ## m \lambda=d \sin(\theta) ##, but will have a spread of ## \Delta \theta ## which also means that two monochromatic lines that are closely spaced in wavelength could blur across each other and be indistinguishable. The minimum spread ## \Delta \lambda ## that can be resolved is given is what this ## \Delta \lambda ## is referring to. That's what questions 3 and 4 are about. Compute ## \Delta \lambda ##. In this case you have other iron lines besides the 587.8 line. They want to know e.g. can you separate 587.9 or would it blend together? They want to know the closest to 587.8002 nm that you can get and be able to tell you have a different wavelength. ## \\ ## For additional info on the diffraction pattern for a grating with ## N ## grooves, you can use the equation ## I(\theta)=I_o \frac{sin^2(N \phi/2)}{sin^2(\phi/2)} ## where ## \phi=\frac{2 \pi}{\lambda}d \sin(\theta) ##. The intensity pattern can be computed from this formula, and the primary maxima occur when the numerator and denominator are both zero. You take the limit at the angle the denominator goes to zero (at these locations, ## m \lambda=d \sin(\theta_{max}) ## and ##limit \, I(\theta_{max})=N^2 I_o ##), and the spread ## \Delta \lambda ## is found by taking the location of the first zero that occurs in the numerator after the denominator is no longer zero. (You compute the wavelength ## \lambda +\Delta \lambda ## that is centered at this position.) This formula for ## I(\theta) ## is exact, but a little complicated to derive. (If you had an advanced Modern Physics course, perhaps they derived it for you when they treated diffraction theory=two-slit and multi-slit, etc.) The formula is also a little tricky to analyze, but once you figure out how to use it, it's not overly complex. You might find this extra info of interest if you want to know the details behind the above formulas. ## \\ ## One additional item about diffraction gratings: Normally, the dffraction theory is shown using slits and Huygens sources at each of the slits. Diffraction gratings are normally reflective rather than transmissive, but each of the lines on a diffraction grating can be considered to be a "Huygens mirror". The same interference equations apply for the reflective case that apply for the transmissive case.

Thanks for the info! So, for part 3,4 and 5 I want to fine Δλ = λ/(2*5000*3.5) ? Then, λ +/- Δλ is the shortest/longest above/below I can resolve and in turn this λ range is indistinguishable from the iron line?

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