What is the wavelength difference based on a mini spectrom.?

[ilittlepotato]

Homework Statement

I'm a bit stuck on how to approach this problem.

A miniature spectrometer used for chemical analysis has a diffraction grating with 800 slits/mm set 25.0 mm in front of the detector "screen." The detector can barely distinguish two bright lines that are 30 μm apart in the first-order spectrum. What is the resolution of the spectrometer at a wavelength of 550 nm ? That is, if two distinct wavelengths can barely be distinguished, one of them being 550 nm , what is the wavelength difference Δ λ between the two?

Homework Equations

dsin(theta)=mlamda

The Attempt at a Solution

I tried to find theta by using tan^-1(30micrometers/25 milimeters) but then the value came out very small so I'm not sure if I'm doing it right. I was going to use the theta to plug into dsin(theta)=mlamda, but the answer wasn't right.

I also tried to do 10^-6(sin(tan-1(30um/25mm)) and got 1.199999136 x 10^-9m as the difference, but that's also not the right approach.

 

[Charles Link]

For this one I think you need to assume the diffraction grating is a transmissive grating, and it does require a lens element immediately following it that focuses the light with a focal length of $f=25 \, mm$. The detector (most likely an array of very small (virtually microscopic) detectors) can measure locations accurate to $\Delta x=30 \, um=f \Delta \theta$ tells you the $\Delta \theta$ (in radians) that can be resolved. $( \Delta \theta=\frac{\Delta x}{f})$. Primary maxima occur at $m \lambda=d \sin(\theta)$, so that gives the location (in angle $\theta$) of the spectral line of wavelength $\lambda$. $\\$ (The $x$ location in the plane of the detector for a given wavelength $\lambda$ is given by $x= f \theta$, for small angles $\theta$. This is the result of using a lens with focal length $f$. The far-field diffraction pattern, which has primary maxima given by the equation $m \lambda=d \sin(\theta)$, is made to occur in the near field at a distance of 25 mm away because parallel lines incident on a lens are brought to a focus in the focal plane of the lens. Parallel rays incident at angle $\theta$ (they would be at angle $\theta$ in the far-field if there was no lens) come to a focus at location $x=f \theta$ in the focal plane of the lens, where the detector is located.). $\\$ They tell you the spectrometer is used with the first-order set of lines, so that $m=1$. This means $\lambda=d \sin(\theta)$ with $d=(1/800)$ mm. You can take a derivative here and assume the $\cos(\theta)$ result is approximately equal to 1. This gives $\Delta \lambda=d \Delta \theta$, where $\Delta \theta$ was computed above. (To be more precise, you could use $\Delta \lambda =d \cos(\theta) \, \Delta \theta$, and compute $\cos(\theta)$ given that $\lambda=d \sin(\theta)$, to get $\cos(\theta)$ in terms of $\lambda$ and $d$. (Remember $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$). Otherwise the result for $\Delta \lambda$ is independent of $\lambda$). $\\$ Note: In this homework problem, it really appears that the necessary background material was not supplied. Even the problem is very poorly stated. I have specialized in diffraction grating spectroscopy so the calculation is rather routine for me, but from what they supplied you with, it really could leave you guessing where to begin with it.